The probability of events occurring. Total probability formula. Bayes formula
Example 1. In the first urn: three red, one white balls. In the second urn: one red, three white balls. A coin is tossed at random: if it is a coat of arms, it is chosen from the first urn, otherwise, from the second.
Solution:
a) the probability that a red ball was drawn
A – got a red ball
P 1 – the coat of arms fell, P 2 - otherwise
b) The red ball is selected. Find the probability that it is taken from the first urn from the second urn.
B 1 – from the first urn, B 2 – from the second urn
,
Example 2. There are 4 balls in a box. Can be: only white, only black or white and black. (Composition unknown).
Solution:
A – probability of a white ball appearing
a) All white:
(the probability that you got one of the three options where there are white ones) 
(probability of a white ball appearing where everyone is white) 
b) Pulled out where everyone is black 
c) pulled out the option where everyone is white and/or black
- at least one of them is white 
P a +P b +P c = 
Example 3. There are 5 white and 4 black balls in an urn. 2 balls are taken out of it in a row. Find the probability that both balls are white.
Solution:
5 white, 4 black balls
P(A 1) – the white ball was taken out 
P(A 2) – probability that the second ball is also white 
P(A) – white balls chosen in a row
Example 3a. There are 2 fake and 8 real in a pack banknotes. 2 bills were pulled out of the pack in a row. Find the probability that both of them are fake.
Solution:
P(2) = 2/10*1/9 = 1/45 = 0.022
Example 4. There are 10 bins. There are 9 urns with 2 black and 2 white balls. There are 5 whites and 1 black in 1 urn. A ball was drawn from an urn taken at random.
Solution:
P(A) - ? a white ball is taken from an urn containing 5 white
B – probability of being drawn from an urn containing 5 whites
, - taken out from others
C 1 – probability of a white ball appearing at level 9. 
C 2 – probability of a white ball appearing, where there are 5 of them
P(A 0)= P(B 1) P(C 1)+P(B 2) P(C 2) 
Example 5. 20 cylindrical rollers and 15 cone-shaped ones. The picker takes 1 roller, and then another.
Solution:
a) both rollers are cylindrical
P(C 1)=; P(Ts 2)=
C 1 – first cylinder, C 2 – second cylinder
P(A)=P(Ts 1)P(Ts 2) =
b) At least one cylinder
K 1 – first cone-shaped.
K 2 - second cone-shaped.
P(B)=P(Ts 1)P(K 2)+P(Ts 2)P(K 1)+P(Ts 1)P(Ts 2)
;
c) the first cylinder, but not the second
P(C)=P(C 1)P(K 2) 
e) Not a single cylinder.
P(D)=P(K 1)P(K 2) 
e) Exactly 1 cylinder
P(E)=P(C 1)P(K 2)+P(K 1)P(K 2)
Example 6. There are 10 standard parts and 5 defective parts in a box.
Three parts are drawn at random
a) One of them is defective
P n (K)=C n k ·p k ·q n-k ,
P – probability of defective products 
q – probability of standard parts
n=3, three parts
b) two out of three parts are defective P(2)
c) at least one standard
P(0) - no defective
P=P(0)+ P(1)+ P(2) - probability that at least one part will be standard
Example 7. The 1st urn contains 3 white and black balls, and the 2nd urn contains 3 white and 4 black balls. 2 balls are transferred from the 1st urn to the 2nd without looking, and then 2 balls are drawn from the 2nd. What is the probability that they are different colors?
Solution:
When moving balls from the first urn, the following options are possible:
a) took out 2 white balls in a row
P BB 1 = 
In the second step there will always be one less ball, since in the first step one ball was already taken out.
b) took out one white and one black ball
The situation when the white ball is drawn first, and then the black one
P warhead = 
The situation when the black ball was drawn first, and then the white one
P BW = 
Total: P warhead 1 =
c) took out 2 black balls in a row
P HH 1 = 
Since 2 balls were transferred from the first urn to the second urn, the total number of balls in the second urn will be 9 (7 + 2). Accordingly, we will look for all possible options:
a) first a white and then a black ball was taken from the second urn
P BB 2 P BB 1 - means the probability that first a white ball was drawn, then a black ball, provided that 2 white balls were drawn from the first urn in a row. That is why the number of white balls in this case is 5 (3+2).
P BC 2 P BC 1 - means the probability that first a white ball was drawn, then a black ball, provided that white and black balls were drawn from the first urn. That is why the number of white balls in this case is 4 (3+1), and the number of black balls is five (4+1).
P BC 2 P BC 1 - means the probability that first a white ball was drawn, then a black ball, provided that both black balls were drawn from the first urn in a row. That is why the number of black balls in this case is 6 (4+2).
The probability that 2 balls drawn will be of different colors is equal to: 
Answer: P = 0.54
Example 7a. From the 1st urn containing 5 white and 3 black balls, 2 balls were randomly transferred to the 2nd urn containing 2 white and 6 black balls. Then 1 ball was drawn at random from the 2nd urn.
1) What is the probability that the ball drawn from the 2nd urn turns out to be white?
2) The ball taken from the 2nd urn turned out to be white. Calculate the probability that balls of different colors were moved from the 1st urn to the 2nd.
Solution.
1) Event A - the ball drawn from the 2nd urn turns out to be white. Let's consider the following options for the occurrence of this event.
a) Two white balls were placed from the first urn into the second: P1(bb) = 5/8*4/7 = 20/56.
There are a total of 4 white balls in the second urn. Then the probability of drawing a white ball from the second urn is P2(4) = 20/56*(2+2)/(6+2) = 80/448
b) White and black balls were placed from the first urn into the second: P1(bch) = 5/8*3/7+3/8*5/7 = 30/56.
There are a total of 3 white balls in the second urn. Then the probability of drawing a white ball from the second urn is P2(3) = 30/56*(2+1)/(6+2) = 90/448
c) Two black balls were placed from the first urn into the second: P1(hh) = 3/8*2/7 = 6/56.
There are a total of 2 white balls in the second urn. Then the probability of drawing a white ball from the second urn is P2(2) = 6/56*2/(6+2) = 12/448
Then the probability that the ball drawn from the 2nd urn turns out to be white is:
P(A) = 80/448 + 90/448 + 12/448 = 13/32
2) The ball taken from the 2nd urn turned out to be white, i.e. the total probability is P(A)=13/32.
Probability that balls of different colors (black and white) were placed in the second urn and white was chosen: P2(3) = 30/56*(2+1)/(6+2) = 90/448
P = P2(3)/ P(A) = 90/448 / 13/32 = 45/91
Example 7b. The first urn contains 8 white and 3 black balls, the second urn contains 5 white and 3 black balls. One ball is chosen at random from the first, and two balls from the second. After this, one ball is taken at random from the selected three balls. This last ball turned out to be black. Find the probability that a white ball is drawn from the first urn.
Solution.
Let's consider all variants of event A - out of three balls, the drawn ball turns out to be black. How could it happen that among the three balls there was a black one?
a) A black ball was taken from the first urn, and two white balls were taken from the second urn.
P1 = (3/11)(5/8*4/7) = 15/154
b) A black ball was taken from the first urn, two black balls were taken from the second urn.
P2 = (3/11)(3/8*2/7) = 9/308
c) A black ball was taken from the first urn, one white and one black ball were taken from the second urn.
P3 = (3/11)(3/8*5/7+5/8*3/7) = 45/308
d) A white ball was taken from the first urn, and two black balls were taken from the second urn.
P4 = (8/11)(3/8*2/7) = 6/77
e) A white ball was taken from the first urn, one white and one black ball were taken from the second urn.
P5 = (8/11)(3/8*5/7+5/8*3/7) = 30/77
The total probability is: P = P1+P2+ P3+P4+P5 = 15/154+9/308+45/308+6/77+30/77 = 57/77
The probability that a white ball is drawn from a white urn is:
Pb(1) = P4 + P5 = 6/77+30/77 = 36/77
Then the probability that a white ball was chosen from the first urn, given that a black ball was chosen from three balls, is equal to:
Pch = Pb(1)/P = 36/77 / 57/77 = 36/57
Example 7c. The first urn contains 12 white and 16 black balls, the second urn contains 8 white and 10 black balls. At the same time, a ball is drawn from the 1st and 2nd urns, mixed and returned one to each urn. Then a ball is drawn from each urn. They turned out to be the same color. Determine the probability that there are as many white balls left in the 1st urn as there were at the beginning.
Solution.
Event A - a ball is drawn simultaneously from the 1st and 2nd urns.
Probability of drawing a white ball from the first urn: P1(B) = 12/(12+16) = 12/28 = 3/7
Probability of drawing a black ball from the first urn: P1(H) = 16/(12+16) = 16/28 = 4/7
Probability of drawing a white ball from the second urn: P2(B) = 8/18 = 4/9
Probability of drawing a black ball from the second urn: P2(H) = 10/18 = 5/9
Event A happened. Event B - a ball is drawn from each urn. After shuffling, the probability of a white or black ball returning to the urn is ½.
Let's consider the options for event B - they turned out to be the same color.
For the first urn
1) a white ball was placed in the first urn and a white ball was drawn, provided that a white ball was previously drawn, P1(BB/A=B) = ½ * 12/28 * 3/7 = 9/98
2) a white ball was placed in the first urn and a white ball was pulled out, provided that a black ball was pulled out earlier, P1(BB/A=H) = ½ * 13/28 * 4/7 = 13/98
3) a white ball was placed in the first urn and a black one was pulled out, provided that a white ball was pulled out earlier, P1(BC/A=B) = ½ * 16/28 * 3/7 = 6/49
4) a white ball was placed in the first urn and a black one was pulled out, provided that a black ball was pulled out earlier, P1(BC/A=H) = ½ * 15/28 * 4/7 = 15/98
5) a black ball was placed in the first urn and a white ball was drawn, provided that a white ball was previously drawn, P1(BW/A=B) = ½ * 11/28 * 3/7 = 33/392
6) a black ball was placed in the first urn and a white ball was drawn, provided that a black ball was drawn earlier, P1(B/A=H) = ½ * 12/28 * 4/7 = 6/49
7) a black ball was placed in the first urn and a black one was pulled out, provided that a white ball was pulled out earlier, P1(HH/A=B) = ½ * 17/28 * 3/7 = 51/392
8) a black ball was placed in the first urn and a black one was pulled out, provided that a black ball was drawn earlier, P1(HH/A=H) = ½ * 16/28 * 4/7 = 8/49
For the second urn
1) a white ball was placed in the first urn and a white ball was drawn, provided that a white ball was previously drawn, P1(BB/A=B) = ½ * 8/18 * 3/7 = 2/21
2) a white ball was placed in the first urn and a white ball was pulled out, provided that a black ball was pulled out earlier, P1(BB/A=H) = ½ * 9/18 * 4/7 = 1/7
3) a white ball was placed in the first urn and a black one was pulled out, provided that a white ball was pulled out earlier, P1(BC/A=B) = ½ * 10/18 * 3/7 = 5/42
4) a white ball was placed in the first urn and a black one was pulled out, provided that a black ball was pulled out earlier, P1(BC/A=H) = ½ * 9/18 * 4/7 = 1/7
5) a black ball was put into the first urn and a white ball was pulled out, provided that a white ball was pulled out earlier, P1(BW/A=B) = ½ * 7/18 * 3/7 = 1/12
6) a black ball was placed in the first urn and a white ball was pulled out, provided that a black ball was pulled out earlier, P1(BW/A=H) = ½ * 8/18 * 4/7 = 8/63
7) a black ball was placed in the first urn and a black one was pulled out, provided that a white ball was pulled out earlier, P1(HH/A=B) = ½ * 11/18 * 3/7 = 11/84
8) a black ball was placed in the first urn and a black one was drawn, provided that a black ball had been drawn earlier, P1(HH/A=H) = ½ * 10/18 * 4/7 = 10/63
The balls turned out to be the same color:
a) white
P1(B) = P1(BB/A=B) + P1(BB/A=B) + P1(BW/A=B) + P1(BW/A=B) = 9/98 + 13/98 + 33 /392 + 6/49 = 169/392
P2(B) = P1(BB/A=B) + P1(BB/A=B) + P1(BW/A=B) + P1(BW/A=B) = 2/21+1/7+1 /12+8/63 = 113/252
b) black
P1(H) = P1(HH/A=B) + P1(HH/A=H) + P1(HH/A=B) + P1(HH/A=H) = 6/49 + 15/98 + 51 /392 + 8/49 = 223/392
P2(H) = P1(HH/A=B) + P1(HH/A=H) + P1(HH/A=B) + P1(HH/A=H) =5/42+1/7+11 /84+10/63 = 139/252
P = P1(B)* P2(B) + P1(H)* P2(H) = 169/392*113/252 + 223/392*139/252 = 5/42
Example 7d. The first box contains 5 white and 4 blue balls, the second contains 3 and 1, and the third contains 4 and 5, respectively. A box was chosen at random and a ball pulled out of it turned out to be blue. What is the probability that this ball is from the second box?
Solution.
A - event of drawing a blue ball. Let's consider all the possible outcomes of such an event.
H1 - the ball drawn from the first box,
H2 - the ball pulled out from the second box,
H3 - a ball drawn from the third box.
P(H1) = P(H2) = P(H3) = 1/3
According to the conditions of the problem, the conditional probabilities of event A are equal to:
P(A|H1) = 4/(5+4) = 4/9
P(A|H2) = 1/(3+1) = 1/4
P(A|H3) = 5/(4+5) = 5/9
P(A) = P(H1)*P(A|H1) + P(H2)*P(A|H2) + P(H3)*P(A|H3) = 1/3*4/9 + 1 /3*1/4 + 1/3*5/9 = 5/12
The probability that this ball is from the second box is:
P2 = P(H2)*P(A|H2) / P(A) = 1/3*1/4 / 5/12 = 1/5 = 0.2
Example 8. Five boxes with 30 balls each contain 5 red balls (this is a box of composition H1), six other boxes with 20 balls each contain 4 red balls (this is a box of composition H2). Find the probability that a red ball taken at random is contained in one of the first five boxes.
Solution: The problem is to apply the total probability formula.
P(H 1) = 5/11
The probability that any the taken ball is contained in one of six boxes:
P(H2) = 6/11
The event happened - the red ball was pulled out. Therefore, this could happen in two cases:
a) pulled out from the first five boxes.
P 5 = 5 red balls * 5 boxes / (30 balls * 5 boxes) = 1/6
P(P 5 /H 1) = 1/6 * 5/11 = 5/66
b) pulled out from six other boxes.
P 6 = 4 red balls * 6 boxes / (20 balls * 6 boxes) = 1/5
P(P 6 /H 2) = 1/5 * 6/11 = 6/55
Total: P(P 5 /H 1) + P(P 6 /H 2) = 5/66 + 6/55 = 61/330
Therefore, the probability that a red ball drawn at random is contained in one of the first five boxes is:
P k.sh. (H1) = P(P 5 /H 1) / (P(P 5 /H 1) + P(P 6 /H 2)) = 5/66 / 61/330 = 25/61
Example 9. The urn contains 2 white, 3 black and 4 red balls. Three balls are drawn at random. What is the probability that at least two balls will be the same color?
Solution. There are three possible outcomes:
a) among the three drawn balls there were at least two white ones.
P b (2) = P 2b
The total number of possible elementary outcomes for these tests is equal to the number of ways in which 3 balls can be extracted from 9: 
Let's find the probability that among the 3 selected balls, 2 are white.
Number of options to choose from 2 white balls: 
Number of options to choose from 7 other balls third ball: 
b) among the three drawn balls there were at least two black ones (i.e. either 2 black or 3 black).
Let's find the probability that among the selected 3 balls, 2 are black.
Number of options to choose from 3 black balls: 
Number of options to choose from 6 other balls of one ball: 
P 2h = 0.214
Let's find the probability that all the selected balls are black. 
P h (2) = 0.214+0.0119 = 0.2259
c) among the three drawn balls there were at least two red ones (i.e., either 2 red or 3 red).
Let's find the probability that among the 3 selected balls, 2 are red.
Number of options to choose from 4 black balls: 
Number of options to choose from: 5 white balls, remaining 1 white: 
Let's find the probability that all the selected balls are red. 
P to (2) = 0.357 + 0.0476 = 0.4046
Then the probability that at least two balls will be the same color is equal to: P = P b (2) + P h (2) + P k (2) = 0.0833 + 0.2259 + 0.4046 = 0.7138
Example 10. The first urn contains 10 balls, 7 of them white; The second urn contains 20 balls, 5 of which are white. One ball is drawn at random from each urn, and then one ball is drawn at random from these two balls. Find the probability that the white ball is drawn.
Solution. The probability that a white ball is drawn from the first urn is P(b)1 = 7/10. Accordingly, the probability of drawing a black ball is P(h)1 = 3/10.
The probability that a white ball is drawn from the second urn is P(b)2 = 5/20 = 1/4. Accordingly, the probability of drawing a black ball is P(h)2 = 15/20 = 3/4.
Event A - a white ball is taken from two balls
Let's consider the possible outcome of event A.
- A white ball was drawn from the first urn, and a white ball was drawn from the second urn. Then a white ball was drawn from these two balls. P1 = 7/10*1/4 = 7/40
- A white ball was drawn from the first urn and a black ball was drawn from the second urn. Then a white ball was drawn from these two balls. P2 = 7/10*3/4 = 21/40
- A black ball was drawn from the first urn, and a white ball was drawn from the second urn. Then a white ball was drawn from these two balls. P3 = 3/10*1/4 = 3/40
P = P1 + P2 + P3 = 7/40 + 21/40 + 3/40 = 31/40
Example 11. There are n tennis balls in the box. Of these, m were played. For the first game, two balls were taken at random and put back after the game. For the second game we also took two balls at random. What is the probability that the second game will be played with new balls?
Solution. Consider event A - the game was played for the second time with new balls. Let's see what events can lead to this.
Let us denote by g = n-m the number of new balls before being pulled out.
a) for the first game two new balls were pulled out.
P1 = g/n*(g-1)/(n-1) = g(g-1)/(n(n-1))
b) for the first game, they pulled out one new ball and one already played one.
P2 = g/n*m/(n-1) + m/n*g/(n-1) = 2mg/(n(n-1))
c) for the first game, two played balls were pulled out.
P3 = m/n*(m-1)/(n-1) = m(m-1)/(n(n-1))
Let's look at the events of the second game.
a) Two new balls were drawn, under condition P1: since new balls had already been drawn for the first game, then for the second game their number decreased by 2, g-2.
P(A/P1) = (g-2)/n*(g-2-1)/(n-1)*P1 = (g-2)/n*(g-2-1)/(n- 1)*g(g-1)/(n(n-1))
b) Two new balls were drawn, under condition P2: since one new ball had already been drawn for the first game, then for the second game their number decreased by 1, g-1.
P(A/P2) =(g-1)/n*(g-2)/(n-1)*P2 = (g-1)/n*(g-2)/(n-1)*2mg /(n(n-1))
c) Two new balls were drawn, under condition P3: since previously no new balls were used for the first game, their number did not change for the second game g.
P(A/P3) = g/n*(g-1)/(n-1)*P3 = g/n*(g-1)/(n-1)*m(m-1)/(n (n-1))
Total probability P(A) = P(A/P1) + P(A/P2) + P(A/P3) = (g-2)/n*(g-2-1)/(n-1)* g(g-1)/(n(n-1)) + (g-1)/n*(g-2)/(n-1)*2mg/(n(n-1)) + g/n *(g-1)/(n-1)*m(m-1)/(n(n-1)) = (n-2)(n-3)(n-m-1)(n-m)/(( n-1)^2*n^2)
Answer: P(A)=(n-2)(n-3)(n-m-1)(n-m)/((n-1)^2*n^2)
Example 12. The first, second and third boxes contain 2 white and 3 black balls, the fourth and fifth boxes contain 1 white and 1 black ball. A box is randomly selected and a ball is drawn from it. What is the conditional probability that the fourth or fifth box is chosen if the ball drawn is white?
Solution.
The probability of choosing each box is P(H) = 1/5.
Let us consider the conditional probabilities of event A - drawing the white ball.
P(A|H=1) = 2/5
P(A|H=2) = 2/5
P(A|H=3) = 2/5
P(A|H=4) = ½
P(A|H=5) = ½
Total probability of drawing a white ball:
P(A) = 2/5*1/5 + 2/5*1/5 +2/5*1/5 +1/2*1/5 +1/2*1/5 = 0.44
Conditional probability that the fourth box is selected
P(H=4|A) = 1/2*1/5 / 0.44 = 0.2273
Conditional probability that the fifth box is selected
P(H=5|A) = 1/2*1/5 / 0.44 = 0.2273
In total, the conditional probability that the fourth or fifth box is selected is
P(H=4, H=5|A) = 0.2273 + 0.2273 = 0.4546
Example 13. There were 7 white and 4 red balls in the urn. Then another ball of white or red or black color was put into the urn and after mixing one ball was taken out. It turned out to be red. What is the probability that a) a red ball was placed? b) black ball?
Solution.
a) red ball
Event A - the red ball is drawn. Event H - the red ball is placed. Probability that a red ball was placed in the urn P(H=K) = 1 / 3
Then P(A|H=K)= 1 / 3 * 5 / 12 = 5 / 36 = 0.139
b) black ball
Event A - the red ball is drawn. Event H - a black ball is placed.
Probability that a black ball was placed in the urn P(H=H) = 1/3
Then P(A|H=H)= 1 / 3 * 4 / 12 = 1 / 9 = 0.111
Example 14. There are two urns with balls. One contains 10 red and 5 blue balls, in the second there are 5 red and 7 blue balls. What is the probability that a red ball will be drawn at random from the first urn and a blue ball from the second?
Solution. Let the event A1 be a red ball drawn from the first urn; A2 - a blue ball is drawn from the second urn:
,
Events A1 and A2 are independent. The probability of the joint occurrence of events A1 and A2 is equal to
Example 15. There is a deck of cards (36 pieces). Two cards are drawn at random in a row. What is the probability that both cards drawn will be red?
Solution. Let event A 1 be the first red card drawn. Event A 2 - the second red card drawn. B - both cards taken out are red. Since both event A 1 and event A 2 must occur, then B = A 1 · A 2 . Events A 1 and A 2 are dependent, therefore, P(B) :
, 
From here 
Example 16. Two urns contain balls that differ only in color, and in the first urn there are 5 white balls, 11 black and 8 red balls, and in the second there are 10, 8, 6 balls, respectively. One ball is drawn at random from both urns. What is the probability that both balls are the same color?
Solution. Let index 1 mean white, index 2 mean black; 3 - red color. Let the event A i be that a ball of the i-th color is drawn from the first urn; event B j - a ball of color j is drawn from the second urn; event A - both balls are the same color.
A = A 1 · B 1 + A 2 · B 2 + A 3 · B 3. Events A i and B j are independent, and A i · B i and A j · B j are incompatible for i ≠ j. Hence,
P(A)=P(A 1) P(B 1)+P(A 2) P(B 2)+P(A 3) P(B 3) =
Example 17. From an urn with 3 white and 2 black balls, balls are drawn one at a time until black appears. Find the probability that 3 balls will be drawn from the urn? 5 balls?
Solution.
1) the probability that 3 balls will be drawn from the urn (i.e. the third ball will be black, and the first two will be white).
P=3/5*2/4*2/3=1/5
2) the probability that 5 balls will be drawn from the urn
This situation is not possible, because only 3 white balls.
P=0
Example 6. The box contains 11 parts, 3 of them are non-standard. One part is taken from the box twice, without returning them back. Find the probability that a standard part will be taken out of the box the second time - event B, if a non-standard part was taken out the first time - event A.
After the first extraction, in a box of 10 parts, 8 standard parts remained, and, therefore, the required probability
Total probability formula. Bayes formula
Example 7. There are three identical-looking urns: the first contains 5 white and 10 black balls; the second contains 9 white and 6 black balls; in the third there are only black balls. One ball is drawn from a randomly selected urn. What is the probability that this ball is black.
Event A – a black ball is taken out. Event A
H 1 – the ball was taken from the first urn;
H 2 – the ball was taken from the second urn;
H 3 – the ball was taken from the third urn.
Since the ballot boxes look identical, then:
A for each hypothesis.
The black ball was taken from the first urn:
Likewise:
1/3*2/3+1/3*2/5+1/3*1=31/45
Example 8. There are two urns: the first contains 5 white and 10 black balls; the second urn contains 9 white and 6 black balls. One ball is transferred from the first urn to the second without looking. After this, one ball is drawn from the second urn. Find the probability that this ball will be black.
Event A– a black ball was taken from the second urn. Event A can happen with one of the incompatible events (hypotheses):
H 1 – a white ball was transferred from the first urn to the second;
H 2 – a black ball was transferred from the first urn to the second.
Probabilities of hypotheses:
Let's find the conditional probabilities of the event A. If a white ball is transferred from the first urn to the second, then the second urn contains 10 white and 6 black balls. This means that the probability of getting a black ball from it is equal to:
Likewise:
According to the total probability formula:
Example 9. There are three urns: the first contains 5 white and 10 black balls; the second contains 9 white and 6 black balls; the third urn contains 15 black balls (no white balls). One ball was taken from a randomly chosen urn. This ball turned out to be black. Find the probability that the ball was drawn from the second urn.
Event A– one ball was taken from a randomly chosen urn.
Event A can happen with one of the incompatible events (hypotheses):
H 1 – the ball was taken from the first urn;
H 2 – the ball was taken from the second urn;
H 3 – the ball was taken from the third urn.
The prior probabilities of the hypotheses are:
In Problem 4 the conditional probabilities of the event are found A and its total probability:
Let's find the posterior probability of the hypothesis using Bayes' formula H 2 .
The black ball is taken from the second urn:
Let's compare and:
Thus, if it is known that a black ball was drawn, then the probability that it was drawn from the second urn decreases (this corresponds to the condition that the second urn contains the least number of black balls).
Bernoulli's formula
Example 10. There are six children in the family. The probability of having a girl is 0.49. Find the probability that among these children one is a girl.
Event A- a girl was born.
P = P(A) = 0,49;
q = 1 – p = 1 – 0,49 = 0,51.
Bernoulli formula:
Only six children, that means n=6.
We need to find the probability that there is exactly one girl among them, which means m = 1.
Example 11. The coin is tossed 6 times. Find the probability that the coat of arms will appear no more than 5 times.
Event A– when tossing a coin, a coat of arms appears.
The coin is tossed 6 times, which means n = 6.
Event B– the coat of arms will appear no more than 5 times.
Opposite event:
– the coat of arms will appear more than 5 times, that is, 6 times.
When a coin is tossed, we can say that it will land heads up, or probability this is 1/2. Of course, this does not mean that if a coin is tossed 10 times, it will necessarily land on heads 5 times. If the coin is "fair" and if it is tossed many times, then heads will land very close half the time. Thus, there are two types of probabilities: experimental And theoretical .
Experimental and theoretical probability
If we flip a coin a large number of times - say 1000 - and count how many times it lands on heads, we can determine the probability that it lands on heads. If heads are thrown 503 times, we can calculate the probability of it landing:
503/1000, or 0.503.
This experimental definition of probability. This definition of probability comes from observation and study of data and is quite common and very useful. Here, for example, are some probabilities that were determined experimentally:
1. The probability that a woman will develop breast cancer is 1/11.
2. If you kiss someone who has a cold, then the probability that you will also get a cold is 0.07.
3. A person who has just been released from prison has an 80% chance of returning to prison.
If we consider tossing a coin and taking into account that it is just as likely that it will come up heads or tails, we can calculate the probability of getting heads: 1/2. This is a theoretical definition of probability. Here are some other probabilities that have been determined theoretically using mathematics:
1. If there are 30 people in a room, the probability that two of them have the same birthday (excluding year) is 0.706.
2. During a trip, you meet someone, and during the conversation you discover that you have a mutual friend. Typical reaction: “This can’t be!” In fact, this phrase is not suitable, because the probability of such an event is quite high - just over 22%.
Thus, experimental probabilities are determined through observation and data collection. Theoretical probabilities are determined through mathematical reasoning. Examples of experimental and theoretical probabilities, such as those discussed above, and especially those that we do not expect, lead us to the importance of studying probability. You may ask, "What is true probability?" In fact, there is no such thing. Probabilities within certain limits can be determined experimentally. They may or may not coincide with the probabilities that we obtain theoretically. There are situations in which it is much easier to determine one type of probability than another. For example, it would be sufficient to find the probability of catching a cold using theoretical probability.
Calculation of experimental probabilities
Let us first consider the experimental definition of probability. The basic principle we use to calculate such probabilities is as follows.
Principle P (experimental)
If in an experiment in which n observations are made, a situation or event E occurs m times in n observations, then the experimental probability of the event is said to be P (E) = m/n.
Example 1 Sociological survey. An experimental study was conducted to determine the number of left-handed people, right-handed people and people whose both hands are equally developed. The results are shown in the graph.
a) Determine the probability that the person is right-handed.
b) Determine the probability that the person is left-handed.
c) Determine the probability that a person is equally fluent in both hands.
d) Most Professional Bowling Association tournaments are limited to 120 players. Based on the data from this experiment, how many players could be left-handed?
Solution
a)The number of people who are right-handed is 82, the number of left-handers is 17, and the number of those who are equally fluent in both hands is 1. The total number of observations is 100. Thus, the probability that a person is right-handed is P
P = 82/100, or 0.82, or 82%.
b) The probability that a person is left-handed is P, where
P = 17/100, or 0.17, or 17%.
c) The probability that a person is equally fluent in both hands is P, where
P = 1/100, or 0.01, or 1%.
d) 120 bowlers, and from (b) we can expect that 17% are left-handed. From here
17% of 120 = 0.17.120 = 20.4,
that is, we can expect about 20 players to be left-handed.
Example 2 Quality control
. It is very important for a manufacturer to maintain the quality of its products at high level. In fact, companies hire quality control inspectors to ensure this process. The goal is to produce the minimum possible number of defective products. But since the company produces thousands of products every day, it cannot afford to test every product to determine whether it is defective or not. To find out what percentage of products are defective, the company tests far fewer products.
Ministry Agriculture The US requires that 80% of the seeds sold by growers must germinate. To determine the quality of the seeds that an agricultural company produces, 500 seeds from those that were produced are planted. After this, it was calculated that 417 seeds sprouted.
a) What is the probability that the seed will germinate?
b) Do the seeds meet government standards?
Solution a) We know that out of 500 seeds that were planted, 417 sprouted. Probability of seed germination P, and
P = 417/500 = 0.834, or 83.4%.
b) Since the percentage of seeds germinated has exceeded 80% as required, the seeds meet government standards.
Example 3 Television ratings. According to statistics, there are 105,500,000 households with televisions in the United States. Every week, information about viewing programs is collected and processed. In one week, 7,815,000 households tuned in to the hit comedy series "Everybody Loves Raymond" on CBS and 8,302,000 households tuned in to the hit series "Law & Order" on NBC (Source: Nielsen Media Research). What is the probability that one household's TV is tuned to "Everybody Loves Raymond" during a given week? to "Law & Order"?
Solution The probability that the TV in one household is tuned to "Everybody Loves Raymond" is P, and
P = 7,815,000/105,500,000 ≈ 0.074 ≈ 7.4%.
The chance that the household's TV was tuned to Law & Order is P, and
P = 8,302,000/105,500,000 ≈ 0.079 ≈ 7.9%.
These percentages are called ratings.
Theoretical probability
Suppose we are conducting an experiment, such as throwing a coin or darts, drawing a card from a deck, or testing products for quality on an assembly line. Each possible result of such an experiment is called Exodus . The set of all possible outcomes is called outcome space . Event it is a set of outcomes, that is, a subset of the space of outcomes.
Example 4 Throwing darts. Suppose that in a dart throwing experiment, a dart hits a target. Find each of the following:
b) Outcome space
Solution
a) The outcomes are: hitting black (B), hitting red (R) and hitting white (B).
b) The space of outcomes is (hitting black, hitting red, hitting white), which can be written simply as (H, K, B).
Example 5 Throwing dice.
A die is a cube with six sides, each with one to six dots on it. 
Suppose we throw dice. Find
a) Outcomes
b) Outcome space
Solution
a) Outcomes: 1, 2, 3, 4, 5, 6.
b) Outcome space (1, 2, 3, 4, 5, 6).
We denote the probability that an event E occurs as P(E). For example, “the coin will land on heads” can be denoted by H. Then P(H) represents the probability that the coin will land on heads. When all outcomes of an experiment have the same probability of occurring, they are said to be equally likely. To see the differences between events that are equally likely and events that are not, consider the target shown below. 
For target A, the events of hitting black, red and white are equally probable, since the black, red and white sectors are the same. However, for target B, the zones with these colors are not the same, that is, hitting them is not equally probable.
Principle P (Theoretical)
If an event E can happen in m ways out of n possible equally probable outcomes from the outcome space S, then theoretical probability
events, P(E) is
P(E) = m/n.
Example 6 What is the probability of rolling a die to get a 3?
Solution There are 6 equally probable outcomes on a dice and there is only one possibility of rolling the number 3. Then the probability P will be P(3) = 1/6.
Example 7 What is the probability of rolling an even number on a die?
Solution The event is the throwing of an even number. This can happen in 3 ways (if you roll a 2, 4 or 6). The number of equally probable outcomes is 6. Then the probability P(even) = 3/6, or 1/2.
We will use a number of examples related to standard deck of 52 cards. This deck consists of the cards shown in the figure below. 
Example 8 What is the probability of drawing an Ace from a well-shuffled deck of cards?
Solution There are 52 outcomes (the number of cards in the deck), they are equally likely (if the deck is well shuffled), and there are 4 ways to draw an Ace, so according to the P principle, the probability
P(draw an ace) = 4/52, or 1/13.
Example 9 Suppose we choose, without looking, one ball from a bag with 3 red balls and 4 green balls. What is the probability of choosing a red ball?
Solution There are 7 equally probable outcomes of drawing any ball, and since the number of ways to draw a red ball is 3, we get
P(red ball selection) = 3/7.
The following statements are results from Principle P.
Properties of Probability
a) If event E cannot happen, then P(E) = 0.
b) If event E is certain to happen then P(E) = 1.
c) The probability that event E will occur is a number from 0 to 1: 0 ≤ P(E) ≤ 1.
For example, in a coin toss, the event that the coin lands on its edge has zero probability. The probability that a coin is either heads or tails has a probability of 1.
Example 10 Let's assume that 2 cards are drawn from a 52-card deck. What is the probability that both of them are peaks?
Solution The number n of ways to draw 2 cards from a well-shuffled deck of 52 cards is 52 C 2 . Since 13 of the 52 cards are spades, the number of ways m to draw 2 spades is 13 C 2 . Then,
P(pulling 2 peaks) = m/n = 13 C 2 / 52 C 2 = 78/1326 = 1/17.
Example 11 Suppose 3 people are randomly selected from a group of 6 men and 4 women. What is the probability that 1 man and 2 women will be selected?
Solution The number of ways to select three people from a group of 10 people is 10 C 3. One man can be chosen in 6 C 1 ways, and 2 women can be chosen in 4 C 2 ways. According to the fundamental principle of counting, the number of ways to choose 1 man and 2 women is 6 C 1. 4 C 2 . Then, the probability that 1 man and 2 women will be selected is
P = 6 C 1 . 4 C 2 / 10 C 3 = 3/10.
Example 12 Throwing dice. What is the probability of rolling a total of 8 on two dice?
Solution Each dice has 6 possible outcomes. The outcomes are doubled, meaning there are 6.6 or 36 possible ways in which the numbers on the two dice can appear. (It’s better if the cubes are different, say one is red and the other is blue - this will help visualize the result.) 
The pairs of numbers that add up to 8 are shown in the figure below. There are 5 possible ways receiving a sum equal to 8, hence the probability is 5/36.
We already know that probability is a numerical measure of the possibility of a random event occurring, i.e. an event that may or may not occur if a certain set of conditions are met. When a set of conditions changes, the probability of a random event may change. As an additional condition, we can consider the occurrence of another event. So, if to the set of conditions under which a random event occurs A, add one more, consisting in the occurrence of a random event IN, then the probability of the event occurring A will be called conditional.
Conditional probability of event A- the probability of occurrence of event A, provided that event B occurs. Conditional probability is denoted by (A).
Example 16. There are 7 white and 5 black balls in the box, differing only in color. The experiment consists of randomly taking out one ball and, without putting it back, take out another ball. What is the probability that the second ball drawn is black if the first ball drawn is white?
Solution.
Before us are two random events: event A– the first ball drawn turned out to be white, IN– the second ball drawn is black. A and B are incompatible events, let’s use the classical definition of probability. The number of elementary outcomes when drawing the first ball is 12, and the number of favorable outcomes of getting the white ball is 7. Therefore, the probability P(A) = 7/12.
If the first ball turns out to be white, then the conditional probability of the event IN- the appearance of the second black ball (provided that the first ball was white) is equal to (IN)= 5/11, since before the second ball is taken out there are 11 balls left, of which 5 are black.
Note that the probability of a black ball appearing on the second draw would not depend on the color of the first ball taken out if, after removing the first ball, we put it back in the box.
Consider two random events A and B. Let the probabilities P(A) and (B) be known. Let us determine the probability of occurrence of both event A and event B, i.e. products of these events.
Probability multiplication theorem. The probability of two events occurring is equal to the product of the probability of one of them and the conditional probability of the other, calculated under the condition that the first event occurred:
P(A×B) = P(A)×(B) .
Since for calculating the probability of a product it does not matter which of the considered events A And IN which was the first and which was the second, we can write:
P(A×B) = P(A) × (B) = P(B) × (A).
The theorem can be extended to a product of n events:
P(A 1 A 2 . A p) = P(A x) P(A 2 /A 1) .. P(A p /A 1 A 2 ... A p-1).
Example 17. For the conditions of the previous example, calculate the probability of drawing two balls: a) the white ball first, and the black ball second; b) two black balls.
Solution.
a) From the previous example, we know the probabilities of getting the white ball out of the box first and the black ball second, provided that the white ball was pulled out first. To calculate the probability of both events occurring together, we use the probability multiplication theorem: P(A×B) = P(A) × (B)= 
.
b) Similarly, we calculate the probability of drawing two black balls. Probability of getting the black ball first .
The probability of drawing a black ball a second time, provided that we do not put the first black ball taken out back into the box
(there are 4 black balls left, and there are 11 balls in total). The resulting probability can be calculated using the formula P(A×B)= P(A) × (B) 
0,152.
The probability multiplication theorem has a simpler form if events A and B are independent.
Event B is said to be independent of event A if the probability of event B does not change whether event A occurs or not. If event B is independent of event A, then its conditional probability (B) is equal to the usual probability P(B):
It turns out that if the event IN will be independent of the event A, then the event A will be independent of IN, i.e. (A)= P(A).
Let's prove it. Let's substitute the equality from the definition of independence of an event IN from the event A to the probability multiplication theorem: P(A×B) = P(A)× (B)= P(A)× (B). But in other way P(A×B)= P(B) × (A). Means P(A) × (B)= P(B) × (A) And (A)= P(A).
Thus, the property of independence (or dependence) of events is always mutual and the following definition can be given: two events are called independent, if the appearance of one of them does not change the probability of the appearance of the other.
It should be noted that the independence of events is based on the independence of the physical nature of their origin. This means that the sets of random factors leading to one or another outcome of testing one and another random event are different. So, for example, hitting a target with one shooter does not affect in any way (unless, of course, you come up with any exotic reasons) on the probability of hitting the target with a second shooter. In practice, independent events occur very often, since the causal relationship of phenomena in many cases is absent or insignificant.
Probability multiplication theorem for independent events. The probability of the product of two independent events is equal to the product of the probability of these events: P(A×B) = P(A) × P(B).
The following corollary follows from the probability multiplication theorem for independent events.
If events A and B are incompatible and P(A)¹0, P(B)¹0, then they are dependent.
Let's prove this by contradiction. Let us assume that incompatible events A And IN independent. Then P(A×B) = P(A) ×P(B). And since P(A)¹0, P(B)¹0, i.e. events A And IN are not impossible, then P(A×B)¹0. But, on the other hand, the event Až IN is impossible as a work incompatible events(this was discussed above). Means P(A×B)=0. got a contradiction. Thus, our initial assumption is incorrect. Events A And IN– dependent.
Example 18. Let us now return to the unsolved problem of two shooters shooting at the same target. Let us recall that if the probability of hitting the target by the first shooter is 0.8, and the second is 0.7, it is necessary to find the probability of hitting the target.
Events A And IN– hitting the target by the first and second shooter, respectively, are joint, therefore, to find the probability of the sum of events A + IN– hitting the target with at least one shooter – you must use the formula: P(A+B)=P(A)+P(B)–P(Až IN). Events A And IN independent, therefore P(A×B) = P(A) × P(B).
So, P(A+B) = P(A) + P(B) - P(A) × P(B).
P(A+B)= 0.8 + 0.7 – 0.8×0.7 = 0.94.
Example 19.
Two independent shots are fired at the same target. The probability of a hit on the first shot is 0.6, and on the second - 0.8. Find the probability of hitting the target with two shots.
1) Let us denote a hit on the first shot as an event
A 1, with the second - as the event A 2.
Hitting the target requires at least one hit: either only with the first shot, or only with the second, or with both the first and second. Therefore, the problem requires determining the probability of the sum of two joint events A 1 and A 2:
P(A 1 + A 2) = P(A 1) + P(A 2) - P(A 1 A 2).
2) Since the events are independent, then P(A 1 A 2) = P(A 1) P(A 2).
3) We get: P(A 1 + A 2) = 0.6 + 0.8 - 0.6 0.8 = 0.92.
If the events are incompatible, then P(A B) = 0 and P(A + B) = = P(A) + P(B).
Example 20.
The urn contains 2 white, 3 red and 5 blue balls of the same size. What is the probability that a ball drawn at random from an urn will be colored (not white)?
1) Let event A be the removal of a red ball from the urn,
event B - drawing the blue ball. Then event (A + B)
there is the extraction of a colored ball from an urn.
2) P(A) = 3/10, P(B) = 5/10.
3) Events A and B are incompatible, since only
one ball. Then: P(A + B) = P(A) + P(B) = 0.3 + 0.5 = 0.8.
Example 21.
The urn contains 7 white and 3 black balls. What is the probability of: 1) drawing a white ball from the urn (event A); 2) removing a white ball from the urn after removing one ball from it, which is white (event B); 3) removing a white ball from the urn after removing one ball from it, which is black (event C)?
1) P(A) = = 0.7 (see classical probability).
2)P B (A) = = 0,(6).
3) R S (A) = | = 0,(7).
Example 22.
The mechanism is assembled from three identical parts and is considered inoperative if all three parts fail. There are 15 parts left in the assembly shop, of which 5 are non-standard (defective). What is the probability that a mechanism assembled from the remaining parts taken at random will be inoperative?
1) Let us denote the desired event by A, the choice of the first non-standard part by A 1, the second by A 2, the third by A 3
2) Event A will occur if both event A 1, event A 2, and event A 3 occur, i.e.
A = A 1 A 2 A 3,
since the logical “and” corresponds to a product (see section “Propositional Algebra. Logical Operations”).
3) Events A 1, A 2, A 3 are dependent, therefore P(A 1 A 2 A 3) =
= P(A 1) P(A 2 /A 1) P(A 3 /A 1 A 2).
4)P(A 1) = ,P(A 2 /A 1) = ,P(A 3 /A 1 A 2)= . Then
P(A 1 A 2 A 3) = 0.022.
For independent events: P(A B) = P(A) P(B).
Based on the above, the criterion for the independence of two events A and B:
P(A) = P B (A) = P (A), P (B) = P A (B) = P (B).
Example 23.
The probability of hitting the target by the first shooter (event A) is 0.9, and the probability of hitting the target by the second shooter (event B) is 0.8. What is the probability that the target will be hit by at least one shooter?
1) Let C be the event of interest to us; the opposite event is that both shooters miss.
3) Since when shooting one shooter does not interfere with the other, the events are independent.
We have: P() = P() P() = =(1 - 0.9) (1 - 0.8) =
0,1 0,2 = 0,02.
4) P(C) = 1 -P() = 1 -0.02 = 0.98.
Total Probability Formula
Let event A can occur as a result of the manifestation of one and only one event H i (i = 1,2,... n) from some complete group of incompatible events H 1, H 2,... H n. Events in this group are usually called hypotheses.
Total probability formula. The probability of event A is equal to the sum of paired products of the probabilities of all hypotheses forming a complete group by the corresponding conditional probabilities of a given event A:
P(A) = 
, where = 1.
Example 24.
There are 3 identical urns. The first urn contains 2 white and 1 black ball, the second urn contains 3 white and 1 black ball, and the third urn contains 2 white and 2 black balls. 1 ball is selected from a randomly selected urn. What is the probability that he will be white?
All urns are considered the same, therefore, the probability of choosing the i-th urn is
Р(H i) = 1/3, where i = 1, 2, 3.
2) Probability of drawing a white ball from the first urn: (A) = .
Probability of drawing a white ball from the second urn: (A) = .
Probability of drawing a white ball from the third urn: (A) = .
3) The required probability:
P(A) = 
=0.63(8)
Example 25.
The store receives products for sale from three factories, the relative shares of which are: I - 50%, II - 30%, III - 20%. For factory products, defects are respectively: I - 2%, P - 2%, III - 5%. What is the probability that a product of this product, accidentally purchased in a store, turns out to be of good quality (event A)?
1) The following three hypotheses are possible here: H 1, H 2, H 3 -
the purchased item was produced respectively at factories I, II, III; the system of these hypotheses is complete.
Probabilities: P(H 1) = 0.5; P(H 2) = 0.3; P(H 3) = 0.2.
2) The corresponding conditional probabilities of event A are: (A) = 1-0.02 = 0.98; (A) = 1-0.03 = 0.97; (A) = = 1-0.05 = 0.95.
3) According to the total probability formula, we have: P(A) = 0.5 0.98 + + 0.3 0.97 + 0.2 0.95 = 0.971.
Posterior probability formula (Bayes formula)
Let's consider the situation.
There is a complete group of inconsistent hypotheses H 1, H 2, ... H n, the probabilities of which (i = 1, 2, ... n) are known before experiment (a priori probabilities). An experiment (test) is carried out, as a result of which the occurrence of event A is recorded, and it is known that our hypotheses assigned certain probabilities to this event (i = 1, 2, ...n). What will be the probabilities of these hypotheses after the experiment (a posteriori probabilities)?
The answer to a similar question is given by the posterior probability formula (Bayes formula):
, where i=1,2, ...p.
Example 26.
The probability of hitting an aircraft with a single shot for the 1st missile system (event A) is 0.2, and for the 2nd (event B) - 0.1. Each of the complexes fires one shot, and one hit is recorded on the plane (event C). What is the probability that the successful shot belongs to the first missile system?
Solution.
1) Before the experiment, four hypotheses are possible:
H 1 = A B - the plane is hit by the 1st complex and the plane is hit by the 2nd complex (the product corresponds to the logical “and”),
H 2 = A B - the plane is hit by the 1st complex and the plane is not hit by the 2nd complex,
H 3 = A B - the plane is not hit by the 1st complex and the plane is hit by the 2nd complex,
H 4 = A B - the plane is not hit by the 1st complex and the plane is not hit by the 2nd complex.
These hypotheses form a complete group of events.
2) Corresponding probabilities (with independent action of complexes):
P(H 1) = 0.2 0.1 = 0.02;
P(H2) = 0.2 (1-0.1) = 0.18;
P(H 3) = (1-0.2) 0.1 = 0.08;
P(H 4) = (1-0.2) (1-0.1) = 0.72.
3) Since the hypotheses form a complete group of events, the equality = 1 must be satisfied.
We check: P(H 1) + P(H 2) + P(H 3) + P(H 4) = 0.02 + 0.18 + + 0.08 + 0.72 = 1, thus the group in question hypothesis is correct.
4) Conditional probabilities for the observed event C under these hypotheses will be: (C) = 0, since according to the conditions of the problem one hit was registered, and hypothesis H 1 assumes two hits:
(C) = 1; (C) = 1.
(C) = 0, since according to the conditions of the problem one hit was registered, and hypothesis H 4 assumes the absence of hits. Therefore, hypotheses H 1 and H 4 are eliminated.
5) We calculate the probabilities of hypotheses H 2 and H 3 using the Bayes formula:
0,7, 
0,3.
Thus, with a probability of approximately 70% (0.7) it can be stated that the successful shot belongs to the first missile system.
5.4. Random variables. Distribution law of a discrete random variable
Quite often in practice, such tests are considered, as a result of which a certain number is randomly obtained. For example, when you throw a dice, you get a number of points from 1 to 6; when you take 6 cards from a deck, you get from 0 to 4 aces. Over a certain period of time (say, a day or a month), a certain number of crimes are registered in the city, a certain number of road accidents occur. A gun is fired. The projectile's flight range also takes on a random value.
In all of the tests listed above, we are faced with so-called random variables.
A numerical quantity that takes on one value or another as a result of a random test is called random variable.
The concept of a random variable plays a very important role in probability theory. If the “classical” theory of probability studied mainly random events, then modern probability theory primarily deals with random variables.
In what follows, we will denote random variables by capital Latin letters X, Y, Z, etc., and their possible values by the corresponding lowercase letters x, y, z. For example, if a random variable has three possible values, then we will denote them as follows: , , .
So, examples of random variables could be:
1) the number of points rolled on the top face of the die:
2) the number of aces, when taking 6 cards from the deck;
3) the number of registered crimes per day or month;
4) the number of hits on the target with four shots from a pistol;
5) the distance that a projectile will travel when fired from a gun;
6) the height of a random person.
You can notice that in the first example the random variable can take one of six possible values: 1, 2, 3, 4, 5 and 6. In the second and fourth examples, the number of possible values of the random variable is five: 0, 1, 2, 3, 4 In the third example, the value of the random variable can be any (theoretically) natural number or 0. In the fifth and sixth examples, the random variable can take any real value from a certain interval ( A, b).
If a random variable can take a finite or countable set of values, then it is called discrete(discretely distributed).
Continuous A random variable is a random variable that can take all values from a certain finite or infinite interval.
To specify a random variable, it is not enough to list its possible values. For example, in the second and third examples, the random variables could take the same values: 0, 1, 2, 3 and 4. However, the probabilities with which these random variables take their values will be completely different. Therefore, to specify a discrete random variable, in addition to the list of all possible values, you also need to indicate their probabilities.
The correspondence between possible values of a random variable and their probabilities is called law of distribution discrete random variable. , …, X=
The distribution polygon, as well as the distribution series, completely characterizes the random variable. It is one of the forms of the law of distribution.
Example 27. A coin is tossed randomly. Construct a row and a polygon for the distribution of the number of dropped coats of arms.
A random variable equal to the number of dropped coats of arms can take two values: 0 and 1. Value 1 corresponds to the event - the loss of a coat of arms, value 0 - the loss of heads. The probabilities of getting a coat of arms and getting a tail are the same and equal. Those. the probabilities with which a random variable takes the values 0 and 1 are equal. The distribution series looks like:
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If you have been playing poker for a long time, you may have noticed that sometimes there are hands at the tables that seem far from reality and do not lend themselves to mathematical laws. In this material we will tell you about poker probabilities of various types.
Probability theory plays a huge role in poker. Poker is a game that is based on chances, probabilities and... Ignoring and not knowing poker math will ultimately lead any unlucky player to financial ruin. Luck can play an important role in the short term, but the longer you play, the more important poker math probabilities become.
In most cases, you can determine your odds using basic arithmetic, as well as special and . Understanding poker odds will allow you to win more often than players who blindly hope for luck.
Preflop probabilities
Players do not play poker “in a vacuum”; each player must build on his opponent’s range and calculate his chances of winning exclusively against a specific opponent. In the table below we give you the probabilities of winning against different hand ranges.
Probabilities of certain preflop situations
Postflop probabilities
Now let's look at the probability of various events when playing various .
Beginners often overestimate the value starting hands, For example, . As you can see, suit cards don't hit flushes often enough. Also, pocket pairs only hit a set 12% of the time, so playing small pocket cards is not always profitable.
Poker hand probabilities
In this part of the material we will tell you about the mathematical probabilities for composing various.
As you can see, the royal flush is the rarest and strongest poker hand. The probability of hitting a royal flush in poker is 1 in 649,740. The chance of catching this combination on the flop with pocket Broadway cards is 0.0008%. If there is a potential royal flush on the board, then the probability that it will hit on the turn is 2%, and before the river - 4%.
Cooler probabilities
A cooler is a situation at the poker table when a player loses a hand not because of his own mistakes, but because of an unfortunate combination of circumstances and a stronger hand of his opponent. This is a classic one that professional poker players use in their vocabulary.
Kings to Aces
With aces you have nothing to fear before the flop, but if you have pocket kings, you can always be wary of your opponents' aces. But will such coolers happen often enough? If you play heads-up, your opponent will only get pocket aces once out of 220 hands. But at a full ring table against 8 opponents, the chances of someone getting aces against your pocket kings are much higher. The probability of this event is 1 in 25.
Queens to Kings (Aces)
Queens are a much more vulnerable hand than kings. More often than not, you'll be ahead with them preflop, but don't discount the possibility that one of your opponents got kings or aces. At a full table, the probability of this happening is 1 in 12. A raise, re-raise and all-in in front of you indicate that one of your opponents has received a monster and you should fold your queens.
Important Probabilities for High Pocket Pairs
Probability of getting set in set
Now let's talk about post-flop cooler situations. As you already know, the probability of flopping a set with a pocket pair is 12% or 1 in 8. But the event that many poker players fear is flopping a set against a stronger set. If two players have a pocket pair, then the situation in which both players hit a set on the flop will happen once every hundred flops.
Probability of getting four of a kind in four of a kind
Let's move on from the set to an even stronger combination. The odds of hitting four of a kind when you have a hole card and a set on the flop are 1 in 123.
If the probability of getting a set in a set is not too high, then the probability of a situation where two players get four of a kind in one hand is 1 in 39,000 in heads-up and 1 in 313,000 hands at a full table. For most poker players, this event will only happen once in their entire career.
A solid knowledge of poker odds and probabilities will help you adjust your tactics to your advantage during the game, and just understanding the mathematical principles will give you the emotional stability to play your best game.
