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, they are 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 occurrence white ball
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. The pack contains 2 fake and 8 real 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 different colors?
Solution:
When transferring balls from the first urn, it is possible the following options:
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. total probability is equal to 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 - retrieval event 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 problem conditions conditional probabilities events A are equal:
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 has 10 red and 5 blue balls, the second has 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 color, index 2 - 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
Problem 174tv
a) 3 white balls;
b) less than 3 white balls;
c) at least one white ball.
Problem 176tv
An urn contains 6 black and 5 white balls. 5 balls are randomly drawn. Find the probability that among them there is:
a) 3 white balls;
b) less than 3 white balls;
c) at least one white ball.
Problem 178tv
An urn contains 4 black and 5 white balls. 4 balls are randomly drawn. Find the probability that among them there is:
a) 2 white balls;
b) less than 2 white balls;
c) at least one white ball.
Problem 180tv
An urn contains 6 black and 7 white balls. 4 balls are randomly drawn. Find the probability that among them there is:
a) 4 white balls;
b) less than 4 white balls;
c) at least one white ball.
Problem 184tv
An urn contains 8 black and 6 white balls. 4 balls are randomly drawn. Find the probability that among them there is:
a) 3 white balls;
b) less than 3 white balls;
c) at least one white ball.
Problem 186tv
An urn contains 4 black and 6 white balls. 4 balls are randomly drawn. Find the probability that among them there is:
a) 3 white balls;
b) less than 3 white balls;
c) at least one white ball.
Problem 188tv
An urn contains 5 black and 6 white balls. 5 balls are randomly drawn. Find the probability that among them there is:
a) 4 white balls;
b) less than 4 white balls;
c) at least one white ball.
Job No. 1
Random Events
Option 6.
Task 1.1. Three coins are thrown. Find the probability that only two coins will have a “coat of arms”.
Event A under study – only two out of three coins will have a coat of arms. The coin has two sides, which means that the total number of events when throwing three coins will be 8. In three cases, only two coins will have a coat of arms. We calculate the probability of event A using the formula:
P(A) = m/n = 3/8.
Answer: probability 3/8.
Problem 1.2. The word EVENT is made up of cards, each of which has one letter written on it. The cards are then mixed and taken out one at a time without returning. Find the probability that the letters are taken out in the order of the given word.
The test consists of taking out cards with letters in a random order without returning. The elementary event is the resulting sequence of letters. Event A consists of receiving the right word EVENT . Elementary events are permutations of 7 letters, which means that according to the formula we have n= 7!
The letters in the word EVENT are not repeated, so permutations are not possible in which the word does not change. Their number is 1.
Thus,
P(A) = 1/7! = 1/5040.
Answer: P(A) = 1/5040.
Problem 1.3. As in previous task, find the corresponding probability of the case when given word is the words of ANTONOV ILYA.
This problem is solved similarly to the previous one.
n= 11!; M = 2!*2! = 4.
P(A) = 4/11 = 4/39916800 = 1/9979200
Answer: P(A) =1/9979200.
Problem 1.4. An urn contains 8 black and 6 white balls. 5 balls are randomly drawn. Find the probability that among them there is:
a) 3 white balls;
b) less than 3 white balls;
c) at least one white ball.
8 hours The test will be randomly drawing 5 balls. Elementary
6 b events are all possible combinations of 5 out of 14 balls. Their number is equal
a) A 1 - among the drawn balls there are 3 white ones. This means that among the drawn balls there are 3 white and 2 black. Using the multiplication rule, we get
P(A 1) = 560/2002 = 280/1001.
b) A 2 - among the drawn balls there are less than 3 white ones. This event consists of three incompatible events:
In 1 - among the drawn balls there are only 2 white and 3 black balls,
In 2 - among the drawn balls there is only one white and 4 black balls
In 3 - among the drawn balls there is not a single white ball, all 5 balls are black:
A 2 = B 1 B 2 B 3.
Since events B 1, B 2 and B 3 are incompatible, you can use the formula:
P(A 2) = P(B 1) + P(B 2) + P(B 3);
P(A 2) = 840/2002 + 70/2002 + 56/2002 = 483/1001.
c) - among the drawn balls there is not a single white one. In this case:
P(A 3) = 1 - P() = 1 - 28/1001 = 973/1001.
Answer: P(A 1) = 280/1001, P(A 2) = 483/1001, P(A 3) = 973/1001.
Problem 1.6. The first urn contains 5 white and 7 black balls, and the second urn contains 6 white and 4 black balls. They take it out of the first urn randomly 2 balls, and from the second - 2 balls. Find the probability that among the drawn balls:
a) all balls are the same color;
b) only three white balls;
c) at least one white ball.
1 urn 2 urn Balls were drawn from both urns independently. Tests
5 b 6 b are drawing two balls from the first urn and two balls
7h 4h from the second urn. Elementary events will be combinations
2 or 2 of 12 or 10 balls respectively.
2 2 a) A 1 - all drawn balls are of the same color, i.e. are they all white?
or all black.
Let us define all possible events for each urn:
In 1 - 2 white balls are drawn from the first urn;
In 2 - 1 white and 1 black ball are drawn from the first urn;
In 3 - 2 black balls are drawn from the first urn;
C 1 - 2 white balls are drawn from the second urn;
C 2 - 1 white and 1 black ball are drawn from the second urn;
From 3 - 2 black balls are drawn from the second urn.
This means A 1 = 
, from where, taking into account the independence and incompatibility of events, we obtain
P(A 1) = P(B 1) * P(C 1) + P(B 3) * P(C 3).
Let's find the number of elementary events n 1 and n 2 for the first and second urns, respectively. We have:
Let's find the number of each element of events that determine the following events:
B 1: m 11 = C 1: m 21 =
B 2: m 12 = C 2: m 22 =
B 3: m 13 = C 3: m 23 =
Hence,
P(A 1) = 10/66 * 15/45 + 21 * 6/45 = 5/99 + 7/165 = 46/495.
b) A 2 - among the drawn balls there are only 3 white ones. In this case
A 2 = (B 1 C 2 (B 2 C 1);
P(A 2) = P(B 1) * P(C 1) + P(B 2) * P(C 2)
P(A 2) = 10/66 * 6/45 + 35/66 * 24/45 = 33/99 = 1/3.
c) A 3 - among the drawn balls there is at least one white one.
There is not a single white ball among the recovered balls. Then
P() = P(B 3) * P(C 3) = 21/66 * 6/45 = 7/165;
P(A 3) = 1 - P() = 1 - 7/165 = 158/165.
Answer: P(A 1) = 46/495, P(A 2) = 1/3, P(A 3) = 158/165.
Problem 1.7. The urn contains 5 black and white balls, 4 white balls are added to them. After this, 3 balls are randomly drawn from the urn. Find the probability that all the drawn balls are white, assuming that all possible suggestions about the original contents of the urn are equally possible.
There are two types of tests here: first, the initial contents of the urn are set and then the 3rd ball is randomly drawn, with the result of the second test depending on the result of the first. Therefore, the total probability formula is used.
event A - 3 white balls are randomly drawn. The probability of this event depends on the initial composition of the balls in the urn.
Let's consider the events:
In 1 - there were 5 white balls in the urn;
In 2 - there were 4 white and 1 black balls in the urn;
In 3 - there were 3 white and 2 black balls in the urn;
In 4 - there were 2 white and 3 black balls in the urn;
At 5 - there were 1 white and 4 black balls in the urn.
At 6 - there were 5 black balls in the urn;
Total number of elementary outcomes
Let us find the conditional probabilities of event A under various conditions.
P(A/B 1) = 1.
P(A/B2) = 56/84 = 2/3.
P(A/B 3) = 35/84 = 5/12.
P(A/B 4) = 5/21.
P(A/B 5) = 5/42.
P(A/B 6) = 1/21.
P(A) = 1 * 1/6 + 2/3 * 1/6 + 5/12 * 1/6 + 5/21 * 1/6 + 5/42 * 1/6 + 1/21 * 1/6 = 209/504.
Answer: P(A) = 209/504.
Problem 1.9. There are 11 rifles in the pyramid, 3 of them with optical sights. A shooter, shooting from a rifle with an optical sight, can hit the target with a probability of 87/100, and shooting from a rifle without optical sight, - with a probability of 52/100. Find the probability that a shooter will hit a target shooting from a random rifle.
Considering that the rifles are selected one at a time, we obtain and, respectively (for B 1) and (for B 2); thus P(B 1) = 3/11, P(B 2) = 8/11.
Conditional probabilities are specified in the problem statement:
P(A/B 1) = 0.87 and P(A.B 2) = 0.52.
Hence,
P(A) = 0.87 * 3/11 + 0.52 * 8/11 = 0.615.
Answer: P(A) =0.615.
Problem 1.10. In the installation shop, an electric motor is connected to the device. Electric motors are supplied by three manufacturers. In stock there are electric motors of these factories, respectively, in quantities M 1 = 13, M 2 = 12, and M 3 = 17 pieces, which can operate without failure until the end of the warranty period with probabilities of 0.91, 0.82, and 0.77, respectively. A worker randomly takes one electric motor and mounts it to the device. Find the probability that an electric motor installed and operating without failure until the end of the warranty period was supplied by the first, second or third manufacturer, respectively.
Conditional probabilities are specified in the problem statement: P(A/B 1) = 0.91, P(A/B 2) = 0.82, P(A/B 3) = 0.77.
Similar to the previous problem, let’s find the probabilities:
P(B 1) = 13/42 = 0.3095; P(B 2) = 12/42 = 0.2857; P(B 3) = 17/42 = 0.4048;
P(A) = 0.91 * 0.3095 + 0.82 * 0.2857 + 0.77 * 0.4048 = 0.8276.
Using the Bayes formula (1.8.), we calculate the conditional probabilities of events (hypotheses) B 1:
P(B 1 /A) = 
P(B 2 /A) = 
P(B 3 /A) = 
Answer: P(B 1 /A) = 0.3403, P(B 2 /A) = 0.2831, P(B 3 /A) = 0.3766
Job No. 2
Random variables.
6 - option.
Task 2.1. In each of n independent tests event A occurs with a constant probability of 0.36. Calculate all probabilities p k, k = 0, 1, 2, ..., 11, where k is the frequency of event A. Draw a graph of probabilities p k. Find the most probable frequency.
Asked by: n = 11, p = 0.36, q = 1 - p = 0.64.
Find: p 0, p 1, p 2, ..., p 11 and k.
Using Bernoulli's formula. The value p 0 is calculated using the first of the formulas, and the remaining probabilities p k - using the second.
For the formula we calculate the constant factor
p/q = 0.36/ 0.64 = 0.5625, p 0 = *0.36 0 * 0.64 11 = 0.0073787.
We write the results of the calculations in Table 1. If the calculations are correct, then the equality must be satisfied
Using the found probability values, we will construct their graph (Fig. 1).
Let's find the most probable frequency according to the given conditions:
np - q = 11 * 0.36 - 0.64 = 3.32.np + k = 4.32
This means that the most likely frequency is k = 4 and, as was obtained earlier, the value of p 3 is the maximum.
Table 1
| k | (n-k-1)/ k | r k | k | (n-k-1)/ k | p k |
| - | 0,9926213 |
Figure 1 Probability graph p k
Problem 2.2. In each of n independent trials, event A occurs with a constant probability of 0.47. Find the probability that event A occurs:
a) exactly 330 times;
b) less than 330 and more than 284 times;
c) more than 330 times.
A) Asked by: n = 760, p = 0.47, M = 330.
Find: R 760 (330).
We use local theorem Moivre - Laplace. We find:
We find the value of the function j(x) from the table:
j(1.98) = 0.0562, P 760 (330) = 0.0562/ 13.76 = 0.00408.
b) Find: R 760 (284 We use the Moivre-Laplace integral theorem. We find: We find the value of the function Ф(х) from the table: R 760 (284 V) Find: R 760 (330 We have: x 1 = -1.98, R 760 (330 Problem 2.4. At a telephone exchange, an incorrect connection occurs with a probability of 1/800. Find the probability that among 5600 connections the following occurs: a) exactly 2 incorrect connections; b) less than 3 incorrect connections; c) more than 8 incorrect connections. a) Given: n = 5600, p = 1/800, k = 2. Find: R 800 (2). We get: l = 5600 * 1/800 = 7. R 800 (2) = b) Set k<3. Find: R 200 (k<3). R 800 (k<3) = Р 800
(0) + Р 800
(1) + Р 800
(2) = 0,0223 + 0,0009 + 0,0064 = 0,0296. V) Set k > 8. Find: P 800 (k > 8). This problem can be solved more simply by finding the probability of the opposite event, since in this case fewer terms need to be calculated. Taking into account the previous case, we have Р 800 (k>8) = 1 – Р 800 (k8) = 1 - Р 800 (k<9) = 1 - 0,7291 = 0,2709. Problem 2.6. The random variable X is given by a distribution series. Find the distribution function F(x) of the random variable X and plot it. Calculate for X its average value EX, dispersion DX and mode Mo. Let's plot the distribution function F(x) . The average EX value is calculated using the formula: EX = 8*0.11 + 12*0.14 + 16*0.5 + 24*0.25 = 16.56. Variance: E(X 2) = 8 2 *0.11 + 12 2 *0.14 + 16 2 *0.5 + 24 2 *0.25 = 299.2 DX = 299.2 – 16.52 2 = 26.2896. Distribution function graph Problem 2.7. The random variable X is specified by the probability density function f(x) = Find the distribution function F(x) of the random variable X. Plot a graph of the function f(x) and F(x). Calculate for X its average value EX, dispersion DX, mode Mo and median Me. K = 8, R = 12. We find the distribution function F(x) of a continuous random variable using the formula: Draw graphs of the functions f(x) and F(x). The average value of X is calculated using the formula: EX = To find the variance of X, we use the formulas: E(X 2) = DX = 40.5 – (4.5) 2. The graph shows that f(x) reaches a maximum at the point x = 1/2 and, therefore, Mo = 12. To find the median Me, you need to solve the equation x 2 / 256 = 1/2, or x 2 = 128. We have x = ± 11.31, Me = 11.31. Graph of the distribution function F(x). Job No. 3. Problem 3.1 Based on samples A and B Create a variation series; Calculate relative frequencies (frequencies) and accumulated frequencies; Construct variation series graphs (polygon and histogram); Create an empirical distribution function and plot it; Calculate the numerical characteristics of the variation series: average , dispersion, standard deviation , median Me. Problem 3.2. Calculate unbiased estimates of the population parameters ,S 2 , Spo samples A and B (using the results obtained in task 3.1.), as well as the first column of sample B. Sample A6 N = 73 Start of first interval: 0 Interval length: 1 Sample B6 N = 237 Start of first interval: 285 Interval length: 7 Problem solving. Task 3.1. First, let's solve the problem for sample A. We find: x min = 0 and x max = 11. The range (11 - 0 + 1 = 12) is quite small, so we will compose a variation series according to the values (Table 1). Table 1 We calculate all relative frequencies with the same accuracy. When constructing graphs, we depict on the x-axis values from 0 to 11 and on the n i / n axis - values from 0 to 0.25 (Fig. 1 and 2). Rice. 1. Polygon of variation series of sample A Rice. 2. Histogram of the variation series of sample A. We find the empirical distribution function F * (x) using the formula and accumulated frequencies from table. 1. We have: When plotting the graph F * (x), we plot the function values in the range from 0 to 1.2 (Fig. 3). Fig.3. Graph of the empirical distribution function of sample A. The calculation of the sums for the arithmetic mean and variance using formulas and variation series (see Table 1) is presented in Table. 2. Based on the maximum frequency, we determine c = 7, and the table step k = 1.
.X
8 12 16 24
R
0,11 0,14 0,50 0,25
4
10
7
6
3
7
8
7
4
7
10
7
3
9
3
1
5
8
10
11
6
5
7
6
3
8
4
3
8
4
10
6
8
7
8
7
7
7
4
6
7
10
4
4
0
5
4
4
8
5
5
10
7
3
8
5
6
6
6
3
5
7
8
5
7
10
9
10
8
2
3
6
9
324
296
313
323
312
321
322
301
337
322
329
307
301
328
312
318
327
315
319
317
309
334
323
340
326
322
314
335
313
322
319
325
312
300
323
335
339
326
298
298
337
322
303
314
315
310
316
321
312
315
331
322
321
336
328
315
338
318
327
323
325
314
297
303
322
314
317
330
318
320
312
333
332
319
325
319
307
305
316
330
318
335
327
321
332
288
322
334
295
318
329
305
310
304
326
319
317
316
316
307
309
309
328
317
317
322
316
304
303
350
309
327
345
329
338
311
316
324
310
306
308
302
315
314
343
320
304
310
345
312
330
324
308
326
313
320
328
309
306
306
308
324
312
309
324
321
313
330
330
315
320
313
302
295
337
346
327
320
307
305
323
331
345
315
318
331
322
315
304
324
317
322
312
314
308
303
333
321
312
323
317
288
317
327
292
316
322
319
313
328
313
309
329
313
334
314
320
301
329
319
332
316
300
300
304
306
314
323
318
337
325
321
322
288
313
314
307
329
302
300
316
321
315
323
331
318
334
316
328
294
288
312
312
315
321
332
319
Standard deviation 
The Mo mode is the value with the maximum frequency, i.e. Mo = 7. The median Me is the 37th value of the variation series: Me = 7.
Now, using sample B, we will find x min = 288 and x max = 350. The range (350 - 288 + 1 = 63) is quite large, so we will compose a variation series according to the intervals of values, using the beginning of the first interval and the length of the interval given during sampling (Table 3 ).
Table 3
Rice. 4. Polygon of the variation series of sample B.
Rice. 5. Histogram of the variation series of sample B.
When constructing graphs, we plot along the x-axis values from 285 to 355 and along the n i / n axis - values from 0 to 0.3 (Fig. 4 and 5).
Next, we take into account that the end of each interval is taken as a representative. Taking the ends of the intervals and the corresponding accumulated frequencies as the coordinates of the points (see Table 3) and connecting these points with straight lines, we will construct a graph of the empirical distribution function (Fig. 6).
Rice. 6. Graph of the empirical distribution function of sample B.
To calculate the arithmetic mean and variance using formulas and tables. 3 we define c = 316 and k = 7. We calculate the amounts using the table. 4 (Table 4).
Using formulas we calculate the arithmetic mean 
and variance 227.8
We find the median using the formula: Me =.
Problem 3.2.
Using the formula, we find unbiased estimates of the variance and standard deviation:
n = 73, S -2 = 5.8143, S 2 = 73/72×5.8143 = 5.8951, S = 2.43.
For sample B we have
393.92, = 177.47, n = 237, S 2 = 237/236×177.47 = 178.222, S = 13.35.
Unbiased estimates for the first column of sample B are obtained in a similar way (if this sample contains few repeating elements, the variation series need not be compiled).
The classical definition of probability reduces the concept of probability to the concept of equiprobability (equal possibility) of events, which is considered basic and is not subject to formal definition. This definition is applicable in cases where it is possible to identify a complete group of incompatible and equally probable events - elementary outcomes. For example, consider an urn with balls.
Let an urn contain 7 identical, thoroughly mixed balls, 2 of them red, 1 blue and 4 white. The test will consist of taking one ball at random from an urn. Every event that can occur in a test is an elementary outcome. In this example, there are seven elementary outcomes, which we will denote E 1 , E 2 ,..., E 7. Outcomes E 1 , E 2 - the appearance of a red ball, E 3 - the appearance of a blue ball, E 4 , E 5 , E 6 , E 7 - the appearance of a white ball. In our example events E 1 , E 2 ,... E 7 - pairwise incompatible. In addition, they are also equally possible in this test. Let the event A lies in the fact that a ball taken at random from an urn turns out to be colored (red or blue).
Those elementary outcomes in which the event of interest to us A comes, they call outcomes favorable event A. In our example, outcomes favorable to the event A, are the outcomes E 1 , E 2 and E 3. Reasonable as a measure of the possibility of an event occurring A, that is, the probabilities R(A), take a number equal to the ratio of outcomes favorable to the occurrence of the event A, to the number of all possible outcomes. In our example
R The example we examined led us to the definition of probability, which is commonly called classic .
Probability of the event A call the number ratio m outcomes favorable to this event to the total number n everyone elementary outcomes:
R(A) = . (1.4.4)
The classical definition of probability serves as a good mathematical model for those random experiments in which the number of outcomes is finite, and the outcomes themselves are equally possible.
EXAMPLE 2. The dice are thrown. Find the probability that no more than four points will be rolled.
Solution. Total number of elementary outcomes n= 6 (can roll 1, 2, 3, 4, 5, 6). Among these outcomes, the event is favorable A(no more than four points will roll) only four outcomes m= 4. Therefore, the required probability
EXAMPLE 3. What is the probability of guessing 4 numbers when filling out a sports lotto card “6” from “49”?
Solution. The total number of elementary outcomes of the experiment is equal to the number of ways in which 6 numbers out of 49 can be crossed out, that is n = C. Let's find the number of outcomes favorable to the event we are interested in
A= (4 numbers guessed), 4 numbers out of 6 winning ones can be crossed out C ways, while the remaining two numbers must not be winning. You can cross out 2 wrong numbers out of 43 non-winning ones C ways. Therefore, the number of favorable outcomes m = C× C. Taking into account that all outcomes of the experiment are inconsistent and equally possible, we find the required probability using the classical probability formula:
P(A) = 
EXAMPLE 4. A randomly selected phone number consists of 5 digits. How likely is it that it contains: 1) all the numbers are different; 2) are all numbers odd?
Solution. 1. Since each of the five places in a five-digit number can contain any of the digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, then all different five-digit numbers will be 10 5 (00000 - 1 -th, 00001 - 2nd, 00002 -3rd, ..., 99998 - 99999th, and finally, 99999 - 100,000th). Numbers in which all numbers are different are arrangements of 10 elements of 5.
Formula for number placements from n elements by k:
K! = = n (n - 1) ... (n - k + 1).
Therefore, the number of favorable cases m= = 10× 9× 8× 7× 6 and the desired probability
P(A) = 
= 0,3024.
2. From 5 odd digits (1, 3, 5, 7, 9) you can form 5 5 different five-digit numbers. 5 5 is the number of favorable outcomes m . Since all equally possible cases n= 10 5 , then the required probability
P (A) = = = = 0.03125.
EXAMPLE 5. A full deck of cards (52 sheets) is divided at random into two equal packs of 26 sheets. Find the probabilities of the following events:
A- each pack will contain two aces;
IN- one of the packs will not contain a single ace, and the other will not have all four;
WITH- one of the packs will have one ace, and the other will have three.
Solution. The total number of possible elementary outcomes of the test is equal to the number of ways in which 26 cards can be extracted from 52, that is, the number of combinations from 52 to 26, n= . Number of favorable events A cases
m= (according to the basic rule of combinatorics), where the first factor shows that two aces out of four can be taken in ways, the second factor shows that the remaining 24 cards are taken from 48 cards that do not contain aces in ways. The required probability is equal to the ratio of the number of outcomes favorable to the event A, to the total number of all outcomes:
Event IN can be accomplished in two equally possible ways: either the first packet will have all four aces, and the second - none, or vice versa:
Likewise:
Note that the classical definition of probability was introduced for the case when the space of elementary events is finite, and all outcomes and trials are equally possible and inconsistent.
Job No. 1
Random Events
Option 6.
Task 1.1. Three coins are thrown. Find the probability that only two coins will have a “coat of arms”.
Event A under study – only two out of three coins will have a coat of arms. The coin has two sides, which means that the total number of events when throwing three coins will be 8. In three cases, only two coins will have a coat of arms. We calculate the probability of event A using the formula:
P(A) = m/n = 3/8.
Answer: probability 3/8.
Problem 1.2. The word EVENT is made up of cards, each of which has one letter written on it. The cards are then mixed and taken out one at a time without returning. Find the probability that the letters are taken out in the order of the given word.
The test consists of taking out cards with letters in a random order without returning. The elementary event is the resulting sequence of letters. Event A consists of receiving the desired word EVENT . Elementary events are permutations of 7 letters, which means that according to the formula we have n= 7!
The letters in the word EVENT are not repeated, so permutations are not possible in which the word does not change. Their number is 1.
Thus,
P(A) = 1/7! = 1/5040.
Answer: P(A) = 1/5040.
Problem 1.3. As in the previous problem, find the corresponding probability of the case when the given word is the word ANTONOV ILYA.
This problem is solved similarly to the previous one.
n= 11!; M = 2!*2! = 4.
P(A) = 4/11 = 4/39916800 = 1/9979200
Answer: P(A) =1/9979200.
Problem 1.4. An urn contains 8 black and 6 white balls. 5 balls are randomly drawn. Find the probability that among them there is:
a) 3 white balls;
b) less than 3 white balls;
c) at least one white ball.
8 hours The test will be randomly drawing 5 balls. Elementary
6 b events are all possible combinations of 5 out of 14 balls. Their number is equal
a) A 1 - among the drawn balls there are 3 white ones. This means that among the drawn balls there are 3 white and 2 black. Using the multiplication rule, we get
P(A 1) = 560/2002 = 280/1001.b) A 2 - among the drawn balls there are less than 3 white ones. This event consists of three incompatible events:
In 1 - among the drawn balls there are only 2 white and 3 black balls,
In 2 - among the drawn balls there is only one white and 4 black balls
In 3 - among the drawn balls there is not a single white ball, all 5 balls are black:
B 2 B 3.Since events B 1, B 2 and B 3 are incompatible, you can use the formula:
P(A 2) = P(B 1) + P(B 2) + P(B 3);
P(A 2) = 840/2002 + 70/2002 + 56/2002 = 483/1001.
- among the drawn balls there is not a single white one. In this case:P(A 3) = 1 - P(
) = 1 - 28/1001 = 973/1001.Answer: P(A 1) = 280/1001, P(A 2) = 483/1001, P(A 3) = 973/1001.
Problem 1.6. The first urn contains 5 white and 7 black balls, and the second urn contains 6 white and 4 black balls. 2 balls are randomly drawn from the first urn, and 2 balls from the second. Find the probability that among the drawn balls:
a) all balls are the same color;
b) only three white balls;
c) at least one white ball.
1 urn 2 urn Balls were drawn from both urns independently. Tests
5 b 6 b are drawing two balls from the first urn and two balls
7h 4h from the second urn. Elementary events will be combinations
2 or 2 of 12 or 10 balls respectively.
2 2 a) A 1 - all drawn balls are of the same color, i.e. are they all white?
or all black.
Let us define all possible events for each urn:
In 1 - 2 white balls are drawn from the first urn;
In 2 - 1 white and 1 black ball are drawn from the first urn;
In 3 - 2 black balls are drawn from the first urn;
C 1 - 2 white balls are drawn from the second urn;
C 2 - 1 white and 1 black ball are drawn from the second urn;
From 3 - 2 black balls are drawn from the second urn.
This means A 1 =
, from where, taking into account the independence and incompatibility of events, we obtain P(A 1) = P(B 1) * P(C 1) + P(B 3) * P(C 3).
Let's find the number of elementary events n 1 and n 2 for the first and second urns, respectively. We have:
Let's find the number of each element of events that determine the following events:
C 1: m 21 = C 2: m 22 = C 3: m 23 =Hence,
P(A 1) = 10/66 * 15/45 + 21 * 6/45 = 5/99 + 7/165 = 46/495.
b) A 2 - among the drawn balls there are only 3 white ones. In this case
C 2 (B 2 C 1);P(A 2) = P(B 1) * P(C 1) + P(B 2) * P(C 2)
P(A 2) = 10/66 * 6/45 + 35/66 * 24/45 = 33/99 = 1/3.
c) A 3 - among the drawn balls there is at least one white one.
- among the extracted balls there is not a single white ball. Then ) = P(B 3) * P(C 3) = 21/66 * 6/45 = 7/165;P(A 3) = 1 - P(
) = 1 - 7/165 = 158/165.Answer: P(A 1) = 46/495, P(A 2) = 1/3, P(A 3) = 158/165.
Problem 1.7. The urn contains 5 black and white balls, 4 white balls are added to them. After this, 3 balls are randomly drawn from the urn. Find the probability that all the drawn balls are white, assuming that all possible proposals about the original contents of the urn are equally possible.
There are two types of tests here: first, the initial contents of the urn are set and then the 3rd ball is randomly drawn, with the result of the second test depending on the result of the first. Therefore, the total probability formula is used.
event A - 3 white balls are randomly drawn. The probability of this event depends on the initial composition of the balls in the urn.
Let's consider the events:
In 1 - there were 5 white balls in the urn;
In 2 - there were 4 white and 1 black balls in the urn;
In 3 - there were 3 white and 2 black balls in the urn;
In 4 - there were 2 white and 3 black balls in the urn;
At 5 - there were 1 white and 4 black balls in the urn.
At 6 - there were 5 black balls in the urn;
Total number of elementary outcomes
Let us find the conditional probabilities of event A under various conditions.
P(A/B 1) = 1. P(A/B 2) = 56/84 = 2/3. P(A/B 3) = 35/84 = 5/12. P(A/B 4) = 5/21. P(A/B 5) = 5/42. P(A/B 6) = 1/21.P(A) = 1 * 1/6 + 2/3 * 1/6 + 5/12 * 1/6 + 5/21 * 1/6 + 5/42 * 1/6 + 1/21 * 1/6 = 209/504.
Problem 1.10. In the installation shop, an electric motor is connected to the device. Electric motors are supplied by three manufacturers. In stock there are electric motors of these factories, respectively, in quantities M 1 = 13, M 2 = 12, and M 3 = 17 pieces, which can operate without failure until the end of the warranty period with probabilities of 0.91, 0.82, and 0.77, respectively. A worker randomly takes one electric motor and mounts it to the device. Find the probability that an electric motor installed and operating without failure until the end of the warranty period was supplied by the first, second or third manufacturer, respectively.