The total probability formula allows you to find the probability of an event A, which can occur only with each of n mutually exclusive events that form a complete system if their probabilities are known, and conditional probabilities events A with respect to each of the events of the system are equal to .
Events are also called hypotheses, they are mutually exclusive. Therefore, in the literature you can also find their designation not by the letter B, but with a letter H(hypothesis).
To solve problems with such conditions, it is necessary to consider 3, 4, 5, or in the general case n the possibility of an event A- with every event.
Using the theorems of addition and multiplication of probabilities, we obtain the sum of the products of the probability of each of the events of the system by conditional probability events A for each event in the system. That is, the probability of an event A can be calculated by the formula
or in general
,
which is called total probability formula .
Total probability formula: examples of problem solving
Example 1 There are three identical-looking urns: in the first one there are 2 white balls and 3 black ones, in the second one - 4 white ones and one black one, in the third one - three white balls. Someone randomly approaches one of the urns and takes one ball out of it. Taking advantage total probability formula, find the probability that the ball is white.
Decision. Event A- the appearance of a white ball. We put forward three hypotheses:
First urn selected;
The second urn is chosen;
The third urn has been chosen.
Conditional event probabilities A for each of the hypotheses:
,
,
.
We apply the total probability formula, as a result - the required probability:
.
Example 2 At the first factory, out of every 100 light bulbs, an average of 90 standard ones are produced, at the second - 95, at the third - 85, and the products of these factories account for 50%, 30% and 20%, respectively, of all electric bulbs supplied to the shops of a certain area. Find the probability of purchasing a standard light bulb.
Decision. Let us denote the probability of acquiring a standard light bulb as A, and the events that the purchased light bulb was manufactured at the first, second and third factories, respectively, through . By condition, the probabilities of these events are known: , , and the conditional probabilities of the event A regarding each of them: 
,
,
. These are the probabilities of acquiring a standard light bulb, provided that it is manufactured at the first, second, and third factories, respectively.
Event A will occur if an event occurs or K- the bulb is made at the first factory and is standard, or an event L- the bulb is made at the second factory and is standard, or an event M- the bulb is manufactured at the third factory and is standard. Other possibilities for the occurrence of the event A no. Therefore, the event A is the sum of events K, L and M that are incompatible. Applying the probability addition theorem, we represent the probability of an event A as
and by the probability multiplication theorem we get
i.e, a special case of the total probability formula.
Substituting the probabilities into the left side of the formula, we obtain the probability of the event A :
Example 3 The aircraft is landing at the airport. If the weather allows, the pilot lands the plane, using, in addition to instruments, also visual observation. In this case, the probability of a successful landing is . If the airfield is overcast with low clouds, then the pilot lands the plane, orienting himself only on instruments. In this case, the probability of a successful landing is ; . Devices that provide blind landing have reliability (probability of failure-free operation) P. In the presence of low cloudiness and failed blind landing instruments, the probability of a successful landing is ; . Statistics show that in k% of landings, the airfield is covered with low clouds. To find full probability of the event A- safe landing of the aircraft.
Decision. Hypotheses:
There is no low cloud cover;
There is low cloud cover.
The probabilities of these hypotheses (events):
;
Conditional Probability.
The conditional probability is again found by the formula for the total probability with hypotheses
Blind landing devices work;
Blind landing instruments failed.
The probabilities of these hypotheses are:
According to the total probability formula
Example 4 The device can operate in two modes: normal and abnormal. Normal mode is observed in 80% of all cases of operation of the device, and abnormal - in 20% of cases. Probability of device failure in a certain time t equal to 0.1; in the abnormal 0.7. To find full probability device failure in time t.
Decision. We again denote the probability of device failure as A. So, regarding the operation of the device in each mode (events), the probabilities are known by condition: for the normal mode it is 80% (), for the abnormal mode - 20% (). Event Probability A(that is, the failure of the device) depending on the first event (normal mode) is 0.1 (); depending on the second event (abnormal mode) - 0.7 ( 
). We substitute these values into the total probability formula (that is, the sum of the products of the probability of each of the events of the system and the conditional probability of the event A regarding each of the events of the system) and we have the required result.
If the event BUT can occur only together with one of the events ,, ..., forming a complete group of incompatible events (these events are called hypotheses), then the probability of the occurrence of event A is calculated by the formula full probability :
. (4.1)
Let the event in the scheme described above BUT happened and it is required to find out the probability that it happened together with one of the hypotheses . This probability is calculated from Bayes formulas :
,
.
(4.2)
Problem Solving Samples
Example1 ‑ There are three identical-looking urns; the first one has 2 white and 3 black balls, the second one has 4 white and 1 black ball, the third one has 3 white balls. One of the urns is chosen at random and one ball is drawn from it. Find the probability that this ball is white.
Decision
Experience suggests three hypotheses:
-selection of the first urn, ;
-selection of the second urn, ;
-selection of the third urn, .
Consider the event of interest A - the drawn ball is white. This event can only occur in conjunction with one of the following hypotheses:
According to the total probability formula (4.1), we obtain
Answer: .
Example2 ‑ Two machines produce the same parts that are fed to a common conveyor. The performance of the first machine is twice that of the second. The first machine produces on average 60% of parts of excellent quality, and the second - 84%. The part taken at random from the assembly line turned out to be of excellent quality. Find the probability that this item was produced by the first machine.
Decision
Two assumptions (hypotheses) can be made: - the part is produced by the first automaton, and (since the first automaton produces twice as many parts as the second); - the part is produced by the second automaton, and .
The conditional probability that the part will be of excellent quality if it is produced by the first machine, if it is produced by the second machine.
The probability that a part taken at random will be of excellent quality, according to the total probability formula (4.1), is equal to:
The desired probability that the excellent part taken is produced by the first automaton, according to the Bayes formula, is equal to:
.
Answer: .
Tasks for independent solution
1 There are 20 skiers, 6 cyclists and 4 runners in the group of athletes. The probability of meeting the qualifying standard is as follows: for a skier - 0.9, for a cyclist - 0.8 and for a runner - 0.75. Find the probability that an athlete, chosen at random, will fulfill the norm.
2 From an urn containing 5 white and 3 black balls, one ball is drawn at random and transferred to another urn, which previously contained 2 white and 7 black balls. The color of the transferred ball is not fixed. One ball is drawn at random from the second urn. What is the probability that this ball is white?
3 There are 5 rifles in the pyramid, three of which are equipped with an optical sight. The probability that the shooter will hit the target when fired from a rifle with a telescopic sight is 0.95; for a rifle with a normal scope, this probability is 0.7. Find the probability that the target will be hit if the shooter fires one shot from a rifle taken at random.
4 In the conditions of the previous task, the shooter hit the target. Determine the probability that he fired: from a rifle with a telescopic sight; from a rifle with a conventional sight.
5 To participate in student qualifying sports competitions, 4 students were selected from the first group of the course, 6 from the second, and 5 from the third. The probabilities that a student of the first, second and third groups gets into the team of the institute, respectively, are 0.9; 0.7 and 0.8. A randomly selected student ended up in the national team as a result of the competition. To which group did this student most likely belong?
6 The first urn contains 10 balls, of which 8 are white; The second urn contains 20 balls, 4 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 a white ball is taken.
7 In a group of 10 students who came to the exam, 3 were excellent, 4 were good, 2 were mediocre, and 1 was bad. There are 20 questions in the exam papers. A well-prepared student knows all 20 questions, a well-prepared student knows 16, a mediocre student knows 10, and a poor student knows 5. A random student answered three randomly asked questions. Find the probability that this student is prepared: excellent; bad.
8 Each of the three urns contains 6 black and 4 white balls. One ball is drawn at random from the first urn and transferred to the second urn, after which one ball is randomly drawn from the second urn and transferred to the third urn. Find the probability that a ball drawn at random from the third urn is white.
9 Three single independent shots are fired at the object. The probability of hitting the first shot is 0.4; at the second - 0.5; with the third - 0.7. Three hits are certainly enough to disable an object; with two hits, it fails with a probability of 0.6; with one - with a probability of 0.2. Find the probability that as a result of three shots the object will be disabled.
10 Three arrows fired a volley, and two bullets hit the target. Find the probability that the third shooter hit the target if the probabilities of hitting the target by the first, second and third shooters, respectively, are 0.6; 0.5 and 0.4.
Homework.
1 Repetition of tests. Bernoulli and Poisson formulas. Local and integral theorems of Laplace.
2 Solve problems.
Task1 . There are two urns. The first urn contains two white and three black balls, and the second urn contains three white and five black. From the first and second urns, without looking, one ball is taken and placed in the third urn. The balls in the third urn are shuffled and one ball is taken at random from it. Find the probability that this ball is white.
Task2 . One of the three shooters is called to the line of fire and fires a shot. The target is hit. The probability of hitting the target with one shot for the first shooter is 0.3, for the second - 0.5, for the third - 0.8. Find the probability that the shot was fired by the second shooter.
Task3 . From the first machine, 40% goes to the assembly, from the second - 35%, from the third - 25% of the parts. Among the parts of the first machine 0.2% are defective, the second - 0.3%, the third - 0.5%. Find the probability that:
a) the part received for assembly is defective;
b) the part that turned out to be defective was made on the second machine.
Task4 . In a group of 20 shooters, five are excellent, nine are good and six are mediocre. With one shot, an excellent shooter hits the target with a probability of 0.9, a good one with a probability of 0.8 and a mediocre one with a probability of 0.7. A randomly chosen shooter fired twice; There was one hit and one miss. Which shooter was most likely to be excellent, good, or mediocre?
Example 1. In the first urn: three red, one white balls. In the second urn: one red, three white balls. A coin is thrown at random: if the coat of arms is chosen from the first urn, otherwise, from the second.
Decision:
a) the probability of drawing a red ball
A - got a red ball
P 1 - coat of arms fell out, P 2 - otherwise
b) A red ball is chosen. 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).
Decision:
A is the probability of a white ball appearing
a) All whites:
(probability that one of the three options where there is white is caught)
(probability of a white ball appearing where all are white)
b) Pulled out where everyone is black
c) pulled out a variant where all are white or/and black
- at least one of them is white
P a + P b + P c =
Example 3 . An urn contains 5 white and 4 black balls. 2 balls are taken out of it in a row. Find the probability that both balls are white.
Decision:
5 white, 4 black balls
P(A 1) - drawn a white ball
P(A 2) is the probability that the second ball is also white
P(A) – White balls selected in a row
Example 3a. There are 2 counterfeit and 8 real banknotes in a pack. 2 banknotes were pulled out of the pack in a row. Find the probability that both are false.
Decision:
P(2) = 2/10*1/9 = 1/45 = 0.022
Example 4. There are 10 urns. 9 urns contain 2 black and 2 white balls. There are 5 whites and 1 black in 1 urn. A ball is drawn from an urn taken at random.
Decision:
P(A)-? a white ball is taken from an urn containing 5 white
B - the probability of being taken out of the urn, where 5 are white
, - taken out from others
C 1 - the probability of the appearance of a white ball in lvl 9.
C 2 - the 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 conical ones. The picker takes 1 roller and then another.
Decision:
a) both rollers are cylindrical
P(C 1)=; P(C 2)=
C 1 - the first cylinder, C 2 - the second cylinder
P(A)=P(C 1)P(C 2) =
b) At least one cylinder
K 1 - the first cone.
K 2 - the second cone.
P(B)=P(C 1)P(K 2)+P(C 2)P(K 1)+P(C 1)P(C 2)
;
c) the first cylinder, and the second is not
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 pieces are drawn at random.
a) One of them is defective
P n (K)=C n k p k q n-k ,
P is the probability of defective products
q is the probability of standard parts
n=3, three parts
b) two of the 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 3 black balls, and the 2nd urn contains 3 white and 4 black. 2 balls are transferred from the 1st urn to the 2nd urn without looking, and then 2 balls are drawn from the 2nd urn. What is the probability that they are different colors?
Decision:
When transferring balls from the first urn, the following options are possible:
a) 2 white balls are drawn in a row
P WB 1 =
There will always be one less ball in the second step, since one ball has already been taken out in the first step.
b) one white and one black ball is drawn
The situation when the white ball was drawn first, and then the black one
P BC =
The situation when the black ball was drawn first, and then the white one
P BW =
Total: P CU 1 =
c) 2 black balls are drawn 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 is drawn from the second urn
P BC 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 a white ball was drawn first, 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 taken out, then a black ball, provided that both black balls were taken out of the first urn in a row. That is why the number of black balls in this case is 6 (4+2).
The probability that the drawn 2 balls 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 are randomly transferred to the 2nd urn, containing 2 white and 6 black balls. Then 1 ball is drawn at random from the 2nd urn.
1) What is the probability that the ball drawn from urn 2 is white?
2) The ball drawn from the 2nd urn turned out to be white. Calculate the probability that balls of different colors were transferred from urn 1 to urn 2.
Decision.
1) Event A - the ball drawn from the 2nd urn turned out to be white. Consider the following options for the occurrence of this event.
a) Two white balls are placed from the first urn into the second one: P1(bb) = 5/8*4/7 = 20/56.
There are 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 are placed from the first urn into the second one: P1(bc) = 5/8*3/7+3/8*5/7 = 30/56.
There are 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 are placed from the first urn into the second one: P1(hh) = 3/8*2/7 = 6/56.
There are 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 turned out to be white is equal to:
P(A) = 80/448 + 90/448 + 12/448 = 13/32
2) The ball drawn from the 2nd urn turned out to be white, i.e. the total probability is P(A)=13/32.
The probability that balls of different colors (black and white) were transferred to 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. One ball is chosen at random from the first, and two balls from the second. After that, one ball is taken at random from the chosen three balls. This last ball turned out to be black. Find the probability that a white ball was chosen from the first urn.
Decision.
Let's consider all variants of the event A - out of three balls, the drawn ball turned out to be black. How could it happen that among the three balls was black?
a) A black ball is drawn from the first urn, and two white balls are drawn from the second urn.
P1 = (3/11)(5/8*4/7) = 15/154
b) A black ball is drawn from the first urn, and two black balls are drawn from the second urn.
P2 = (3/11)(3/8*2/7) = 9/308
c) A black ball is drawn from the first urn, and one white and one black ball are drawn from the second urn.
P3 = (3/11)(3/8*5/7+5/8*3/7) = 45/308
d) A white ball is drawn from the first urn, and two black balls are taken from the second urn.
P4 = (8/11)(3/8*2/7) = 6/77
e) A white ball was taken out of the first urn, and one white and one black ball were taken out of 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 was chosen 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, provided that a black one was chosen from three balls, is equal to:
Pch \u003d Pb (1) / P \u003d 36/77 / 57/77 \u003d 36/57
Example 7c. The first urn contains 12 white and 16 black balls, the second urn contains 8 white and 10 black. At the same time, a ball is drawn from the 1st and 2nd urn, mixed and returned one at a time 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.
Decision.
Event A - at the same time, a ball is drawn 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 returning the ball to the urn of a white or black ball is ½.
Consider the variants of event B - they turned out to be of the same color.
For the first urn
1) a white ball was placed in the first urn, and a white one 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 drawn, provided that a black ball was drawn earlier, P1(BB/A=W) = ½ * 13/28 * 4/7 = 13/98
3) a white ball was placed in the first urn and a black one was drawn, provided that a white ball was previously drawn, 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 drawn, provided that a black ball was drawn earlier, P1(BC/A=Ch) = 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 one was drawn, provided that a black ball was previously drawn, P1(BW/A=W) = ½ * 12/28 * 4/7 = 6/49
7) a black ball was placed in the first urn, and a black one was drawn, provided that a white ball was previously drawn, 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 drawn, 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 one 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 drawn, provided that a black ball was drawn earlier, P1(BB/A=W) = ½ * 9/18 * 4/7 = 1/7
3) a white ball was placed in the first urn and a black one was drawn, provided that a white ball was drawn 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 drawn, provided that a black ball was drawn earlier, P1(BC/A=Ch) = ½ * 9/18 * 4/7 = 1/7
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 * 7/18 * 3/7 = 1/12
6) a black ball was placed in the first urn and a white one was drawn, provided that a black ball was previously drawn, P1(BW/A=W) = ½ * 8/18 * 4/7 = 8/63
7) a black ball was placed in the first urn, and a black one was drawn, provided that a white ball was previously drawn, 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 was 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(BH/A=B) + P1(BH/A=B) + P1(BH/A=B) + P1(BH/A=B) = 6/49 + 15/98 + 51 /392 + 8/49 = 223/392
P2(H) = P1(WB/A=B) + P1(BH/A=B) + P1(BH/A=B) + P1(BH/A=B) =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 7g. The first box contains 5 white and 4 blue balls, the second 3 and 1, and the third 4 and 5, respectively. A box is chosen at random and a ball pulled out of it turns out to be blue. What is the probability that this ball is from the second box?
Decision.
A - blue balloon extraction event. Consider all options for the outcome of such an event.
H1 - drawn ball from the first box,
H2 - drawn ball from the second box,
H3 - the drawn ball from the third box.
P(H1) = P(H2) = P(H3) = 1/3
According to the condition of the problem, the conditional probabilities of event A are:
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 the H1 composition box), six other boxes with 20 balls each contain 4 red balls (this is the H2 composition box). Find the probability that a randomly drawn red ball is contained in one of the first five boxes.
Solution: The task of applying the total probability formula.
P(H 1) = 5/11
The probability that any The taken ball is contained in one of six boxes:
P(H 2) = 6/11
The event happened - a red ball was drawn. Therefore, this could happen in two cases:
a) pulled out of the first five boxes.
P 5 = 5 red balls * 5 boxes / (30 balls * 5 boxes) = 1/6
P(P 5 / H 1) \u003d 1/6 * 5/11 \u003d 5/66
b) pulled out of six other boxes.
P 6 = 4 red balls * 6 boxes / (20 balls * 6 boxes) = 1/5
P (P 6 / H 2) \u003d 1/5 * 6/11 \u003d 6/55
Total: P(P 5 /H 1) + P(P 6 /H 2) = 5/66 + 6/55 = 61/330
Therefore, the probability that a randomly drawn red ball 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 . An 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 are of the same color?
Decision. There are three possible outcomes of events:
a) among the three balls drawn, at least two are white.
P b (2) = P 2b
The total number of possible elementary outcomes for these trials is equal to the number of ways in which 3 balls can be drawn out of 9:
Find the probability that 2 of the 3 balls 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 balls drawn, at least two are black (i.e. either 2 black or 3 black).
Find the probability that 2 of the 3 balls 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
Find the probability that all the chosen balls are black.
P h (2) = 0.214+0.0119 = 0.2259
c) among the three balls drawn, at least two are red (i.e. either 2 red or 3 red).
Let's find the probability that among the chosen 3 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:
Find the probability that all the chosen balls are red.
P to (2) = 0.357 + 0.0476 = 0.4046
Then the probability that at least two balls will be of the same color is: P = P b (2) + P h (2) + P c (2) = 0.0833 + 0.2259 + 0.4046 = 0.7138
Example 10 . The first urn contains 10 balls, of which 7 are 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 a white ball is taken.
Decision. The probability that a white ball was 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 was 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
Consider the outcome of event A.
- A white ball is drawn from the first urn, and a white ball is 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 is drawn from the first urn, and a black ball is 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 is drawn from the first urn, and a white ball is 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 a box. Of them played m . For the first game, they took two balls at random and put them back after the game. For the second game, they also took two balls at random. What is the probability that the second game will be played with new balls?
Decision. Consider event A - the game was played for the second time with new balls. Let's see what events can lead to this.
Denote by g = n-m, the number of new balls before pulling out.
a) Two new balls are drawn for the first game.
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.
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))
Consider the events of the second game.
a) Two new balls were drawn, provided P1: since new balls were already 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, subject to P2: since one new ball was already 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) They pulled out two new balls, provided P3: since 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 each 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 selected if the drawn ball is white?
Decision.
The probability of choosing each box is P(H) = 1/5.
Consider the conditional probabilities of the event A - drawing a 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
So, the conditional probability that the fourth or fifth box is chosen is
P(H=4, H=5|A) = 0.2273 + 0.2273 = 0.4546
Example 13 . An urn contains 7 white and 4 red balls. Then another ball of white or red or black color was placed in the urn and after mixing one ball was taken out. He turned out to be red. What is the probability that a) a red ball was placed? b) black ball?
Decision.
a) red ball
Event A - a red ball is drawn. Event H - put a red ball. 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 - a red ball is drawn. Event H - put a black ball.
The probability that a black ball was placed in the urn is 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 other 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 one from the second?
Decision. Let the event A1 - a red ball is drawn from the first urn; A2 - a blue ball is drawn from the second urn:
,
Events A1 and A2 are independent. The probability of 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. What is the probability that both cards drawn are red?
Decision. Let the event A 1 be the first drawn card of the red suit. Event A 2 - the second drawn card of the red suit. B - both drawn cards of red suit. Since both the event A 1 and the event A 2 must occur, then B = A 1 · A 2 . Events A 1 and A 2 are dependent, hence 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, and in the second 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?
Decision. Let index 1 mean white, index 2 black; 3 - red color. Let the event A i - a ball of the i-th color is drawn from the first urn; event B j - a ball of j -th color was taken from the second urn; event A - both balls are of the same color.
A \u003d A 1 B 1 + A 2 B 2 + A 3 B 3. The events A i and B j are independent, while 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 are drawn one at a time until black appears. What is the probability that 3 balls will be drawn from the urn? 5 balls?
Decision.
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
such a situation is not possible, because only 3 white balls.
P=0
The corollary of both main theorems - the probability addition theorem and the probability multiplication theorem - is the so-called total probability formula.
Let it be required to determine the probability of some event that can occur together with one of the events:
forming a complete group of incompatible events. We will call these events hypotheses.
Let us prove that in this case
, (3.4.1)
those. the probability of an event is calculated as the sum of the products of the probability of each hypothesis and the probability of the event under this hypothesis.
Formula (3.4.1) is called the total probability formula.
Proof. Since the hypotheses form a complete group, the event can only appear in combination with any of these hypotheses:
Since the hypotheses are inconsistent, the combinations 
also incompatible; applying the addition theorem to them, we get:
Applying the multiplication theorem to the event, we get:
,
Q.E.D.
Example 1. There are three identical-looking urns; the first urn contains two white and one black ball; in the second - three white and one black; in the third - two white and two black balls. Someone chooses one of the urns at random and draws a ball from it. Find the probability that this ball is white.
Decision. Let's consider three hypotheses:
Choice of the first urn,
Choice of the second urn,
Choice of the third urn
and the event is the appearance of a white ball.
Since the hypotheses, according to the condition of the problem, are equally probable, then
.
The conditional probabilities of the event under these hypotheses are respectively equal:
According to the total probability formula
.
Example 2. Three single shots are fired at an aircraft. The probability of hitting with the first shot is 0.4, with the second - 0.5, with the third 0.7. Three hits are obviously enough to disable an aircraft; with one hit, the aircraft fails with a probability of 0.2, with two hits, with a probability of 0.6. Find the probability that as a result of three shots the aircraft will be put out of action.
Decision. Let's consider four hypotheses:
Not a single shell hit the plane,
One shell hit the plane
The plane was hit by two shells.
Three shells hit the plane.
Using the addition and multiplication theorems, we find the probabilities of these hypotheses:
The conditional probabilities of the event (aircraft failure) under these hypotheses are:
Applying the total probability formula, we get:
Note that the first hypothesis could not be introduced into consideration, since the corresponding term in the total probability formula vanishes. This is usually done when applying the total probability formula, considering not the complete group of inconsistent hypotheses, but only those of them under which a given event is possible.
Example 3. Engine operation is controlled by two regulators. A certain period of time is considered, during which it is desirable to ensure trouble-free operation of the engine. If both regulators are present, the engine fails with probability , if only the first of them is working, with probability , if only the second one is working, if both regulators fail, with probability . The first of the regulators has reliability, the second -. All elements fail independently of each other. Find the total reliability (probability of failure-free operation) of the engine.
