Probability Class 12 Important Questions with Solutions Previous Year Questions
Question 1.
If P(not A) = 0.7, P(B) = 0.7 and P(B/A) = 0.5, then find P(A/B). (All India 2019)
Answer:
Question 2.
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event ‘number is even’ and B be the event ‘number is marked red’. Find whether the events A and B are independent or not. (Delhi 2019)
Or
A die, whose faces are marked 1,2, 3 in red and 4, 5, 6 in green , is tossed. Let A be the event “number obtained is even” and B be the event “number obtained is red”. Find if A and B are independent events. (All India 2017)
Answer:
When a die is thrown, the sample space is
S = {1, 2, 3, 4, 5, 6}
⇒ n(S) = 6
Also, A: number is even and B: number is red.
∴ A = {2, 4, 6} and B = {1, 2, 3} and A ∩ B = {2}
⇒ n(A) = 3, n(B) = 3 and n(A ∩ B) = 1
Thus, A and B are not independent events.
Question 3.
A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. (CBSE 2018)
Answer:
Let us denote the numbers on black die by B1, B2, ……., B6 and the numbers on red die by R1, R2, ….., R6.
Then, we get the following sample space.
s = {(B1, R1) ,(B1, R2), ……., (B1, R6), (B2, R2), ………, (B6, B1), (B6, B2), ……., (B6,R6)
Clearly, n(S) = 36
Now, let A be the event that sum of number obtained on the die is 8 and B be the event that red die shows a number less than 4.
Then, A = {(B2, R6), (B6,R2), (B3,R5), (B5,R3), (B4,R4)}
and B = {(B1, R1), (B1,R2), (B1,R3), (B2,R1), (B2,R2), (B2,R3) ,…….., (B6,R1), (B6,R2), (B6,R3)}
⇒ A ∩ B = {(B6, R2), (B5, R3)}
Now, required probability,
p
Question 4.
Evaluate P(A ∪ B), if 2P (A) = P(B) =
Answer:
Question 5.
Prove that if E and F are independent events, then the events E and F’ are also independent. (Delhi 2017)
Answer:
Given, E and F are independent events, therefore
⇒ PE( ∩ F) = P(E) P(F) …….. (i)
Now, we have,
P(E ∩ F’) + P(E ∩ F) = P(E)
P(E ∩ F’) = P(E) – P(E ∩ F)
P(E ∩ F’) = P(E) – P(E ) P(F) [using Eq. (i))
P(E ∩ F’) = P(E) [1 – P(F)]
P (E ∩ F’) = P(E ) P(F’)
∴ E and F ‘are also independent events.
Hence proved.
Question 6.
A and B throw a pair of dice alternately. A wins the game, if he gets a total of 7 and B wins the game, if he gets a total of 10. If A starts the game, then find the probability that B wins. (Delhi 2016)
Answer:
Here, n(S) = 6 × 6 = 36
Let A = Event of getting a sum of 7 in pair of dice = {(1, 6), (2, 5), (3, 4), (6, 1), (5, 2), (4, 3)}
⇒ n(A) = 6
and B = Event of getting a sum of 10 in pair of dice = {(4, 6), (5, 5), (6, 4)} ⇒ n(B) = 3
Now, the probability that if A start the game, then B wins
P(B wins) = P(Ā ∩ B) + P (Ā ∩ B̄ ∩ Ā ∩ B) + P(Ā ∩ B̄ ∩ Ā ∩ B̄ ∩ Ā ∩ B) + …
= P(Ā) P(B) + P(Ā)P(B̄)P(Ā)P(B) + P(Ā)P(B̄)P(Ā)P(B̄)P(Ā) P(B) + ….. (1)
[∵ events are independent]
Question 7.
A and B throw a pair of dice alternately, till one of them gets a total of 10 and wins the game. Find their respective probabilities of winning, if A starts first. (All India 2016)
Answer:
Here, n(s) = 6 × 6 = 36
Let E = Event of getting a total 10
= {(4, 6), (5, 5), (6,4)}
∴ n(E) = 3
∴ P(getting a total of 10) = P(E) =
and P(not getting a total of 10) = P(Ē)
1 – P(E) = 1 –
Thus, P(A getting 10) =
Now, P(A winning) = = P(Ā ∩ B) + P (Ā ∩ B̄ ∩ Ā) + P(Ā ∩ B̄ ∩ Ā ∩ B̄ ∩ Ā) + …
= P(Ā) + P(Ā)P(B̄)P(A) + P(Ā)P(B̄)P(Ā)P(B̄)P(A) + ……..
Question 8.
Probabilities of solving a specific problem independently by A and B are
(i) problem is solved.
(ii) exactly one of them solves the problem. (All India 2015C, Delhi 2011)
Answer:
The problem is solved means atleast one of them solve it. Also, use the concept A and B are independent events, then their complements are also independent.
Let P(A) = Probability that A solves the problem
P(B) = Probability that B solves the problem
P(Ā) = Probability that A does not solve the problem
and P(Ā) = Probability that B does not solve the problem.
According to the question, we have
(i) P (problem is solved)
= P (A ∩ B̄) + P(Ā ∩ B) + P (A ∩ B)
= P(A) ∙ P(B̄) + P(Ā) ∙ P(B) + P(A) ∙ P(B)
[∵ A and B are independent events]
=
=
Hence, probability that the problem is solved, is
(ii) P (exactly one of them solve the problem)
= P (A solve but B do not solve) + P (A do not solve but B solve)
= P(A ∩ B̄) + P(Ā ∩ B)
= P(A) ∙ P(B̄) + P(Ā) ∙ P(B)
=
Alternate Method
P (problem is solved)
= 1 – P (none of them solve the problem)
= 1 – P(Ā ∩ B̄)
= l – P(Ā) ∙ P(B̄) = 1 –
∵ P(Ā) =
= 1 –
(ii) P (exactly one of them solve the problem)
= P(A) + P(B) – 2P(A ∩ B)
= P(A) + P(B) – 2P(A) × P(A)
=
Question 9.
A couple has 2 children. Find the probability that both are boys, if it is known that
(i) one of them is a boy.
(ii) the older child is a boy. (Delhi 2014C, All India 2014, 2010)
Answer:
Firstly, write the sample space of given data. Then, use concept of conditional probability
P(A / B) =
Let B and b represent older and younger boy child. Also, let G and g represent older and younger girl child. The sample space of the given question is S = {Bb, Bg, Gg, Gb}
∴ n(S) = 4
Let A be the event that both children are boys.
Then, A = {Bb}
∴ n(A) = 1
(i) Let B : Atleast one of the children is a boy
∴ B = {Bb, Bg, Gb} and n(B) = 3
(ii) Let C : The older child is a boy.
Then, C = {Bb, Bg}
∴ n(C) = 2
Question 10.
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, then what is the conditional probability that both are girls? Given that
(i) the youngest is a girl?
(ii) atleast one is a girl? (Delhi 2014)
Answer:
Let B and b represent elder and younger boy child. Also, G and g represent elder and younger girl child. If a family has two children, then all possible cases are
S = {Bb, Bg, Gg, Gb}
∴ n(S) = 4
Let us define event A : Both children are girls, then A = {Gg} ⇒ n(A) = 1
(i) Let E1 : The event that youngest child is a girl.
Then, E1 = {Bg, Gg} and n(E1) = 2
(ii) Let E2: The event that atleast one is girl.
Then, E2 = {Eg, Gg, Gb} ⇒ n(E2) = 3,
Question 11.
A speaks truth in 75% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact? Do you think that statement of B is true? (All India 2013)
Answer:
Let AT: Event that A speaks truth
and BT: Event that B speaks truth.
Given, P(AT) =
[∵ P(Ā) = 1 – P(A)]
=
and P(BT) =
Then, P(B̄T) = 1 –
Now, P (A and B are contradict to each other)
∴ Percentage of P (A and B are contradict to each other) =
Since, B speaks truth in only 90% (i.e. not 100%) of the cases, therefore we think, the statement of B may be false.
Question 12.
P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of cases are they likely to agree in stating the same fact? Do you think, when they agree, means both are speaking truth? (All India 2013)
Answer:
Let pT: Event that P speaks truth
and QT: Event that Q speaks truth.
No, agree does not mean that they are speaking truth.
Question 13.
A speaks truth in 60% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact? In the cases of contradiction do you think, the statement of B will carry more weight as he speaks truth in more number of cases than A? (Delhi 2013)
Answer:
42% Yes
Question 14.
If A and B are two independent events such that P(Ā ∩ B) =
Answer:
Given, A and B are two independent events with
P(Ā ∩ B) =
We know that, if A and B are independent, then Ā, B and A, B̄ are independent events.
Question 15.
Consider the experiment of tossing a coin. If the coin shows head, toss it again, but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’, given that ‘there is atleast one tail’. (Delhi 2014C)
Answer:
The sample space S of the experiment is given as
S = {(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
The probabilities of these elementary events are
The outcomes of the experiment can be represented in the following tree diagram.
Consider the following events:
A = the die shows a number greater than 4 and
B = there is atleast one tail.
We have, A = {(T, 5), (T, 6)},
B = {(H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
and A ∩ B = {(T, 5), (T, 6)}
∴ P(B) = P((H, T)) + P((T, 1)) + P((T, 2)) + P((T, 3)) + P((T, 4)) + P((T, 5)) + P((T, 6))
⇒ P(B) =
and P(A ∩ B) = P((T, 5)) + P((T, 6)) =
∴ Required probability
=
Topic 2: Baye’s Theorem and Probability Distributions
Question 1.
Find the probability distribution of X, the number of heads in a simultaneous toss of two coins. (All India 2019)
Answer:
When two coins are tossed, there may be 1 head, 2 heads or no head at all. Thus, the possible values of X are 0, 1, 2.
Now, P(X = 0) = P (Getting no head) = P(TT) =
P(X = 1) = P (Getting one head) = P(HT or TH) =
P(X = 2) = P (getting two heads) = P(HH) =
Thus, the required probability distribution of X is
Question 2.
The random variable X has a probability distribution P(X) of the following form, where ‘k’ is some number.
Determine the value of ‘k’. (Delhi 2019)
Answer:
Given,
Making it in tabular format, we get the following
Since, sum of all probabilities is equal to 1. (112)
ΣP(X = x) = 1
⇒ P(X = 0) + P(X = 1) + P(X = 2) + 0 + 0 + ……. = 1
⇒ k + 2k + 3k = 1
⇒ 6k = 1 ⇒ k =
Question 3.
Suppose a girl throws a the. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3, 4, 5 or 6 with the die? (C8SE 2019)
Answer:
Let E1 be the event that the girl gets 1 or 2.
E2 be the event that the girl gets 3, 4, 5 or 6
and A be the event that the girl gets exactly a ‘tail’.
Then, P(E1) =
and P(E2) =
Now, required probability
Question 4.
Two numbers are selected at random (without replacement) from the first five positive integers. Let X denotes the larger of the two numbers obtained. Find the mean and variance of X. (CBSE 2018)
Answer:
Total number of possible outcomes
= 5P2 =
Here, X denotes the larger of two numbers obtained.
∴ X can take values 2, 3, 4 and 5.
Now, P(X = 2) = P (getting ( 1. 2) or (2, 1)) =
P(X = 3) = P (getting (1, 3) or (3, 1) or (2, 3) or (3, 2) =
P(X = 4) = P(getting(1,4) or (4, 1)or (2, 4) or (4, 2) or (3, 4) or (4, 3)) =
and P(X = 5) = P (getting (1, 5) or (5, 1) or (2, 5) or (5, 2) or (3, 5) or (5, 3) or (4, 5) or (5, 4))
=
Thus, the probability distribution of X is
Now, mean of X is E(X) = Σ X ∙ P(X)
Question 5.
Two groups are competing for the positions of the Board of Directors of a corporation. The probabilities that the first and second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced way by the second group. (CBSE 2018C)
Answer:
Let E1 and E2 denote the events that first and second group will win. Then,
P(E1) = 0.6 and P(E2) = 0.4
Let E be the event of introducing the new product.
Then, P
Now, we have to find the probability that new product is introduced by second event.
Question 6.
The random variable X can take only the values 0, 1, 2, 3. Given that
P(X = 0) = P(X = 1) = p and
P(X = 2) = P (X = 3) such that
Σpixi2 = 2Σpixi, find the value of P. (Delhi 2017)
Answer:
Given, X = 0, 1, 2, 3 and
P(X = 0) = P(X = 1) = p, P(X = 2) = P(X = 3)
such that Σpixi2 = 2Σpixi
Now, Σpi = 1
⇒ p0 + p1 + p2 + p3 = 1
⇒ p + p + x + x = 1
[let P(X = 2) = P(X = 3) = x]
⇒ 2p + 2x = 1
⇒ 2x = 1 – 2p
⇒ x =
The probability distribution of X is given by
Question 7.
There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and variance of X. (All India 2017)
Answer:
Here, S = {(1, 3), (1, 5), 1, 7), (3, 1), (3, 5), (3, 7), (5, 1), (5, 3), (5, 7), (7,1), (7, 3), (7, 5)}
⇒ n(S) = 12
Let random variable X denotes the sum of the numbers on two cards drawn. So, the random variables X may have values 4, 6, 8, 10 and 12.
At X = 4, P(X) =
At X = 6, P(X) =
At X = 8, P(X) =
At X = 10, P(X) =
At X = 12, P(X) =
Therefore, the required probability distribution is as follows
Question 8.
Three persons A, B and C apply for a job of Manager in a private company. Chances of their selection (A, B and C) are in the ratio 1:2:4. The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3, respectively. If the change does not take place, find the probability that it is due to the appointment of C. (Delhi 2016)
Answer:
Let us define the following events
A = selecting person A
B = selecting person B
C = selecting person C
Question 9.
A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y. (All India 2016)
Answer:
Let us define the following events:
E1 : Bag X is selected
E2 : Bag Y is selected
and E: Getting one white and one black ball in a draw of two balls.
Here, P(E1) = P(E2) =
[∵ probability of selecting each bag is equal]
Now,
=
and
=
∴ The probability that the one white and one black balls are drawn from bag Y,
Question 10.
In a game, a man wins ₹ 5 for getting a number greater than 4 and loses ₹ 1 otherwise, when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a number greater than 4. Find the expected value of the amount he wins/loses. (All India 2016)
Answer:
Let X be a random variable that denotes the amount received by the man. Then, X can take values 5, 4, 3 and – 3.
Now, P(X = 5) = P (getting a number greater than 4 in the first throw) =
P(X = 4) = P (getting a number less than or equal to 4 in the first throw and getting a number greater than 4 in the second throw) =
P(X = 3) = P (getting a number less than or equal to 4 in first two throws and getting a number greater than 4 in the third throw) =
and P(X = – 3) = P (getting a number less than or equal to 4 in all three throws)
=
Thus, the probability distribution of X is
Question 11.
A bag contains 4 balls. Two balls are drawn at random (without replacement) and are found to be white. What is the probability that all the balls in the bag are white? (All India 2016, 2014C)
Answer:
Let A : Two drawn balls are white
E1 : All the balls are white
E2 : Three balls are white
E3 : Two balls are white
Since, E1, E2 and E3 are mutually exclusive and exhaustive events.
∴ P(E1) = P(E2) = P(E3) =
Question 12.
Let X denote the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that
where k is a positive constant. Find the value of k. Also, find the probability that you will get admission in
(i) exactly one college
(ii) at most 2 colleges
(iii) at least 2 colleges. (Foreign 2016)
Answer:
We have,
We know that, sum of ah the probabilities of a distribution is always 1.
∴ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + …….. = 1
⇒ 0 + k + 4k + 2k + k + 0 + 0 + ………. = 1
⇒ 8k = 1 ⇒ k =
∴ (i) P (getting admission in exactly one college)
= P(X = 1) =
(ii) P(getting admission in at most 2 colleges)
= P(X ≤ 2)
= P(X= 0) + P(X = 1) + P(X = 2)
= 0 +
(iii) P (getting admission in at least 2 colleges)
= P(X ≥ 2) = 1 – P(X < 2)
= 1 – [P(X = 0) + P(X = 1))
= 1 – [0 +
Question 13.
A bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 red balls. A die is thrown. If 1 or 2 appears on it, then bag A is chosen, otherwise bag B. If two balls are drawn at random (without replacement) from the selected bag, find the probability of one of them being red and another black. (Delhi 2015)
Answer:
Given, bag A = 4 black and 6 red balls
bag B = 7 black and 3 red balls.
Let E1 = The event that die show 1 or 2
E2 = The event that die show 3 or 4 or 5 or 6
E = The event that among two drawn balls, one of them is red and other is black
Question 14.
Three machines E1, E2 and E3 in a certain factory producing electric bulbs, produce 50%, 25% and 25% respectively, of the total daily output of electric bulbs. It is known that 4% of the bulbs produced by each of machines E1 and E2 are defective and that 5% of those produced by machine E3 are defective. If one bulb is picked up at random from a day’s production, calculate the probability that it is defective. (Foreign 2015)
Answer:
Let A1 : Event that the bulb is produced by machine E1
A2 : Event that the bulb is produced by machine E2
A3 : Event Chat the bulb is produced by machine E3
A: Event that the picked up bulb is defective
Question 15.
Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X. (Foreign 2015)
Answer:
Given, X denotes larger of the two numbers be obtained.
Obviously X can values 3, 4, 5, 6 and 7.
Now,
Therefore, required probability distribution is as follows
Question 16.
From a lot of 15 bulbs which include 5 defectives, a sample of 2 bulbs is drawn at random (without replacement). Find the probability distribution of the number of defective bulbs. (Delhi 2015C)
Answer:
It is given that out of 15 bulbs, 5 are defective.
∴ Number of non-defective bulbs = 15 – 5 = 10
Let X be the random variable which denotes the defective bulbs. So, X may take values 0, 1, 2
P(X = 0) = P (No defective bulb)
=
[∵ bulb is drawn without replacement]
P(X = 1) = P (One defective bulb and one non defective bulb)
Question 17.
Three cards are drawn at random (without replacement) from a well-shuffled pack of 52 playing cards. Find the probability distribution of number of red cards. Hence, find the mean of the distribution. (Foreign 2014)
Answer:
Let X be a random variable that denotes number of red cards in three draws.
Here, X can take values 0, 1, 2, 3.
Now, P(X = 0) = P (getting all black cards)
=
P(X = 1) = P (getting one red card and two black cards)
Question 18.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable XI Find the mean of X. (All India 2014C)
Answer:
Firstly, find the probability of respective ages and make a probability distribution tabLe then using the formula
Mean (X) = ΣX ∙ P(X), calculate mean.
Here, total students = 15
The ages of students in ascending order are
14, 14, 15, 16, 16, 17, 17, 17, 18, 19, 19, 20, 20, 20 and 21.
Now,
P(X = 14) =
P(X = 15) =
P(X = 16) =
P(X = 17) =
P(X = 18) =
P(X = 19) =
P(X = 20) =
P(X = 21) =
Therefore, the probability distribution of random variable X is as follows
∴ Mean(X) = ΣX ∙ P(X)
Question 19.
A bag contains 3 red and 7 black balls. Two balls are selected at random one by one without replacement. If the second selected ball happens to be red, what is the probability that the first selected ball is also red? (Delhi 2014C)
Answer:
Let E1: First ball is red, E2: First ball is black
A: Second ball is red (1)
Total number of balls is 10.
Then, P(E1) =
∴
Then, by Baye’s theorem, probability of second selected ball is red when first selected ball is also red is given by
Hence, the probability that the first selected ball is red, is
Question 20.
An urn contains 4 white and 6 red balls. Four balls are drawn at random (without replacement) from the urn. Find the probability distribution of the number of white balls. (Delhi 2012c)
Answer:
Let X be a random variable that denotes the number of white balls in a draw of 4 balls. Then, X can take values of 0, 1, 2, 3 and 4.
Clearly, P (X = 0) = P (getting no white ball)
Question 21.
Two cards are drawn simultaneously (without replacement) from a well-shuffled deck of 52 cards. Find the mean and variance of number of red cards. (All India 2012)
Answer:
Firstly, find the pobability distribution table of number of red cards. Then, using this table, find the mean and variance.
Let X be the number of red cards. Then, X can take values 0, 1 and 2.
Now, P(X = 0) = p (having no red card)
Now, we know that, mean = Σx ∙ P(x)
and variance = Σx2 ∙ P(x) – [Σx ∙ P(X)]2
Question 22.
Find the mean number of heads in three tosses of a coin. (Foreign 2011)
Answer:
Firstly, we write the probability distribution table for the given experiment. Then, we find mean by using formula
Mean = ΣXiP(xi).
Let X = Number of heads when a corn is tossed three times.
Sample space of given experiment is
S = {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT}
X can take values 0, 1, 2 and 3.
Now, P(X = 0) = P (no head occur) =
P(X = 1) = P (one head occur) =