Chapter 13 Probability | class 12th | Important Question for Maths

Probability Class 12 Important Questions with Solutions

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(AB) = P(A∩B)P(B)=2361836=218=19

Question 4.
Evaluate P(A ∪ B), if 2P (A) = P(B) = 513 and P(A/ B) = 25. (CBSE 2018C)
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) = n(E)n(S)=336=112
and P(not getting a total of 10) = P(Ē)
1 – P(E) = 1 – 112 = 111
Thus, P(A getting 10) = 112 and P(B getting 10) and P(A is not getting 10) = P(B is not getting 10) = 1112
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 12 and 13, respectively. If both try to solve problem independently, then find the probability that
(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]
= (12×23)+(12×13)+(12×13)
= 26+16+16=46=23
Hence, probability that the problem is solved, is 23.

(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)
= (12×23) = (12×13) = 26+16=36=12

Alternate Method
P (problem is solved)
= 1 – P (none of them solve the problem)
= 1 – P(Ā ∩ B̄)
= l – P(Ā) ∙ P(B̄) = 1 – (12×23)
∵ P(Ā) = 12 and P(B̄) = 23
= 1 – 13 = 23

(ii) P (exactly one of them solve the problem)
= P(A) + P(B) – 2P(A ∩ B)
= P(A) + P(B) – 2P(A) × P(A)
= 12+13−2×12×13=12

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) = P(A∩B)P(B) to get the desired result.

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 A_T: Event that A speaks truth
and B_T: Event that B speaks truth.
Given, P(A_T) = 75100, then P(Ā_T) = 1 – 75100
[∵ P(Ā) = 1 – P(A)]
= 25100
and P(B_T) = 90100
Then, P(B̄_T) = 1 – 90100 = 10100
Now, P (A and B are contradict to each other)

∴ Percentage of P (A and B are contradict to each other) = 310 × 100 = 30%
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 p_T: Event that P speaks truth
and Q_T: 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) = 215 and (A ∩ B̄) = 16, then find P (A) and P (B). (Delhi 2015)
Answer:
Given, A and B are two independent events with
P(Ā ∩ B) = 215 and P(A ∩ B) = 16.
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) = 14+112+112+112+112+112+112=34
and P(A ∩ B) = P((T, 5)) + P((T, 6)) = 112+112=16
∴ Required probability
= P(AB)=P(A∩B)P(B)=1/63/4=418=29

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) = 14
P(X = 1) = P (Getting one head) = P(HT or TH) = 24=12
P(X = 2) = P (getting two heads) = P(HH) = 14
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 = 16

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) = 26=13
and P(E₂) = 46=23
P(AE1) = P (getting exactly one tail when a coin is tossed three times) = 38
P(AE2) = P (getting exactly a tail when a coin is tossed once) = 12
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 = 5!3! = 5 × 4 = 20
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)) = 220=110
P(X = 3) = P (getting (1, 3) or (3, 1) or (2, 3) or (3, 2) = 420=210
P(X = 4) = P(getting(1,4) or (4, 1)or (2, 4) or (4, 2) or (3, 4) or (4, 3)) = 620=310
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))
= 820=410
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(EE1) = 0.7 and P(EE2) = 0.3
Now, we have to find the probability that new product is introduced by second event.