CBSE Class 12 Mathematics Probability Important Questions Set B

Question. Three machines \( E_1, E_2, E_3 \) in a certain factory produce 50%, 25% and 25% respectively, of the total daily output of electric tubes. It is known that 4% of the tube produced on each of machines \( E_1 \) and \( E_2 \) are defective and that 5% of those produced on \( E_3 \), are defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.
Answer: Let \( A \) be the event that the picked up tube is defective.
Let \( A_1, A_2, A_3 \) be events such that
\( A_1 = \) event of producing tube by machine \( E_1 \)
\( A_2 = \) event of producing tube by machine \( E_2 \)
\( A_3 = \) event of producing tube by machine \( E_3 \)
\( P(A_1) = \frac{50}{100} = \frac{1}{2}, P(A_2) = \frac{25}{100} = \frac{1}{4}, P(A_3) = \frac{25}{100} = \frac{1}{4} \)
Also, \( P(A/A_1) = \frac{4}{100} = \frac{1}{25}; P(A/A_2) = \frac{4}{100} = \frac{1}{25} \) and \( P(A/A_3) = \frac{5}{100} = \frac{1}{20} \)
Now, \( P(A) \) is required.
From concept of total probability,
\( P(A) = P(A_1) \cdot P(A/A_1) + P(A_2) \cdot P(A/A_2) + P(A_3) \cdot P(A/A_3) \)
\( = \frac{1}{2} \times \frac{1}{25} + \frac{1}{4} \times \frac{1}{25} + \frac{1}{4} \times \frac{1}{20} = \frac{1}{50} + \frac{1}{100} + \frac{1}{80} \)
\( = \frac{8 + 4 + 5}{400} = \frac{17}{400} = 0.0425 \)

Question. Four bad oranges are accidentally mixed with 16 good ones. Find the probability distribution of the number of bad oranges when two oranges are drawn at random from this lot. Find the mean and variance of the distribution.
Answer: Let \( X \) be the number of bad oranges in two draws of orange from the lot. Here, \( X \) is random variable and may have value 0, 1, 2.
Now, \( P(X = 0) = \frac{{}^{16}C_2}{{}^{20}C_2} = \frac{16!}{2! \cdot 14!} \times \frac{2! \cdot 18!}{20!} = \frac{18 \times 17 \times 16 \times 15}{20 \times 19 \times 18 \times 17} = \frac{60}{95} \)
\( P(X = 1) = \frac{{}^4C_1 \times {}^{16}C_1}{{}^{20}C_2} = \frac{4 \times 16 \times 2}{20 \times 19} = \frac{32}{95} \); \( P(X = 2) = \frac{{}^4C_2}{{}^{20}C_2} = \frac{4 \times 3}{20 \times 19} = \frac{3}{95} \)
Now, required probability distribution is

Now, \( \text{Mean} = \sum X_i P(X_i) = 0 \times \frac{60}{95} + 1 \times \frac{32}{95} + 2 \times \frac{3}{95} = \frac{32}{95} + \frac{6}{95} = \frac{38}{95} = \frac{2}{5} \)
\( \text{Variance} = \sum X_i^2 P(X_i) - (\sum X_i P(X_i))^2 = 1 \times \frac{32}{95} + 4 \times \frac{3}{95} - \left( \frac{2}{5} \right)^2 = \frac{44}{95} - \frac{4}{25} = \frac{144}{475} \)

Question. A coin is biased so that the head is three times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails. Hence find the mean of the number of tails.
Answer: A coin is biased such that
\( P(H) = \frac{3}{4} \) and \( P(T) = \frac{1}{4} \)
Let \( X = \) No. of tails when coin is tossed twice.
\( \therefore \) When \( X = 0 \) (i.e. No Tail)
\( P(X = 0) = P(H) \cdot P(H) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \)
when \( X = 1 \) (i.e., only one tail)
\( \Rightarrow P(X = 1) = P(T) \cdot P(H) + P(H) \cdot P(T) \)
\( \Rightarrow P(X = 1) = \frac{1}{4} \times \frac{3}{4} + \frac{3}{4} \times \frac{1}{4} = \frac{6}{16} \)
when \( X = 2 \) (i.e., both tails).
\( P(X = 2) = P(T) \cdot P(T) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \)
Probability distribution is given by

We have, \( \text{mean} = \sum p_i \cdot X_i \)
\( \therefore \text{Mean} = 0 \times \frac{9}{16} + 1 \times \frac{6}{16} + 2 \times \frac{1}{16} \)
\( = \frac{6}{16} + \frac{2}{16} = \frac{8}{16} = \frac{1}{2} \)

Question. Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only three cards are spades?
Answer: Let \( X \) denote the number of spade card
and, \( p = \) probability that drawn card is spade \( = \frac{13}{52} = \frac{1}{4} \)
\( \Rightarrow p = \frac{1}{4} \)
\( \therefore q = \) probability that drawn card is not spade \( = 1 - \frac{1}{4} = \frac{3}{4} \)
\( \Rightarrow q = \frac{3}{4} \)
and \( n = 5 \).
(i) \( P(X = 5) = {}^5C_5 \left( \frac{1}{4} \right)^5 \times \left( \frac{3}{4} \right)^{5-5} = 1 \times \left( \frac{1}{4} \right)^5 \times 1 = \left( \frac{1}{4} \right)^5 = \frac{1}{1024} \)
(ii) \( P(X = 3) = {}^5C_3 \left( \frac{1}{4} \right)^3 \times \left( \frac{3}{4} \right)^{5-3} = \frac{5!}{3! \cdot 2!} \times \left( \frac{1}{4} \right)^3 \times \left( \frac{3}{4} \right)^2 \)
\( = 10 \times \frac{9}{4^5} = \frac{90}{1024} = \frac{45}{512} \)

Long Answer Questions 

Question. There are two boxes I and II. Box I contains 3 red and 6 black balls. Box II contains 5 red and ‘n’ black balls. One of the two boxes, box I and box II is selected at random and a ball is drawn at random. The ball drawn is found to be red. If the probability that this red ball comes out from box II is \( \frac{3}{5} \), find the value of ‘n’.
Answer: We have events
\( E_1 \): bag I is selected; \( E_2 \): bag II is selected; \( A \): getting a red ball
So, \( P(E_1) = \frac{1}{2}, P(E_2) = \frac{1}{2}, P(A/E_1) = \frac{3}{9} = \frac{1}{3} \) and \( P(A/E_2) = \frac{5}{5+n} \)
Now, \( P(E_2/A) = \frac{P(E_2) \cdot P(A/E_2)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( \Rightarrow \frac{3}{5} = \frac{\frac{1}{2} \times \frac{5}{5+n}}{\frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{5}{5+n}} = \frac{\frac{5}{5+n}}{\frac{1}{3} + \frac{5}{5+n}} \)
\( \Rightarrow \frac{3}{5} = \frac{5}{5+n} \times \frac{3(5+n)}{(5+n) + 15} = \frac{15}{20+n} \)
\( \Rightarrow 60 + 3n = 75 \Rightarrow 3n = 15 \Rightarrow n = 5 \)

Question. A card from a pack of 52 playing cards is lost. From the remaining cards of the pack three cards are drawn at random (without replacement) and are found to be all spades. Find the probability of the lost card being a spade.
Answer: Let \( E_1, E_2, E_3, E_4 \) and \( A \) be event defined as
\( E_1 = \) the lost card is a spade card, \( E_2 = \) the lost card is a non spade card
and \( A = \) drawing three spade cards from the remaining cards.
Now, \( P(E_1) = \frac{13}{52} = \frac{1}{4}, P(E_2) = \frac{39}{52} = \frac{3}{4} \)
\( P(A/E_1) = \frac{{}^{12}C_3}{{}^{51}C_3} = \frac{220}{20825}; P(A/E_2) = \frac{{}^{13}C_3}{{}^{51}C_3} = \frac{286}{20825} \)
Here, required probability \( = P(E_1/A) \)
\( \therefore P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( = \frac{\frac{1}{4} \times \frac{220}{20825}}{\frac{1}{4} \times \frac{220}{20825} + \frac{3}{4} \times \frac{286}{20825}} = \frac{220}{220 + 3 \times 286} = \frac{220}{1078} = \frac{10}{49} \)

Question. Bag I contains 3 red and 4 black balls and bag II contains 4 red and 5 black balls. Two balls are transferred at random from bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred balls were both black.
Answer: Let \( E_1, E_2, E_3 \) and \( A \) be events such that
\( E_1 = \) both transferred balls from bag I to bag II are red.
\( E_2 = \) both transferred balls from bag I to bag II are black.
\( E_3 = \) out of two transferred balls one is red and other is black.
\( A = \) drawing a red ball from bag II.
Here, \( P(E_2/A) \) is required.
Now, \( P(E_1) = \frac{{}^3C_2}{{}^7C_2} = \frac{3!}{2! \cdot 1!} \times \frac{2! \cdot 5!}{7!} = \frac{3}{21} = \frac{1}{7} \); \( P(E_2) = \frac{{}^4C_2}{{}^7C_2} = \frac{6}{21} = \frac{2}{7} \)
\( P(E_3) = \frac{{}^3C_1 \times {}^4C_1}{{}^7C_2} = \frac{3 \times 4}{21} = \frac{12}{21} = \frac{4}{7} \)
\( P(A/E_1) = \frac{6}{11}, P(A/E_2) = \frac{4}{11}, P(A/E_3) = \frac{5}{11} \)
\( \therefore P(E_2/A) = \frac{P(E_2) \cdot P(A/E_2)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) + P(E_3) \cdot P(A/E_3)} \)
\( = \frac{\frac{2}{7} \times \frac{4}{11}}{\frac{1}{7} \times \frac{6}{11} + \frac{2}{7} \times \frac{4}{11} + \frac{4}{7} \times \frac{5}{11}} = \frac{\frac{8}{77}}{\frac{6}{77} + \frac{8}{77} + \frac{20}{77}} = \frac{8}{34} = \frac{4}{17} \)

Question. There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up head 75% of the times and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Answer: Let \( E_1, E_2, E_3 \) and \( A \) be event defined as:
\( E_1 = \) selection of a two headed coin; \( E_2 = \) selection of a biased coin
\( E_3 = \) selection of an unbiased coin; \( A = \) coin shows head after tossing
Now, \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \)
\( P(A/E_1) = 1, P(A/E_2) = \frac{75}{100} = \frac{3}{4}, P(A/E_3) = \frac{1}{2} \)
Here, required probability \( = P(E_1/A) \)
By using Baye’s theorem,
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) + P(E_3) \cdot P(A/E_3)} \)
\( = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}} = \frac{1}{1 + \frac{3}{4} + \frac{1}{2}} = \frac{1}{\frac{4+3+2}{4}} = \frac{4}{9} \)

Question. An urn contains 4 white and 3 red balls. Let X be the number of red balls in a random draw of three balls. Find the mean and variance of X.
Answer: Let \( X \) be the number of red balls in a random draw of three balls.
As there are 3 red balls, possible values of \( X \) are 0, 1, 2, 3.
\( P(0) = \frac{{}^4C_3}{{}^7C_3} = \frac{4 \times 3 \times 2}{7 \times 6 \times 5} = \frac{4}{35} \); \( P(1) = \frac{{}^3C_1 \times {}^4C_2}{{}^7C_3} = \frac{3 \times 6}{35} = \frac{18}{35} \)
\( P(2) = \frac{{}^3C_2 \times {}^4C_1}{{}^7C_3} = \frac{3 \times 4}{35} = \frac{12}{35} \); \( P(3) = \frac{{}^3C_3 \times {}^4C_0}{{}^7C_3} = \frac{1 \times 1}{35} = \frac{1}{35} \)
For calculation of Mean & Variance

\( \text{Mean} = \sum XP(X) = \frac{9}{7} \)
\( \text{Variance} = \sum X^2 \cdot P(X) - (\sum X \cdot P(X))^2 = \frac{15}{7} - \frac{81}{49} = \frac{105 - 81}{49} = \frac{24}{49} \)

Question. A bag contains 5 red and 3 black balls and another bag contains 2 red and 6 black balls. Two balls are drawn at random (without replacement) from one of the bags and both are found to be red. Find the probability that balls are drawn from first bag.
Answer: Let \( E \) be the event of drawing two red balls.
Let \( B_1 \) be the event that first bag is chosen and \( B_2 \) be the event that second bag is chosen.
\( P(B_1) = P(B_2) = \frac{1}{2}, P(E/B_1) = \frac{5 \times 4}{8 \times 7} = \frac{20}{56} \)
\( P(E/B_2) = \frac{2 \times 1}{8 \times 7} = \frac{2}{56} \)
\( \therefore P(B_1/E) = \frac{P(B_1) \times P(E/B_1)}{P(B_1) \times P(E/B_1) + P(B_2) \times P(E/B_2)} \)
\( = \frac{\frac{1}{2} \times \frac{20}{56}}{\frac{1}{2} \times \frac{20}{56} + \frac{1}{2} \times \frac{2}{56}} = \frac{20}{20 + 2} = \frac{20}{22} = \frac{10}{11} \)

Question. In a certain college, 4% of boys and 1% of girls are taller than 1.75 metres. Furthermore, 60% of the students in the college are girls. A student is selected at random from the college and is found to be taller than 1.75 metres. Find the probability that the selected student is a girl.
Answer: Let \( E_1, E_2, A \) be events such that
\( E_1 = \) student selected is girl; \( E_2 = \) student selected is boy
\( A = \) student selected is taller than 1.75 metres.
Here \( P(E_1/A) \) is required.
Now, \( P(E_1) = \frac{60}{100} = \frac{3}{5}, P(E_2) = \frac{40}{100} = \frac{2}{5}, P(A/E_1) = \frac{1}{100}, P(A/E_2) = \frac{4}{100} \)
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( = \frac{\frac{3}{5} \times \frac{1}{100}}{\frac{3}{5} \times \frac{1}{100} + \frac{2}{5} \times \frac{4}{100}} = \frac{3/500}{3/500 + 8/500} = \frac{3}{11} \)

Question. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Answer: Let \( E_1, E_2 \) and \( A \) be event such that
\( E_1 = \) Production of items by machine A
\( E_2 = \) Production of items by machine B
\( A = \) Selection of defective items.
\( P(E_1) = \frac{60}{100} = \frac{3}{5}, P(E_2) = \frac{40}{100} = \frac{2}{5}, P(A/E_1) = \frac{2}{100} = \frac{1}{50}, P(A/E_2) = \frac{1}{100} \)
\( P(E_2/A) \) is required
By Baye’s theorem
\( P(E_2/A) = \frac{P(E_2) \cdot P(A/E_2)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( = \frac{\frac{2}{5} \times \frac{1}{100}}{\frac{3}{5} \times \frac{1}{50} + \frac{2}{5} \times \frac{1}{100}} = \frac{2/500}{3/250 + 2/500} = \frac{2}{6+2} = \frac{2}{8} = \frac{1}{4} \)

Objective Type Questions:

Question. Choose and write the correct option in the following questions.
(i) If \( A \) and \( B \) are two independent events with \( P(A) = \frac{3}{5} \) and \( P(B) = \frac{4}{9} \), then \( P(A' \cap B') \) equals
(a) \( \frac{4}{15} \)
(b) \( \frac{8}{45} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{2}{9} \)
Answer: (d)

(ii) If \( A \) and \( B \) are two events and \( A \neq \phi, B \neq \phi \), then
(a) \( P(A | B) = P(A) \cdot P(B) \)
(b) \( P(A | B) = \frac{P(A \cap B)}{P(B)} \)
(c) \( P(A | B) \cdot P(B | A) = 1 \)
(d) \( P(A | B) = P(A) | P(B) \)
Answer: (b)

(iii) If two events are independent then
(a) they must be mutually exclusive
(b) the sum of their probabilities must be equal to 1
(c) both (a) and (b) are correct.
(d) None of the above
Answer: (d)

(iv) If the events \( A \) and \( B \) are independent, then \( P(A \cap B) \) equals
(a) \( P(A) + P(B) \)
(b) \( P(A) - P(B) \)
(c) \( P(A) \cdot P(B) \)
(d) \( P(A)/P(B) \)
Answer: (c)

Question. For the following probability distribution:

\( E(X^2) \) is equal to
(a) 3
(b) 5
(c) 7
(d) 10
Answer: (d)

Question. Eight coins are tossed together. The probability of getting exactly 3 heads is
(a) \( \frac{1}{256} \)
(b) \( \frac{7}{32} \)
(c) \( \frac{5}{32} \)
(d) \( \frac{3}{32} \)
Answer: (b)

Question. A card is picked at random from a pack of 52 playing cards. Given that the picked card is a queen, the probability of this card to be a card of spade is 
(a) \( \frac{1}{3} \)
(b) \( \frac{4}{13} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{2} \)
Answer: (c)

Question. A die is thrown once. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then \( P(A \cup B) \) is  
(a) \( \frac{2}{5} \)
(b) \( \frac{3}{5} \)
(c) 0
(d) 1
Answer: (d)

Fill in the blanks.

Question. If \( P(A) = 0.8, P(B) = 0.5, P(B/A) = 0.4 \) then the value of \( P(A/B) \) is equal to ____________.
Answer: 0.64

Question. The probability distribution of a random variable X, is given below:

then the value of \( k \) is ____________.
Answer: 0.3

Question. If the mean and variance of a binomial variate X are 2 and 1 respectively, then \( P(X \geq 1) = \) ____________.
Answer: \( \frac{15}{16} \)

Very Short Answer Questions:

Question. Two coins are tossed. What is the probability of coming up two heads if it is known that at least one head comes up?
Answer: \( \frac{1}{3} \)

Question. Four cards are drawn from 52 cards with replacement. Find the probability of getting at least 3 aces.
Answer: \( \frac{49}{28561} \)

Question. A bag contains 5 white and 4 red balls. 2 balls are drawn from the bag. Find the probability that both balls are white.
Answer: \( \frac{5}{18} \)

Question. The probability that Hari hits a target is \( \frac{1}{4} \). He fires 64 times. Find the expected number (\( \mu \)) of times he will hit the target and also the variance (\( \sigma^2 \)).
Answer: 16, 12

Question. If \( P(A) = 0.4, P(B) = p, P(A \cup B) = 0.6 \) and A and B are given to be independent events, find the value of ‘\( p \)’. 
Answer: \( \frac{1}{3} \)

Short Answer Questions–I:

Question. Find the binomial distribution of which the mean is 4 and variance is 3.
Answer: \( B(16, \frac{1}{4}) \)

Question. Events A and B are such that \( P(A) = \frac{1}{2}, P(B) = \frac{7}{12} \) and \( P(\text{not A or not B}) = \frac{1}{4} \). State whether A and B are independent.
Answer: No, A and B are not independent.

Question. Find the mean of the binomial distribution \( B(4, \frac{1}{3}) \).
Answer: \( \frac{4}{3} \)

Question. If \( P(A) = \frac{3}{10}, P(B) = \frac{2}{5} \) and \( P(A \cup B) = \frac{3}{5} \), then find \( P(A/B) + P(B/A) \).
Answer: \( \frac{7}{12} \)

Question. In a girl’s hostel’s, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspaper. A student is selected at random. If she reads English newspaper, then find the probability that she reads Hindi newspaper also.
Answer: \( \frac{1}{2} \)

Question. The probability of simultaneous occurrence of atleast one of two events A and B is \( p \). If the probability that exactly one of A, B occurs is \( q \), then prove that \( P(A) + P(B) = 2p - q \).
Answer: Given \( P(A \cup B) = p \) and \( P(A \cap \bar{B}) + P(\bar{A} \cap B) = q \).
\( \because P(A \cup B) = P(A \cap \bar{B}) + P(\bar{A} \cap B) + P(A \cap B) \)
\( \Rightarrow p = q + P(A \cap B) \Rightarrow P(A \cap B) = p - q \).
Now, \( P(A) + P(B) = P(A \cup B) + P(A \cap B) = p + (p - q) = 2p - q \). Hence proved.