CBSE Class 12 Mathematics Probability Previous Year Questions Set 01

Short Answer Questions (2 Marks)

Question 1. 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. [CBSE 2017, 2M]
Answer:
Given A: number obtained is even, i.e. A = {2, 4, 6}
and B: number obtained is red, i.e. B = {1, 2, 3}
⇒ A ∩ B = {2}
Now, \( P(A) = \frac{3}{6} = \frac{1}{2} \), \( P(B) = \frac{3}{6} = \frac{1}{2} \) and \( P(A \cap B) = \frac{1}{6} \) [1]
Here \( P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \) [½]
∵ P(A ∩ B) ≠ P(A) . P(B)
So A and B are not independent events. [½]

Question 2. 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, 2M]
Answer:
Let E: obtaining the sum 8 on the dice
F: red die resulted in a number less than 4.
i.e. E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
⇒ n(E) = 5
F = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}
⇒ n(F) = 18
and E ∩ F = {(5, 3), (6, 2)} [1]
⇒ n(E ∩ F) = 2
Here, \( P(F) = \frac{18}{36} = \frac{1}{2} \)
and \( P(E \cap F) = \frac{2}{36} = \frac{1}{18} \)
∴ Required probability = P(E/F)
\( = \frac{P(E \cap F)}{P(F)} = \frac{1/18}{1/2} = \frac{1}{9} \) [1]

Question 3. If P(not A) = 0.7, P(B) = 0.7 and P(B/A) = 0.5, then find P(A/B). [CBSE 2019, 2M]
Answer:
Given \( P(\overline{A}) = 0.7 \) ⇒ 1 − P(A) = 0.7 ⇒ P(A) = 0.3 [½]
and P(B) = 0.7
P(B/A) = 0.5; P(A/B) = ?
We know that \( P(B/A) = \frac{P(B \cap A)}{P(A)} \)
⇒ \( 0.5 = \frac{P(B \cap A)}{0.3} \)
⇒ P(B ∩ A) = 0.5 × 0.3 = 0.15 [½]
Now, P(A ∩ B) = P(B ∩ A) = 0.15
⇒ \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.15}{0.70} = \frac{15}{70} = \frac{3}{14} \) [1]

Question 4. A purse contains 3 silver and 6 copper coins and a second purse contains 4 silver and 3 copper coins. If a coin is drawn at random from one of the two purses, find the probability that it is a silver coin. [CBSE 2020, 2M]
Answer:
Let \( E_1 \): select first purse and \( E_2 \): select second purse
A: silver coin is drawn
⇒ \( P(E_1) = \frac{1}{2},\ P(E_2) = \frac{1}{2},\ P(A/E_1) = \frac{3}{9},\ P(A/E_2) = \frac{4}{7} \) [1]
∴ \( P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2) \)   (by total probability theorem)
\( = \frac{1}{2} \times \frac{3}{9} + \frac{1}{2} \times \frac{4}{7} \)
or \( P(A) = \frac{1}{6} + \frac{2}{7} = \frac{19}{42} \) [1]

Long Answer Questions I (4 Marks)

Question 5. 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. [CBSE 2016, 4M]
OR
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.
Answer:
Let \( E_1 \) = event of drawing bag X
\( E_2 \) = event of drawing bag Y
E = event of drawing one white and one black ball.
⇒ \( P(E_1) = P(E_2) = \frac{1}{2} \)
Also, \( P\left(\frac{E}{E_1}\right) = \left(\frac{4}{6}\right)\left(\frac{2}{5}\right) + \left(\frac{2}{6}\right)\left(\frac{4}{5}\right) = \frac{16}{30} \)
\( P\left(\frac{E}{E_2}\right) = \left(\frac{3}{6}\right)\left(\frac{3}{5}\right) + \left(\frac{3}{6}\right)\left(\frac{3}{5}\right) = \frac{18}{30} \) [2]
Required probability \( = P(E_2/E) = \frac{P(E_2) \times P(E/E_2)}{P(E_1) \times P(E/E_1) + P(E_2) \times P(E/E_2)} \) (using Bayes' theorem)
\( = \frac{\frac{1}{2} \times \frac{18}{30}}{\left(\frac{1}{2} \times \frac{16}{30}\right) + \left(\frac{1}{2} \times \frac{18}{30}\right)} = \frac{9}{17} \) [2]
OR
Let \( A_i \) and \( B_i \) be the events of throwing 10 by A and B respectively.
\( P(A_i) = P(B_i) = \frac{3}{36} = \frac{1}{12} \)
\( P(\overline{A_i}) = P(\overline{B_i}) = \frac{11}{12} \) [1½]
P(A wins) = \( P(A) + P(\overline{A})P(\overline{B})P(A) + \ldots \)
\( = \frac{1}{12} + \frac{1}{12} \times \left(\frac{11}{12}\right)^2 + \frac{1}{12} \times \left(\frac{11}{12}\right)^4 + \ldots \) [1]
\( = \frac{1}{12}\left[1 + \left(\frac{11}{12}\right)^2 + \left(\frac{11}{12}\right)^4 + \ldots\right] = \frac{1}{12}\left[\frac{1}{1 - \frac{11^2}{12^2}}\right] \)
\( P(\text{A wins}) = \frac{1}{12} \times \left[\frac{12^2}{144 - 121}\right] = \frac{12}{23} \) [1]
\( P(\text{B wins}) = 1 - P(\text{A wins}) = 1 - \frac{12}{23} = \frac{11}{23} \) [½]

Question 6. Of the students in a school, it is known that 30% have 100% attendance and 70% students are irregular. Previous year results report that 70% of all students who have 100% attendance attain A grade and 10% irregular students attain A grade in their annual examination. At the end of the year, one student is chosen at random from the school and he was found to have an A grade. What is the probability that the student has 100% attendance? Is regularity required only in school? Justify your answer. [CBSE 2017, 4M]
Answer:
Let \( E_1 \): student has 100% attendance
\( E_2 \): irregular students
E: student has A grade
\( P(E_1) = \frac{30}{100} \) and \( P(E_2) = \frac{70}{100} \)
\( P(E/E_1) = \frac{70}{100} \); \( P(E/E_2) = \frac{10}{100} \) [1]
Required probability \( = P(E_1/E) = \frac{P(E_1) \times P(E/E_1)}{P(E_1) \times P(E/E_1) + P(E_2) \times P(E/E_2)} \) (using Bayes' theorem) [1]
\( = \frac{\frac{30}{100} \times \frac{70}{100}}{\left(\frac{30}{100} \times \frac{70}{100}\right) + \left(\frac{70}{100} \times \frac{10}{100}\right)} \) [1]
\( = \frac{2100}{2100 + 700} = \frac{21}{28} = \frac{3}{4} \)
Regularity is required everywhere, as it is a value relevant in every sphere of life. [1]

Question 7. Suppose a girl throws a die. 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? [CBSE 2015, 2018, 4M]
Answer:
When a die is thrown, the sample space contains 6 outcomes, i.e. S = {1, 2, 3, 4, 5, 6}
Let \( E_1 \): she gets 1 or 2 on the die
\( E_2 \): she gets 3, 4, 5 or 6 on the die
A: she gets exactly one tail
Here, \( P(E_1) = \frac{2}{6} = \frac{1}{3} \), \( P(E_2) = \frac{4}{6} = \frac{2}{3} \)
and \( P(A/E_1) = \frac{3}{8} \), \( P(A/E_2) = \frac{1}{2} \) [1]
∵ when she gets 1 or 2, she throws a coin three times, and the outcomes are {HHH, TTT, HHT, THH, HTH, TTH, HTT, THT}
Required probability = \( P(E_2/A) \)
\( = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \)   (using Bayes' theorem) [1]
\( = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} = \frac{8}{11} \) [2]

Long Answer Questions II (6 Marks)

Question 8. Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold? [CBSE 2011, 6M]
Answer:
Let \( E_1 \) = box I is chosen,
\( E_2 \) = box II is chosen,
\( E_3 \) = box III is chosen,
A = the coin drawn is of gold.
Here, \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \)
⇒ \( P(A/E_1) = \frac{2}{2} = 1,\ P(A/E_2) = 0,\ P(A/E_3) = \frac{1}{2} \) [2]
Required probability = \( P(E_1/A) \)
\( = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)} \)   (using Bayes' theorem) [2]
\( = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times 0 + \frac{1}{3} \times \frac{1}{2}} = \frac{2}{3} \) [2]

Question 9. In a hockey match, both teams A and B scored the same number of goals up to the end of the game, so to decide the winner, the referee asked both the captains to throw a die alternately and decided that the team whose captain gets a six first will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match. [CBSE 2013, 6M]
Answer:
Let E = captain A gets a six
F = captain B gets a six
\( P(E) = \frac{1}{6} \)    \( P(F) = \frac{1}{6} \)
\( P(\overline{E}) = \frac{5}{6} \)    \( P(\overline{F}) = \frac{5}{6} \) [1]
A wins if he throws a six in the 1st or 3rd or 5th throw ……
Winning probability in the first throw = \( P(E) = \frac{1}{6} \)
Winning probability in the third throw = \( P(E')P(F')P(E) = \left(\frac{5}{6}\right)^2 \times \frac{1}{6} \)
Winning probability in the fifth throw = \( P(E')P(F')P(E')P(F')P(E) = \left(\frac{5}{6}\right)^4 \times \frac{1}{6} \) [2]
Hence the probability of winning of A is
\( P(A) = \frac{1}{6} + \left(\frac{5}{6}\right)^2 \frac{1}{6} + \left(\frac{5}{6}\right)^4 \cdot \frac{1}{6} + \ldots \) [1]
⇒ \( P(A) = \frac{1/6}{1 - (5/6)^2} = \frac{6}{11} \) [1]
∴ Probability of winning of B, \( P(B) = 1 - P(A) = 1 - \frac{6}{11} = \frac{5}{11} \) [1]

Question 10. There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and the third is also a biased coin that comes up tails 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin? [CBSE 2014, 6M]
Answer:
Let \( E_1 \) = two-headed coin
\( E_2 \) = biased coin that comes up heads (75%)
\( E_3 \) = biased coin that comes up tails (40%)
E = heads comes up
Here \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \)
and \( P(E/E_1) = 1;\ P(E/E_2) = 3/4;\ P(E/E_3) = 1 - \frac{2}{5} = \frac{3}{5} \) [2]
Required probability \( = P(E_1/E) = \frac{P(E_1) \times P(E/E_1)}{P(E_1) \times P(E/E_1) + P(E_2) \times P(E/E_2) + P(E_3) \times P(E/E_3)} \) (by Bayes' theorem) [2]
\( = \frac{\frac{1}{3} \times 1}{\left(\frac{1}{3} \times 1\right) + \left(\frac{1}{3} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{3}{5}\right)} = \frac{20}{47} \) [2]

Question 11. A bag contains 5 red and 4 black balls, a second bag contains 3 red and 6 black balls. One of the two bags is selected at random and two balls are drawn at random (without replacement), both of which are found to be red. Find the probability that the balls are drawn from the second bag. [CBSE 2019, 6M]
Answer:
Let \( E_1 \): first bag is selected
\( E_2 \): second bag is selected
E: both drawn balls are red
Here, \( P(E_1) = P(E_2) = \frac{1}{2} \)
⇒ \( P(E/E_1) = \frac{{}^{5}C_{2}}{{}^{9}C_{2}} = \frac{5}{18},\ P(E/E_2) = \frac{{}^{3}C_{2}}{{}^{9}C_{2}} = \frac{1}{12} \) [2]
∴ Required probability = \( P(E_2/E) \)
\( = \frac{P(E/E_2) \cdot P(E_2)}{P(E/E_1) \cdot P(E_1) + P(E/E_2) \cdot P(E_2)} \) (using Bayes' theorem) [2]
\( = \frac{\frac{1}{12} \times \frac{1}{2}}{\frac{5}{18} \times \frac{1}{2} + \frac{1}{12} \times \frac{1}{2}} \)
\( = \frac{\frac{1}{12}}{\frac{5}{18} + \frac{1}{12}} = \frac{3}{13} \) [2]