CBSE Class 12 Mathematics Probability Class Test Set 01

Class Test Questions with Solutions

Question 1. If A and B are any two events such that P(A) + P(B) − P(A and B) = P(A), then find P(A|B). [1]
Answer:
Given P(A) + P(B) − P(A ∩ B) = P(A)
∴ P(A ∩ B) = P(B)
Now \( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
\( = \frac{P(B)}{P(B)} = 1 \)

Question 2. If 2P(A) = P(B) = 5/13 and \( P(A|B) = \frac{2}{5} \), then find P(A ∪ B). [1]
Answer:
Given \( 2P(A) = P(B) = \frac{5}{13} \)
∴ \( P(A) = \frac{5}{26},\ P(B) = \frac{5}{13} \)
Now \( P\left(\frac{A}{B}\right) = \frac{P(A \cap B)}{P(B)} \)
\( \frac{2}{5} = \frac{P(A \cap B)}{5/13} \Rightarrow P(A \cap B) = \frac{2}{13} \)
Now P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
\( = \frac{5}{26} + \frac{5}{13} - \frac{2}{13} \)
\( = \frac{11}{26} \)

Question 3. An unbiased die is thrown twice. Let the event A be 'odd number on the first throw' and B the event 'odd number on the second throw'. Check the independence of the events A and B. [1]
Answer:
If all the 36 elementary events of the experiment are considered to be equally likely, we have
\( P(A) = \frac{18}{36} = \frac{1}{2} \) and \( P(B) = \frac{18}{36} = \frac{1}{2} \)
Also P(A ∩ B) = P(odd number on both throws) \( = \frac{9}{36} = \frac{1}{4} \)
Now \( P(A)P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
Clearly P(A ∩ B) = P(A) × P(B)
Thus, A and B are independent events.

Question 4. Determine P(E|F) if a die is thrown three times such that
E: 4 appears on the third toss
F: 6 and 5 appear respectively on the first two tosses [2]
Answer:
If a die is thrown 3 times, then the number of elements in the sample space will be 6 × 6 × 6 = 216
E = {(1, 1, 4), (1, 2, 4), …, (6, 6, 4)}
F = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}   ∴ E ∩ F = {(6, 5, 4)}
Now \( P(F) = \frac{6}{216},\ P(E \cap F) = \frac{1}{216} \)
∴ \( P\left(\frac{E}{F}\right) = \frac{P(E \cap F)}{P(F)} = \frac{1}{6} \)

Question 5. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred 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 ball is black. [3]
Answer:
Let \( E_1 \): ball transferred from bag I to bag II is red
\( E_2 \): ball transferred from bag I to bag II is black
A: ball drawn from bag II is red
Now, \( P(E_1) = \frac{3}{7},\ P(E_2) = \frac{4}{7} \)
When a red ball is added to bag II,
\( P(A/E_1) = \frac{5}{10} = \frac{1}{2} \)
When a black ball is added to bag II,
\( P(A/E_2) = \frac{4}{10} = \frac{2}{5} \)
∴ Required probability \( P\left(\frac{E_2}{A}\right) \)
\( = \frac{P(E_2) \times P(A/E_2)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( = \frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2} + \frac{4}{7} \times \frac{2}{5}} = \frac{16}{31} \)

Question 6. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. [3]
Answer:
Let \( E_1 \) = man speaks truth
\( E_2 \) = man speaks false
A = event that 6 appears
⇒ \( P(E_1) = \frac{3}{4} \); \( P(E_2) = \frac{1}{4} \)
and \( P(A/E_1) = \frac{1}{6} \); \( P(A/E_2) = \frac{5}{6} \)
∴ Required probability is
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( = \frac{\frac{3}{4} \times \frac{1}{6}}{\frac{3}{4} \times \frac{1}{6} + \frac{1}{4} \times \frac{5}{6}} = \frac{3}{8} \)

Question 7. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let \( \frac{3}{4} \) be the probability that he knows the answer and \( \frac{1}{4} \) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \( \frac{1}{4} \), what is the probability that the student knows the answer given that he answered it correctly? [3]
Answer:
Let \( E_1 \) = student knows the answer
\( E_2 \) = student guesses the answer
A = answer is correct
⇒ \( P(E_1) = \frac{3}{4},\ P(E_2) = \frac{1}{4} \)
and \( P(A/E_1) = 1,\ P(A/E_2) = \frac{1}{4} \)
∴ Required probability
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \cdot 1 + \frac{1}{4} \cdot \frac{1}{4}} = \frac{12}{13} \)

Question 8. A manufacturing firm produces steel pipes in three plants A, B and C with daily production of 500, 1000 and 2000 units, respectively. The fraction of defective steel pipes output produced by the plants A, B and C are respectively 0.005, 0.008 and 0.010. If a pipe is selected from a day's total production and found to be defective, then find out the probability that it came from the first plant. [5]
Answer:
Let \( E_1 \): pipe selected from plant A
\( E_2 \): pipe selected from plant B
\( E_3 \): pipe selected from plant C
A: selected pipe is defective
\( P(E_1) = \frac{500}{3500} = \frac{1}{7},\ P(E_2) = \frac{1000}{3500} = \frac{2}{7},\ P(E_3) = \frac{2000}{3500} = \frac{4}{7} \)
Also P(A/\( E_1 \)) = .005, P(A/\( E_2 \)) = .008, P(A/\( E_3 \)) = .010
Required probability is
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) + P(E_3) \cdot P(A/E_3)} \)
\( = \frac{\frac{1}{7} \times .005}{\frac{1}{7} \times .005 + \frac{2}{7} \times .008 + \frac{4}{7} \times .010} = \frac{5}{61} \)

Question 9. A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is \( \frac{31}{42} \), then determine the value of n. [5]
Answer:
A coin picked up at random from the bag may be a fair coin or a coin having head on both sides. The tossed coin may result in a head in the following mutually exclusive ways.
(a) A fair coin is taken out of the bag and then it is tossed to get a head.
(b) A two-headed coin is taken out of the bag and then it is tossed to get a head.
So, let us define the following events.
\( E_1 \) = taking out a fair coin from the bag
\( E_2 \) = taking out a coin having head on both sides
A = tossed coin results in a head

Clearly, \( P(E_1) = \frac{n + 1}{2n + 1} \) and \( P(E_2) = \frac{n}{2n + 1} \)
Also, \( P(A/E_1) = \frac{1}{2} \) and \( P(A/E_2) = 1 \)
By total probability theorem,
\( P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2) \)
\( = \frac{n + 1}{2n + 1} \times \frac{1}{2} + \frac{n}{2n + 1} \times 1 = \frac{3n + 1}{2(2n + 1)} \)
But it is given that \( P(A) = \frac{31}{42} \)
∴ \( \frac{31}{42} = \frac{3n + 1}{2(2n + 1)} \Rightarrow 62n + 31 = 63n + 21 \)
⇒ n = 10