CBSE Class 12 Mathematics Probability Important Questions Set 03

Question. 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 is correct
Answer: (d)

Question. Three person \( A, B \) and \( C \), fire at a target in turn, starting with \( A \). Their probability of hitting the target are 0.4, 0.3 and 0.2, respectively. The probability of two hits is
(a) 0.025
(b) 0.188
(c) 0.339
(d) 0.475
Answer: (b)

Question. In a college, 30% students fail in Physics, 25% fail in Mathematics and 10% fail in both. One student is chosen at random. The probability that she fails in Physics, if she has failed in Mathematics is
(a) \( \frac{3}{10} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{7}{90} \)
(d) \( \frac{1}{3} \)
Answer: (b)

Question. In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
(a) \( 10^{-1} \)
(b) \( \left(\frac{1}{2}\right)^5 \)
(c) \( \left(\frac{9}{10}\right)^5 \)
(d) \( \frac{9}{10} \)
Answer: (c)

Question. Two dice are thrown. The probability of getting an odd number on first and a multiple of 3 on the other die, is ____________.
Answer: \( \frac{1}{6} \)

Question. If \( A \) and \( B' \) are independent events then \( P(A' \cup B) = 1 - \) ____________.
Answer: \( P(A) \cdot P(B') \)

Question. Given that \( E \) and \( F \) are events such that \( P(E) = 0.6, P(F) = 0.3 \) and \( P(E \cap F) = 0.2 \), find \( P(E/F) \) and \( P(F/E) \).
Answer: \( P(E/F) = \frac{0.2}{0.3} = \frac{2}{3} \) and \( P(F/E) = \frac{0.2}{0.6} = \frac{1}{3} \)

Question. Compute \( P(A/B) \) if \( P(B) = 0.5 \) and \( P(A \cap B) = 0.32 \).
Answer: \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = \frac{32}{50} = \frac{16}{25} \)

Question. Given that \( P(\bar{A}) = 0.4, P(B) = 0.2 \) and \( P\left(\frac{A}{B}\right) = 0.5 \). Find \( P(A \cup B) \).
Answer: \( P(A) = 1 - 0.4 = 0.6 \). Given \( P(A/B) = \frac{P(A \cap B)}{P(B)} = 0.5 \Rightarrow P(A \cap B) = 0.5 \times 0.2 = 0.1 \). Now, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.2 - 0.1 = 0.7 \)

Question. 10% of the bulbs produced in a factory are of red colour and 2% are red and defective. If one bulb is picked up at random, determine the probability of its being defective if it is red.
Answer: Let \( R \) be the event that the bulb is red and \( D \) be the event that it is defective. \( P(R) = \frac{10}{100} = 0.1 \) and \( P(R \cap D) = \frac{2}{100} = 0.02 \). Required probability \( P(D/R) = \frac{P(R \cap D)}{P(R)} = \frac{0.02}{0.1} = \frac{2}{10} = \frac{1}{5} \)

Question. Events \( A \) and \( B \) are such that \( P(A) = \frac{1}{2}, P(B) = \frac{7}{12} \) and \( P(\text{not } A \text{ or not } B) = \frac{1}{4} \). State whether \( A \) and \( B \) are independent.
Answer: \( P(\bar{A} \cup \bar{B}) = P(\overline{A \cap B}) = 1 - P(A \cap B) = \frac{1}{4} \Rightarrow P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4} \). Now \( P(A) \cdot P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24} \). Since \( P(A \cap B) \neq P(A) \cdot P(B) \), the events are not independent.

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: \( P(A \cup B) = p \). Probability of exactly one of \( A, B \) occurs is \( P(A) + P(B) - 2P(A \cap B) = q \). We know \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = p \Rightarrow P(A) + P(B) = p + P(A \cap B) \). Substituting this in the second equation: \( p + P(A \cap B) - 2P(A \cap B) = q \Rightarrow p - P(A \cap B) = q \Rightarrow P(A \cap B) = p - q \). Now, \( P(A) + P(B) = p + (p - q) = 2p - q \).

Question. \( A \) 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: Probability of getting 10 is \( \frac{3}{36} = \frac{1}{12} \). \( P(A \text{ wins}) = \frac{12}{23} \) and \( P(B \text{ wins}) = \frac{11}{23} \)

Question. 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 \( X \).
Answer: Mean = 8, Variance = \( \frac{20}{3} \)

Question. 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 has A grade. What is the probability that the student has 100% attendance?
Answer: Let \( E_1 \) be 100% attendance and \( E_2 \) be irregular. \( P(E_1) = 0.3, P(E_2) = 0.7 \). Let \( A \) be getting A grade. \( P(A/E_1) = 0.7, P(A/E_2) = 0.1 \). By Bayes' theorem, \( P(E_1/A) = \frac{0.3 \times 0.7}{(0.3 \times 0.7) + (0.7 \times 0.1)} = \frac{0.21}{0.21 + 0.07} = \frac{0.21}{0.28} = \frac{3}{4} \)

Question. Five bad oranges are accidently mixed with 20 good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence find the mean and variance of the distribution.
Answer: Total oranges = 25. Probability of bad orange \( p = \frac{5}{25} = \frac{1}{5} \), \( q = \frac{4}{5} \). Number of trials \( n = 4 \).
Probability Distribution:
\( X = 0, P(X) = {}^4C_0 (\frac{1}{5})^0 (\frac{4}{5})^4 = \frac{256}{625} \)
\( X = 1, P(X) = {}^4C_1 (\frac{1}{5})^1 (\frac{4}{5})^3 = \frac{256}{625} \)
\( X = 2, P(X) = {}^4C_2 (\frac{1}{5})^2 (\frac{4}{5})^2 = \frac{96}{625} \)
\( X = 3, P(X) = {}^4C_3 (\frac{1}{5})^3 (\frac{4}{5})^1 = \frac{16}{625} \)
\( X = 4, P(X) = {}^4C_4 (\frac{1}{5})^4 (\frac{4}{5})^0 = \frac{1}{625} \)
Mean \( = np = 4 \times \frac{1}{5} = \frac{4}{5} \). Variance \( = npq = 4 \times \frac{1}{5} \times \frac{4}{5} = \frac{16}{25} \)