Probability
Probability
star_batch_jee_advanced_2025
Grade 12
Question:
Observe the following columns:
(A) The probability that $A, B, C$ solve a problem is $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$. If the probability that the problem will be solved is $\lambda$ and that the problem is solved by only one of them is $\mu$, then
(B) The probability of hitting a target by three men is $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively. If the probability that exactly two of them will hit the target is $\lambda$ and that at least two of them hit the target is $\mu$, then
(C) A bag contains 4 white and 2 black balls. Another contains 3 white and 5 black balls. One ball is drawn from each bag. If the probability that both are black is $\lambda$ and that both are white is $\mu$, then
Step-by-Step Solution
Key Concept: Use conditional probability and the addition principle for mutually exclusive events to find relationships between $\lambda$ and $\mu$.
For option (A) with $\lambda = 1$, the problem cannot be solved. For option (B), given $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, $P(C) = \frac{1}{4}$, we calculate $\lambda = P(A \cap B \cap C) + P(A \cap B \cap \overline{C}) + P(A \cap \overline{B} \cap C) = \frac{1}{24}$ and $\mu = \lambda + P(A) \cdot P(B) \cdot P(C) = \frac{7}{24}$, giving $\mu - \lambda = \frac{1}{24}$ and $\mu + \lambda = \frac{13}{24}$ (P,T). For option (C), $\lambda = \frac{5}{24}$, $\mu = \frac{6}{24}$, so $\mu - \lambda = \frac{1}{24}$ (T) and $\mu + \lambda = \frac{11}{24}$ (R).
Correct Answer: [A-q,s][B-p,t][C-r,t]