A problem is given to 3 students A,B and C, whose probabilities of so…

The correct answer is Option (3) → (A) and (B) only

The students work independently, so the probabilities multiply. The probabilities of failing are \( \frac{1}{2},\ \frac{2}{3},\ \frac{3}{4} \).

    \( \displaystyle P(\text{not solved})=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}=\frac{1}{4} \)

    \( \displaystyle P(\text{solved})=1-\frac{1}{4}=\frac{3}{4} \)

Exactly one student solves it when one succeeds and the other two fail:

    \( \displaystyle P(\text{exactly one})=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}+\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{3}{4}+\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{4} \)

    \( \displaystyle =\frac{1}{4}+\frac{1}{8}+\frac{1}{12} \)

    \( \displaystyle =\frac{11}{24} \)

    \( \displaystyle P(\text{at least two})=1-\frac{1}{4}-\frac{11}{24} \)

    \( \displaystyle =\frac{7}{24} \)

So P(problem solved) = \( \frac{3}{4} \), P(problem not solved) = \( \frac{1}{4} \), P(exactly one) = \( \frac{11}{24} \) and P(at least two) = \( \frac{7}{24} \).

Checking the statements

(A) P(Problem solved) = \( \frac{3}{4} \) — Correct. \( 1-\frac{1}{4}=\frac{3}{4} \).

(B) P(Problem not solved) = \( \frac{1}{4} \) — Correct. \( \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}=\frac{1}{4} \).

(C) P(Problem solved by exactly one student) = \( \frac{7}{24} \) — Incorrect. The value is \( \frac{11}{24} \); \( \frac{7}{24} \) is the probability of at least two.

(D) P(Problem solved by at least 2 students) = \( \frac{17}{24} \) — Incorrect. The value is \( \frac{7}{24} \), not \( \frac{17}{24} \).