Probability of solving a specific problem independently by A and B ar…

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

Let E and F be the events that A and B solve the problem. They are independent, so

    \( \displaystyle P(E)=\frac{1}{2},\ P(\bar{E})=\frac{1}{2} \)

    \( \displaystyle P(F)=\frac{1}{3},\ P(\bar{F})=\frac{2}{3} \)

The problem is solved unless both fail:

    \( \displaystyle P(E\cup F)=1-P(\bar{E})\,P(\bar{F}) \)

    \( \displaystyle =1-\frac{1}{2}\times\frac{2}{3} \)

    \( \displaystyle =\frac{2}{3} \)

Exactly one solves it when one succeeds and the other fails:

    \( \displaystyle P(E)\,P(\bar{F})+P(\bar{E})\,P(F)=\frac{1}{2}\times\frac{2}{3}+\frac{1}{2}\times\frac{1}{3} \)

    \( \displaystyle =\frac{1}{3}+\frac{1}{6} \)

    \( \displaystyle =\frac{1}{2} \)

Both solve it with probability

    \( \displaystyle P(E\cap F)=\frac{1}{2}\times\frac{1}{3} \)

    \( \displaystyle =\frac{1}{6} \)

“At most one” is the complement of “both”:

    \( \displaystyle 1-P(E\cap F)=1-\frac{1}{6} \)

    \( \displaystyle =\frac{5}{6} \)

So P(solved) = \( \frac{2}{3} \), P(exactly one) = \( \frac{1}{2} \), P(both) = \( \frac{1}{6} \) and P(at most one) = \( \frac{5}{6} \).

Checking the statements

(A) P(Problem is solved) = \( \frac{2}{3} \) — Correct. \( 1-\frac{1}{2}\times\frac{2}{3}=\frac{2}{3} \).

(B) P(Exactly one of them solves the problem) = \( \frac{1}{2} \) — Correct. \( \frac{1}{3}+\frac{1}{6}=\frac{1}{2} \).

(C) P(Both of them solve the problem) = \( \frac{5}{6} \) — Incorrect. P(both) is \( \frac{1}{2}\times\frac{1}{3}=\frac{1}{6} \).

(D) P(At most one solves the problem) = \( \frac{5}{6} \) — Correct. \( 1-\frac{1}{6}=\frac{5}{6} \).