A, B and C fire one shot each at a target, independently. The probabilities that A, B and C hit the target are \(\displaystyle\frac{1}{2}\), \(\displaystyle\frac{2}{3}\) and \(\displaystyle\frac{3}{5}\) respectively. The probability that exactly one of them hits the target is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3}{10}\)
The probabilities of missing are
\( \displaystyle P(A') = \frac{1}{2},\quad P(B') = \frac{1}{3},\quad P(C') = \frac{2}{5} \)
“Exactly one hits” is made of three cases that cannot happen together, so their probabilities are added. In each case the shots are independent, so the probabilities are multiplied:
\( \displaystyle P(\text{only A}) = \frac{1}{2}\times\frac{1}{3}\times\frac{2}{5} = \frac{2}{30} \)
\( \displaystyle P(\text{only B}) = \frac{1}{2}\times\frac{2}{3}\times\frac{2}{5} = \frac{4}{30} \)
\( \displaystyle P(\text{only C}) = \frac{1}{2}\times\frac{1}{3}\times\frac{3}{5} = \frac{3}{30} \)
Adding:
\( \displaystyle P(\text{exactly one}) = \frac{2+4+3}{30} = \frac{9}{30} = \frac{3}{10} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{13}{30}\) — Incorrect. This is the probability that exactly two of them hit the target.
Option (3) \(\displaystyle\frac{1}{15}\) — Incorrect. This is \(\frac{1}{2}\times\frac{1}{3}\times\frac{2}{5}\), the probability that all three miss.
Option (4) \(\displaystyle\frac{1}{5}\) — Incorrect. This is \(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{5}\), the probability that all three hit.