Three persons A, B and C fire at a target independently of each other. The probabilities that A, B and C hit the target are \(\displaystyle\frac{1}{2}\), \(\displaystyle\frac{2}{3}\) and \(\displaystyle\frac{3}{4}\) respectively. The probability that the target is hit by at least two of them is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{17}{24}\)
The probabilities of missing are \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{1}{4}\) for A, B and C. Since the persons fire independently, the probability of each case is a product.
“At least two hit” means exactly two hit or all three hit. The four cases cannot happen together, so their probabilities are added.
Exactly two hit (the third misses):
\( \displaystyle A, B \text{ hit, } C \text{ misses: } \frac{1}{2} \times \frac{2}{3} \times \frac{1}{4} = \frac{2}{24} \)
\( \displaystyle A, C \text{ hit, } B \text{ misses: } \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{3}{24} \)
\( \displaystyle B, C \text{ hit, } A \text{ misses: } \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} \)
All three hit:
\( \displaystyle \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} \)
Adding the four cases:
\( \displaystyle P(\text{at least two hit}) = \frac{2 + 3 + 6 + 6}{24} \)
\( \displaystyle = \frac{17}{24} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{11}{24}\) — Incorrect. This is \(\frac{2 + 3 + 6}{24}\), the probability that exactly two hit; it leaves out the case where all three hit.
Option (3) \(\displaystyle\frac{1}{4}\) — Incorrect. This is \(\frac{6}{24}\), the probability that all three hit.
Option (4) \(\displaystyle\frac{7}{24}\) — Incorrect. This is \(1 - \frac{17}{24}\), the probability that at most one of them hits.