Three shots are fired at the same target. The probabilities…
Three shots are fired at the same target. The probabilities of hitting the target with the first, second and third shot are $p_1=0.4$, $p_2=0.5$, $p_3=0.7$. Determine the probability that there will be exactly one hit in the three shots.
Explanation of Method
"Exactly one hit" happens in three mutually exclusive ways: only the first shot hits, only the second, or only the third. Each way is a product of independent events (multiplication rule), and the three ways are added (addition rule).
Step 1: Events
$A_k$ = "shot $k$ hits", independent, with $q_k=1-p_k$: $q_1=0.6$, $q_2=0.5$, $q_3=0.3$.
Step 2: The three ways
| Way | Event | Probability |
|---|---|---|
| only shot 1 hits | $A_1\overline{A_2}\,\overline{A_3}$ | $0.4\cdot0.5\cdot0.3=0.06$ |
| only shot 2 hits | $\overline{A_1}A_2\overline{A_3}$ | $0.6\cdot0.5\cdot0.3=0.09$ |
| only shot 3 hits | $\overline{A_1}\,\overline{A_2}A_3$ | $0.6\cdot0.5\cdot0.7=0.21$ |
Step 3: Addition rule
\[ \boxed{P(\text{exactly one hit})=p_1q_2q_3+q_1p_2q_3+q_1q_2p_3=0.06+0.09+0.21=0.36.} \]
Step 4: The full distribution of the number of hits
| Hits | Probability |
|---|---|
| $0$ | $0.6\cdot0.5\cdot0.3=0.09$ |
| $1$ | $0.36$ |
| $2$ | $0.4\cdot0.5\cdot0.3+0.4\cdot0.5\cdot0.7+0.6\cdot0.5\cdot0.7=0.41$ |
| $3$ | $0.4\cdot0.5\cdot0.7=0.14$ |
| total | $1$ ✓ |
A quick way to get all of them: expand $(q_1+p_1z)(q_2+p_2z)(q_3+p_3z)$; the coefficient of $z^k$ is the probability of $k$ hits.
Result
| Answer | |
|---|---|
| Probability of exactly one hit | $0.06+0.09+0.21=0.36$ |