Two shooters, who hit the target with probabilities 0.7 and…
Two shooters, who hit the target with probabilities $0.7$ and $0.8$ respectively, each fire one shot. Determine the probability that the target will be hit.
Explanation of Method
"The target is hit" means at least one shooter hits. The two hits are independent but not mutually exclusive (both may hit), so we cannot simply add $0.7+0.8$. Either use the opposite event (both miss) or the addition rule for compatible events $P(A+B)=P(A)+P(B)-P(AB)$.
Step 1: Events
$A$ = "the first shooter hits", $P(A)=0.7$; $B$ = "the second hits", $P(B)=0.8$; independent.
Step 2: Method 1 — the opposite event
The target is not hit only if both miss:
\[ P(\overline A\,\overline B) = 0.3\cdot0.2=0.06, \qquad \boxed{P(A+B)=1-0.06=0.94.} \]
Step 3: Method 2 — addition rule
\[ P(A+B)=P(A)+P(B)-P(A)P(B)=0.7+0.8-0.56=0.94. \ \checkmark \]
Step 4: A common mistake
$0.7+0.8=1.5\gt1$ is not a probability: adding counts the case "both hit" twice. Subtracting $P(AB)=0.56$ once corrects this.
Step 5: All outcomes
| Outcome | Probability |
|---|---|
| both hit | $0.7\cdot0.8=0.56$ |
| only the first hits | $0.7\cdot0.2=0.14$ |
| only the second hits | $0.3\cdot0.8=0.24$ |
| both miss | $0.3\cdot0.2=0.06$ |
| total | $1$ ✓ |
Result
| Answer | |
|---|---|
| Probability the target is hit | $1-0.3\cdot0.2=0.94$ |