A circular target consists of three zones: I, II and III.…
A circular target consists of three zones: I, II and III. The probability of hitting the first zone with one shot is $0.15$, the second $0.23$, the third $0.17$. Find the probability of a miss.
Explanation of Method
A single shot can land in at most one zone, so the events "hit zone I", "hit zone II", "hit zone III" are mutually exclusive, and the probability of hitting the target is their sum (addition rule). A miss is the opposite event of hitting the target, so its probability is $1$ minus that sum.
- zone I: $0.15$
- zone II: $0.23$
- zone III: $0.17$
Step 1: Events
$A_1,A_2,A_3$ = "the shot hits zone I, II, III". They are mutually exclusive, and "the target is hit" $=A_1+A_2+A_3$.
Step 2: Probability of a hit
\[ P(A_1+A_2+A_3)=0.15+0.23+0.17=0.55. \]
Step 3: Probability of a miss
A miss is the opposite event:
\[ \boxed{P(\text{miss})=1-0.55=0.45.} \]
Step 4: Remark
The zones are rings (and a central disc), so they do not overlap. If the events could overlap, we could not simply add their probabilities.
Result
| Answer | |
|---|---|
| Hit | $0.15+0.23+0.17=0.55$ |
| Miss | $1-0.55=0.45$ |