An airplane under fire consists of three parts of different…
An airplane under fire consists of three parts of different vulnerability. To destroy the airplane it is sufficient to hit the first part once, or the second part twice, or the third part three times. The conditional probability that a hit falls on a given part is proportional to its area; the parts occupy $0.1$, $0.2$ and $0.7$ of the total area. Suppose the airplane was hit by two shells. Find the probability that it is destroyed.
Explanation of Method
Each of the two hits independently falls on part $1$, $2$ or $3$ with probabilities $0.1$, $0.2$, $0.7$. With only two hits, part $3$ can never be hit three times, so the airplane is destroyed iff at least one hit falls on part $1$, or both hits fall on part $2$. It is easiest to find the probability of survival: no hit on part $1$ and not both hits on part $2$.
Step 1: The model
Each hit lands on part $1$, $2$, $3$ with probabilities $a_1=0.1$, $a_2=0.2$, $a_3=0.7$, independently of the other hit.
Step 2: When does the airplane survive?
No hit on part $1$ (both hits on parts $2$ or $3$), and not both hits on part $2$. The possible (ordered) pairs:
| Hits on parts | Probability | Destroyed? |
|---|---|---|
| $(2,3)$ or $(3,2)$ | $2\cdot0.2\cdot0.7=0.28$ | no |
| $(3,3)$ | $0.7^2=0.49$ | no |
| $(2,2)$ | $0.2^2=0.04$ | yes |
| at least one on part 1 | $1-0.9^2=0.19$ | yes |
Step 3: The answer
\[ \boxed{P(\text{destroyed})=1-(0.28+0.49)=0.23.} \]
Directly: $P(\text{at least one hit on part }1)+P(\text{both on part }2)=0.19+0.04=0.23$ ✓ (these two events are incompatible).
Result
| Answer | |
|---|---|
| Destroyed | $0.19+0.04=0.23$ |
| Survives | $0.28+0.49=0.77$ |