Under the conditions of the previous problem, find the…
Under the conditions of the previous problem, find the probability that the airplane will be destroyed if it is hit by three shells.
Explanation of Method
Again it is easier to describe survival. The airplane survives three hits iff part $1$ is not hit, part $2$ is hit at most once, and part $3$ at most twice. Since the three hits must land somewhere, the only possibility is: exactly one hit on part $2$ and two hits on part $3$. The number of hits on each part follows a multinomial distribution.
Step 1: When does the airplane survive?
Let $k_1,k_2,k_3$ be the numbers of hits on the three parts, $k_1+k_2+k_3=3$. Survival requires $k_1=0$, $k_2\le1$, $k_3\le2$. With $k_1=0$: $k_2+k_3=3$, and $k_3\le2$ forces $k_2\ge1$; together with $k_2\le1$ this gives
\[ k_1=0,\quad k_2=1,\quad k_3=2. \]
Step 2: Its probability
The hit on part $2$ can be the first, second or third shell ($3$ orders):
\[ P(\text{survive})=3\cdot0.2\cdot0.7^2=3\cdot0.2\cdot0.49=0.294. \]
Step 3: The answer
\[ \boxed{P(\text{destroyed})=1-0.294=0.706.} \]
Step 4: Comparison
| Number of hits | P(destroyed) |
|---|---|
| $1$ | $0.1$ |
| $2$ | $0.23$ (Problem 128) |
| $3$ | $0.706$ |
The third hit makes a big difference, because it can complete a destruction of part $3$ ($0.7^3=0.343$) or of part $2$.
Result
| Answer | |
|---|---|
| Survival | $k=(0,1,2)$: $3\cdot0.2\cdot0.7^2=0.294$ |
| Destroyed | $1-0.294=0.706$ |