Three ammunition warehouses are bombed, and only one bomb is…
Three ammunition warehouses are bombed, and only one bomb is dropped. The probability that it falls on the first warehouse is $0.01$, on the second $0.008$, on the third $0.025$. If the bomb falls on any one of the warehouses, it destroys all three. Find the probability that the warehouses will be destroyed.
Explanation of Method
One bomb can fall on at most one warehouse, so the events "falls on warehouse 1", "on warehouse 2", "on warehouse 3" are mutually exclusive. The warehouses are destroyed iff one of these events occurs, so by the addition rule for incompatible events we add the three probabilities.
Step 1: Events
$A_k$ = "the bomb falls on warehouse $k$": $P(A_1)=0.01$, $P(A_2)=0.008$, $P(A_3)=0.025$. One bomb cannot fall on two warehouses, so $A_iA_j=\varnothing$.
Step 2: Addition rule
\[ \boxed{P(A_1+A_2+A_3)=0.01+0.008+0.025=0.043.} \]
Step 3: A common mistake
Treating the events as independent and computing $1-(0.99)(0.992)(0.975)\approx0.0425$ is wrong: these events are not independent but mutually exclusive (if the bomb fell on warehouse $1$, it certainly did not fall on warehouse $2$). For incompatible events the probabilities simply add.
Result
| Answer | |
|---|---|
| Probability | $0.01+0.008+0.025=0.043$ |