Three guns fire independently at a target with hit probabilities 0.2, 0.3 and 0.5. Find (1) the probability that at least one gun hits the target, and (2) the probability that exactly one gun hits it.

Step-by-step solution

1. 1

   Name the events and their misses. Let $A_1,A_2,A_3$ be the hits, with $P(A_1)=0.2$, $P(A_2)=0.3$, $P(A_3)=0.5$. Independence means probabilities multiply, and the miss probabilities are the complements $0.8$, $0.7$ and $0.5$.

2. 2

   (1) Use the complement rather than a union of three events. Inclusion–exclusion on three events needs seven terms; the complement needs one product:

   $P(\text{at least one hit})=1-P(\text{all three miss})=1-(0.8)(0.7)(0.5)$

3. 3

   Evaluate.

   $(0.8)(0.7)(0.5)=0.28\quad\Longrightarrow\quad P=1-0.28=0.72$

4. 4

   (2) List the three mutually exclusive ways to get exactly one hit. Exactly one hit means one gun hits and the other two miss:

   $P=\underbrace{(0.2)(0.7)(0.5)}_{\text{only gun 1}}+\underbrace{(0.8)(0.3)(0.5)}_{\text{only gun 2}}+\underbrace{(0.8)(0.7)(0.5)}_{\text{only gun 3}}$

   Note each gun's own probability appears once and the other two enter as misses — a very common place to slip.

5. 5

   Add the three terms.

   $0.07+0.12+0.28=0.47$

   $\boxed{P(\text{at least one})=0.72,\qquad P(\text{exactly one})=0.47}$

6. 6

   Check with the full distribution. The probabilities of $0,1,2,3$ hits are $0.28$, $0.47$, $0.22$ and $0.03$ (for two hits: $0.2\cdot 0.3\cdot 0.5+0.2\cdot 0.7\cdot 0.5+0.8\cdot 0.3\cdot 0.5=0.03+0.07+0.12=0.22$; for three: $0.2\cdot 0.3\cdot 0.5=0.03$). They sum to $1.00$ ✓, and $0.47+0.22+0.03=0.72$ reproduces part (1).