Problem 126
Under the conditions of the previous problem (p_1=0.4…
Under the conditions of the previous problem ($p_1=0.4$, $p_2=0.5$, $p_3=0.7$), find the probability that the target will be hit at least once.
Explanation of Method
"At least one hit" is the opposite of "all three shots miss". The probability that all miss is the product of the miss probabilities (independence); subtracting from $1$ gives the answer. We check it with the distribution from Problem 125.
Step 1: The opposite event
\[ P(\text{no hit})=q_1q_2q_3=0.6\cdot0.5\cdot0.3=0.09. \]
Step 2: The answer
\[ \boxed{P(\text{at least one hit})=1-0.09=0.91.} \]
Step 3: Check
From Problem 125: $P(1)+P(2)+P(3)=0.36+0.41+0.14=0.91$. ✓
The complement method needs only one product instead of seven.
Result
| Answer | |
|---|---|
| Probability | $1-0.6\cdot0.5\cdot0.3=0.91$ |