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Question
Three persons, A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is ______.
Options
0.024
0.188
0.336
0.452
Solution
Three persons, A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is 0.188.
Explanation:
Given that: P(A) = 0.4, P(B) = 0.3 and P(C) = 0.2
Also `"P"(bar"A")` = 1 – 0.4 = 0.6
`"P"(bar"B")` = 1 – 0.3 = 0.7
And `"P"(bar"C")` = 1 – 0.2 = 0.8
∴ Probabilities of two hits
= `"P"("A")."P"("B")."P"(bar"C") + "P"("A")."P"(bar"B")."P"("C") + "P"(bar"A")."P"("B")."P"("C")`
= 0.4 × 0.3 × 0.8 + 0.4 × 0.7 × 0.2 + 0.6 × 0.3 × 0.2
= 0.096 + 0.056 + 0.036
= 0.188
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