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 ______.

विकल्प

• 0.024

• 0.188

• 0.336

• 0.452

Answer 1

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