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

A. 0.024
B. 0.188
C. 0.336
D. 0.452

Answer

B

Explanation

Here, `P(A)=0.4,P(barA)=0.6,P(B)=0.3,P(barB)=0.7`
`P(C )=0.2andP(barC)=0.8`
`therefore` Probability of two hits=`P_(A)cdotP_(B)cdotP_(barC)+P_(A)cdotP_(barB)cdotP_(C)+P_(barA)cdotP_(B)cdotP_(C)`
`=0.4xx0.3xx0.8+0.4xx0.7xx0.2+0.6xx0.3xx0.2`
`=0.096+0.056+0.036=0.188`