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Three persons,A,B and C,fire at a target in turn,starting with A.Thier probability of hitting the target are 0.4,0.3 and 0.2 respectively.The probability of two hits is

$\begin{array}{1 1}(A)\;0.024\\(B)\;0.0188\\(C)\;0.336\\(D)\;0.452\end{array} $

1 Answer

 A,B,C let Target  let\(S_{i}\) defined as  hitting let  \(F_{i}\) defined as not hetTwo hits can take placeas\(S_{1}\;S_{2}\;S_{3}\)or\(S_{1}\;F_{2}\;S_{3}\)or\(F_{1}\;S_{2}\;S_{3}\)

p(two hits)=0.096+0.056+0.036
'B'is the correct choice


answered Feb 21, 2013 by poojasapani_1
edited Feb 21, 2013 by poojasapani_1

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