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Home  >>  CBSE XII  >>  Math  >>  Probability
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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} $

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1 Answer

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 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\(F_{1}\;S_{2}\;S_{3}\)

\(p(S_{1}\;S_{2}\;F_{3})\)=0.4x0.3x0.8=0.096
\(p(S_{1}\;F_{2}\;S_{3})\)=0.4x0.7x0.2=0.056
\(p(F_{1}\;S_{2}\;S_{3})\)=0.6x0.3x0.2=0.036
p(two hits)=0.096+0.056+0.036
=0.0188
'B'is the correct choice

 

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

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