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Questions  >>  CBSE XII  >>  Math  >>  Probability
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Q)

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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A)
Solution :
A,B,C let Target let $S_{i}$ defined as hitting let $F_{i}$ defined as nothet Two hits can take place as $S_{1}\;S_{2}\;S_{3}$ or $S_{1}\;F_{2}\;S_{3} \;or \;F_{1}\;S_{2}\;S_{3}$
$p(S_{1}\;S_{2}\;F_{3})=0.4x0.3x0.8=0.096$
$p(S_{1}\;F_{2}\;S_{3})=0.4x0.7x0.2=0.056$
$p(F_{1}\;S_{2}\;S_{3})=0.6x0.3x0.2=0.036$
$p(two \;hits)=0.096+0.056+0.036$
$\qquad=0.0188$
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