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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 isA. 0.024B. 0.188C. 0.336D. 0.452 |
Answer» 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` |
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