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 pr

Question

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 pr

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

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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`

5 views Answered 2023-02-05 15:29