Question

Three rifle man take one shot each at the same target. The probability of the first rifle man hitting the target is 0.4, the probability of the 2^rd rifle man hitting target is 0.5 and the probability of third rifle man hitting the target is 0.8. Then the probability that exactly two of them hit the target is

• A. 0.92
• B. 0.44
• C. 0.94
• D. None of these

Answer 1

Answer: (B)

0.44

Answer 2

P(1).. P(3) + P(1) . P(2) . $\mathrm { P } ( \overline { \bar { C } } )$ + $\mathrm { P } ( \overline { \mathrm { A } } )$ . P(2). P(3)

= (0.4) (0.5) (0.8) + (0.4).(0.5)(0.2) + (0.6).(0.5)(0.8)

= 0.160 + 0.04 + 0.240 = 0.44