Question

In a combat b/n 3 persons A, B, C the probabilities of A, B, C shooting at the target are 0.3, 0.5, 1. They will get the gun in the order A, B, C, A, B, C .... Every person targets to hit the relatively dangerous person and the combat continues until one person is left unhit. Then probability that A is left unhit. (If a person is lit, he is out of combat)

• A. 9/130
• B. 693/2600
• C. 9/130 + 693/2600
• D. 9/2600

Answer 1

Answer: (C)

9/130 + 693/2600

Answer 2

(i) If A hits C then,

= (0.3) (0.5) [P₁ + P₃ + P₅ + .........

= (0.15) [0.3 + (0.7) (0.5) (0.3) + ........]

= (0.15) $\left[ \frac { 0.3 } { 1 - 0.35 } \right] = \frac { 0.15 \times 0.3 } { 0.65 } = \frac { 9 } { 130 }$

ii) If A does not hit C,

= (0.7) [(0.5) (P₁ + P₃ + P₅ + .....) + 0.3 x 0.5]

= 0.35 $\left[ \frac { 9 } { 13 } + \frac { 3 } { 10 } \right] = \frac { 7 } { 20 } \times \frac { 60 + 39 } { 130 } = \frac { 693 } { 2600 }$

Required probability = $\frac { 9 } { 130 } + \frac { 693 } { 2600 }$.