Question

Two people C and D are shooting at a target one after the other, until one of them hits the target. The probability of C and D hitting the target are respectively given by pC and pD. Assume that the success or failure of hitting the target is statistically independent for each shot. Evaluate the probability that C hits the target before D if C starts shooting first.

Answer 1

Probability of C hitting the target = pC

Probability of C not hitting the target = 1 - pC

Probability of D hitting the target = pD

Probability of D not hitting the target = 1 - pD

Since, the probabilities are independent the success or failure of one won't affect the success or failure or second.

Let, C starts

P[ C hits the target before D ] = P[ hits the target first time ] + P[ hits the target third time ] + P[ hits the target fifth time ] + .........

P[ hits the target first time ] = pC

P[ hits the target third time ] = P[ C doesn't hit ] * P[ D doesn't hit ] * P[ C hits ] = ( 1 - pC )*( 1 - pD )*pC

P[ hits the target fifth time ] = P[ C doesn't hit ] * P[ D doesn't hit ] * P[ C doesn't hit ] * P[ D doesn't hit ] * P[ C hits ] = ( 1 - pC )*( 1 - pD )*( 1 - pC )*( 1 - pD )*pC

P[ C hits the target before D ] = pC + ( 1 - pC )*( 1 - pD )*pC +  ( 1 - pC )*( 1 - pD )*( 1 - pC )*( 1 - pD )*pC + ..............

This becomes an infinite GP with a = pC and r = ( 1 - pC )*( 1 - pD )

Sum of infinite GP = a / ( 1 - r )

P[ C hits the target before D ] = pC / ( 1 - ( 1 - pC )*( 1 - pD ) )

P[ C hits the target before D ] = pC / ( 1 - ( 1 - pC - pD + pC*pD) )

P[ C hits the target before D ] = pC / ( pC + pD - pC*pD )