A and B take part in a rifle shooting match. The probability of A hitting the target is 0.4, while the probability of B hitting the target is 0.6. If A has the first shot, post which both strike alternately, then the probability that A hits the target before B hits it is

According to question, probability of A hitting the target=$$P(A)=0.4$$

probability of B hitting the target=$$P(B)=0.6$$

So, probability of A missing the target=$$P(A^c)=0.6$$

probability of B missing the target=$$P(B^c)=0.4$$

Let, E denote the event of A hitting the target first.

The event E can occur in the following ways:

i.) A hits the target in the first shot

ii.) A and B both miss the target in first shot and after that A hits in the second shot

iii.) A and B both miss the target in first and second shot and after that A hits in the third shot and so on....

So basically, $$P\left(E\right)=P\left(A\right)+P\left(A^c∩\ B^c∩\ A\right)+P\left(A^c∩\ B^c∩\ A^c\ ∩\ B^c∩\ A\right)+......$$

or, $$P\left(E\right)=P\left(A\right)+P\left(A^c\right)\cdot P\ \left(B^c\right)\cdot P\left(A\right)+P\left(A^c\right)\cdot P\left(B^c\right)\cdot P\left(A^c\right)\cdot P\left(B^c\right)\cdot P\ \left(A\right)+......$$

or, $$P\left(E\right)=0.4+0.6\times\ 0.4\times\ 0.4+0.6\times\ 0.4\times\ 0.6\times\ 0.4\times\ 0.4+.....$$

or, $$P\left(E\right)=0.4\left(1+0.6\times\ 0.4+0.6\times\ 0.4\times\ 0.6\times\ 0.4+.....\right)$$

or, $$P\left(E\right)=0.4\left(1+0.24+0.24^2+.....\right)$$

Now, the term in bracket is an infinite G.P. series with common ratio 0.24

or, $$P\left(E\right)=\dfrac{0.4}{1-0.24}=\dfrac{0.4}{0.76}=\dfrac{40}{76}=\dfrac{10}{19}$$