A simple tricky probability

They will need to both miss their first $n$ shots, where $n$ is a non-negative integer, after which $A$ hits his target and $B$ misses. Thus the probability will be

$\displaystyle\sum_{n=0}^{\infty} \left[\left(\dfrac{2}{5} \right)^{n} \left(\dfrac{2}{7} \right)^{n} \right] * \left(\dfrac{3}{5}\right) * \left(\dfrac{2}{7} \right) = \dfrac{1}{1 - \dfrac{4}{35}} * \dfrac{6}{35} = \dfrac{6}{31} = \boxed{0.194}$

to 3 decimal places.

This is actually simpler than it seems. Using infinite probabilities:

p= 2/7 ( 3/5 + 2/5 * p) Solving this yields p=6/31.

A rather tedious solution ... :-)

The case that " $B$ requires more shots than $A$ " will arise whenever $B$ hits target after missing $n$ shots whereas A hits the target in $n$ or lesser shots.

Let $M_a^nH_a = MISS_a ... n$ times $... MISS_a HIT_a$ (i.e. $A$ Misses the target $n$ times and Hits in the $n+1$ shot).

For the event $M_b^nH_b$ ( $n \geq 1$ ), the corresponding events such that $B$ requires more shot than $A$ are:

$H_a, M_aH_a, ..., M_a^{n-1}H_a$ (i.e. all the cases where $A$ hits in at most $n$ shots.)

The probability that $B$ hits in $n+1$ shots and $A$ hits in at most $n$ shoots $=$

$P(M_b^nH_b) \times (P(H_a) + P(M_aH_a) + ... + P(M_a^{n-1}H_a) =$

$\left(\dfrac{2}{7}\right)^n \times \dfrac{5}{7} \times \left( \dfrac{3}{5} + \dfrac{2}{5} \times \dfrac{3}{5} + \left(\dfrac{2}{5}\right)^2 \times \dfrac{3}{5} + ... + \left(\dfrac{2}{5}\right)^{n-1} \times \dfrac{3}{5} \right)$

Probability that $B$ will require more shots than $A$ $=$

$\displaystyle\sum_{n=1}^\infty \left[\left(\dfrac{2}{7}\right)^n \times \dfrac{5}{7} \times \left( \dfrac{3}{5} + \dfrac{2}{5} \times \dfrac{3}{5} + \left(\dfrac{2}{5}\right)^2 \times \dfrac{3}{5} + ... + \left(\dfrac{2}{5}\right)^{n-1} \times \dfrac{3}{5} \right) \right] =$

$\dfrac{3}{7} \times \displaystyle\sum_{n=1}^\infty \left[\left(\dfrac{2}{7}\right)^n \times \left( 1 + \dfrac{2}{5} + \left(\dfrac{2}{5}\right)^2 + ... + \left(\dfrac{2}{5}\right)^{n-1} \right) \right] =$

$\dfrac{3}{7} \times \left[\left(\dfrac{2}{7} + \left(\dfrac{2}{7}\right)^2 + ...\right) + \dfrac{2}{5} \times \left( \left(\dfrac{2}{7}\right)^2 + \left(\dfrac{2}{7}\right)^3 + ... \right) + \left(\dfrac{2}{5}\right)^2 \times \left( \left(\dfrac{2}{7}\right)^3 + \left(\dfrac{2}{7}\right)^4 + ... \right) \right] =$

$\dfrac{3}{7} \times \left[\left(\dfrac{2}{5}\right) + \left( \dfrac{2}{7} \right) \times \left( \dfrac{2}{5} \right)^2 + \left( \dfrac{2}{7} \right)^2 \times \left( \dfrac{2}{5} \right)^3 + \left( \dfrac{2}{7} \right)^3 \times \left( \dfrac{2}{5} \right)^4 + ... \right] =$

$\dfrac{3}{7} \times \left( \dfrac{\frac{2}{5}}{1-\frac{2}{7} \times \frac{2}{5}} \right) = \dfrac{6}{31} = 0.194$