Who shoots the target first?

Let $a,b$ be the number of shots $A,B$ respectively takes until they hit the target. Then since their results are independent of one another, the desired probability will be

$S = \displaystyle\sum_{n=1}^{\infty} (P(a = n) \times P(b \gt n))$ .

Now for $P(a = n)$ , $A$ must miss their first $n - 1$ shots and then hit on their $n$ th shot. Thus

$P(a = n) = \left(1 - \dfrac{3}{5}\right)^{n - 1} \times \dfrac{3}{5} = \dfrac{3}{2} \times \left(\dfrac{2}{5}\right)^{n}$ . Next, we have that

$P(b \gt n) = \displaystyle\sum_{k=n+1}^{\infty} P(b = k) = \sum_{k=n+1}^{\infty} \left(\dfrac{2}{7}\right)^{k - 1} \times \dfrac{5}{7} = \dfrac{5}{7} \sum_{k=n + 1}^{\infty} \left(\dfrac{2}{7}\right)^{k-1} = \dfrac{5}{7} \times \dfrac{\left(\dfrac{2}{7}\right)^{n}}{1 - \dfrac{2}{7}} = \left(\dfrac{2}{7}\right)^{n}$ .

Thus $S = \dfrac{3}{2} \displaystyle\sum_{n=1}^{\infty} \left(\dfrac{2}{5} \times \dfrac{2}{7}\right)^{n} = \dfrac{3}{2} \times \dfrac{\dfrac{4}{35}}{1 - \dfrac{4}{35}} = \dfrac{3}{2} \times \dfrac{4}{31} = \dfrac{6}{31}$ .

So finally $a + b = 6 + 31 = \boxed{37}$ .

Comments: The probability that $A,B$ require the same number of shots will be

$\displaystyle\sum_{n=1}^{\infty} (P(A = n) \times P(B = n)) = \sum_{n=1}^{\infty} \left(\left(\dfrac{4}{35}\right)^{n-1} \times \dfrac{3}{5} \times \dfrac{5}{7}\right) = \dfrac{3}{7} \times \dfrac{1}{1 - \dfrac{4}{35}} = \dfrac{3}{7} \times \dfrac{35}{31} = \dfrac{15}{31}$ .

The probability that $B$ requires fewer shots will then be $1 - \dfrac{6}{31} - \dfrac{15}{31} = \dfrac{10}{31}$ .