Question:

Step 1: Define the probability that $A$ needs exactly $n$ shots: $A$ misses the first $n-1$ shots and hits on the $n$-th. $$P(A = n) = \left(\frac{2}{5}\right)^{n-1} \cdot \frac{3}{5}$$ Similarly for $B$: $$P(B = n) = \left(\frac{2}{7}\right)^{n-1} \cdot \frac{5}{7}$$ Step 2: Compute the probability that $A$ and $B$ require the same number of shots: $$P(A = B) = \sum_{n=1}^{\infty} P(A=n) \cdot P(B=n) = \sum_{n=1}^{\infty} \left(\frac{2}{5}\right)^{n-1}\frac{3}{5} \cdot \left(\frac{2}{7}\right)^{n-1}\frac{5}{7}$$ $$= \frac{3}{5} \cdot \frac{5}{7} \sum_{n=1}^{\infty} \left(\frac{4}{35}\right)^{n-1} = \frac{3}{7} \cdot \frac{1}{1 - \frac{4}{35}} = \frac{3}{7} \cdot \frac{35}{31} = \frac{15}{31}$$ So option (c) is correct. Step 3: Compute the probability that $B$ requires more shots than $A$: $$P(B > A) = \sum_{n=1}^{\infty} P(A=n) \cdot P(B > n) = \sum_{n=1}^{\infty} \left(\frac{2}{5}\right)^{n-1}\frac{3}{5} \cdot \left(\frac{2}{7}\right)^{n}$$ $$= \frac{3}{5} \cdot \frac{2}{7} \sum_{n=1}^{\infty} \left(\frac{4}{35}\right)^{n-1} = \frac{6}{35} \cdot \frac{35}{31} = \frac{6}{31}$$ So option (a) is correct. Step 4: Since $P(A=B) + P(B>A) + P(A>B) = 1$: $$P(A > B) = 1 - \frac{15}{31} - \frac{6}{31} = \frac{10}{31}$$ This is the probability that $B$ requires less shots than $A$. So option (b) is correct. Step 5: Since $P(B > A) = \frac{6}{31} \neq \frac{10}{31} = P(A > B)$, option (d) is incorrect. Step 6: The correct options are (a), (b), and (c), i.e., options 1, 2, 3.