IPMAT Indore 2025 · Quantitative Ability · Question 36

Step 1 — list the ways A can win. A wins on his 1st shot, or his 2nd, or his 3rd, …

A wins on shot 1: 0.4

A wins on shot 2: A misses \left( 0.6 \right), B misses \left( 0.4 \right), A hits \left( 0.4 \right) → 0.6\times 0.4\times 0.4

A wins on shot 3: both miss twice, then A hits → {\left( 0.6\times 0.4 \right)}^{2}\times 0.4

Step 2 — spot the geometric series. Each extra round multiplies by the “both miss” factor

r=0.6\times 0.4=0.24

So

P\left( A \right)=0.4\left( 1+0.24+{0.24}^{2}+\cdots \right)

Step 3 — sum the infinite GP with \frac{1}{1-r}:

P\left( A \right)=\frac{0.4}{1-0.24}=\frac{0.4}{0.76}=\frac{40}{76}=\frac{10}{19}

Sense check: 10/19\approx 0.53 — just over half, which fits: A is the weaker shot but goes first.

Answer: Option 2 — \frac{10}{19}