Ten cards are numbered from 1 to 10. A and B are playing a game by drawing a card, alternately, with a replacement, till one gets a prime number then the probability of B’s winning if A started the game is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{3}{8} \)
The primes from 1 to 10 are 2, 3, 5, 7, so on each draw
\( \displaystyle P(\text{prime})=p=\frac{4}{10}=\frac{2}{5} \)
\( \displaystyle P(\text{not prime})=q=\frac{3}{5} \)
The cards are replaced, so the draws are independent. B wins on the 2nd, 4th, 6th, … draw, each time after all earlier draws fail:
\( \displaystyle P(\text{B wins})=qp+q^3p+q^5p+\ldots \)
This is a geometric series with first term qp and common ratio \( q^2 \):
\( \displaystyle P(\text{B wins})=\frac{qp}{1-q^2} \)
\( \displaystyle =\frac{\frac{3}{5}\cdot\frac{2}{5}}{1-\frac{9}{25}} \)
\( \displaystyle =\frac{\frac{6}{25}}{\frac{16}{25}}=\frac{3}{8} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{2}{5} \) — Incorrect. \( \frac{2}{5} \) is the probability of drawing a prime on a single draw.
Option (2) \( \displaystyle\frac{3}{5} \) — Incorrect. \( \frac{3}{5} \) is the probability of not drawing a prime on a single draw.
Option (3) \( \displaystyle\frac{5}{8} \) — Incorrect. \( \frac{5}{8} \) is the probability that A wins, \( \frac{p}{1-q^2} \); B’s probability is \( 1-\frac{5}{8} \).