Question

Two persons \(A\) and \(B\) take part in a shooting competition. \(A\) can hit the target with probability \(0.6\), and \(B\) can hit the target with probability \(0.8\). \(A\) has the first shot, after which they strike alternately. Then, the probability that \(A\) wins the competition is:

• A. \(\frac{7}{10}\)
• B. \(\frac{15}{23}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{11}{17}\)

Hint:

In alternate shooting problems, write all winning cases as an infinite geometric series. The common ratio is usually the probability that both players miss in one full round.

Answer 1

The Correct Option is B

Step 1: State the individual hit and miss probabilities.
A hits with probability P(A) = 0.6 = 3/5, so A misses with probability 2/5. B hits with probability P(B) = 0.8 = 4/5, so B misses with probability 1/5.

Step 2: Model A's winning scenarios as an infinite sequence.
Since A shoots first, A can win on the first shot, or after both miss once and A then hits, or after both miss twice and A then hits, and so on. Thus P(A wins) = (3/5) + (2/5)(1/5)(3/5) + (2/5)²(1/5)²(3/5) + ...

Step 3: Recognize the geometric series.
This is a geometric progression with first term a = 3/5 and common ratio r = (2/5)(1/5) = 2/25.

Step 4: Sum the infinite geometric series.
P(A wins) = a/(1 - r) = (3/5)/[1 - 2/25] = (3/5)/(23/25) = (3/5) × (25/23) = 15/23.

Step 5: Final conclusion.
The probability that A wins the game is 15/23.