Question

A and B take part in a rifle shooting match. The probability of A hitting the target is 0.4, while the probability of B hitting the target is 0.6. If A has the first shot, post which both strike alternately, then the probability that A hits the target before B hits it is

• A. \( \frac{9}{19} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{10}{19} \)

Hint:

In alternate-turn probability problems, identify the event that constitutes one full 'round' where the state resets (e.g., both players miss). The probability of this event is the common ratio 'r' of the geometric series. The first term 'a' is the probability of the player winning on their very first turn.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a probability problem involving an infinite geometric series. A wins if he hits the target on his turn, provided no one has hit the target before. We need to sum the probabilities of all the disjoint events in which A can win.
Step 2: Key Formula or Approach:
Let P(A) be the probability of A hitting and P(B) be the probability of B hitting. P(A') is A missing, P(B') is B missing. A can win on his 1st shot, or 2nd shot (which is the 3rd shot of the game), or 3rd shot (5th shot of the game), and so on. The total probability is the sum of these probabilities: \( P(\text{A wins}) = P(A) + P(A' \cap B' \cap A) + P(A' \cap B' \cap A' \cap B' \cap A) + \dots \) This forms an infinite geometric progression. The sum is given by \( S = \frac{a}{1-r} \), where 'a' is the first term and 'r' is the common ratio.
Step 3: Detailed Explanation:
We are given: P(A hits) = P(A) = 0.4 P(A misses) = P(A') = 1 - 0.4 = 0.6 P(B hits) = P(B) = 0.6 P(B misses) = P(B') = 1 - 0.6 = 0.4 A shoots first. The sequence of events for A to win are:

A hits on the 1st shot: The probability is P(A) = 0.4.
A wins on the 3rd shot: This means A misses (1st), B misses (2nd), and A hits (3rd). The probability is \( P(A') \times P(B') \times P(A) = 0.6 \times 0.4 \times 0.4 = 0.096 \).
A wins on the 5th shot: This means (A misses, B misses) happens twice, then A hits. The probability is \( (P(A') \times P(B'))^2 \times P(A) = (0.6 \times 0.4)^2 \times 0.4 = (0.24)^2 \times 0.4 \).
The total probability of A winning is the sum of this infinite series: \[ P(\text{A wins}) = 0.4 + (0.6 \times 0.4 \times 0.4) + ((0.6 \times 0.4)^2 \times 0.4) + \dots \] \[ P(\text{A wins}) = 0.4 + (0.24 \times 0.4) + (0.24^2 \times 0.4) + \dots \] This is a geometric progression with: First term, \( a = 0.4 \) Common ratio, \( r = 0.6 \times 0.4 = 0.24 \) The sum of an infinite geometric progression is \( S = \frac{a}{1-r} \). \[ P(\text{A wins}) = \frac{0.4}{1 - 0.24} = \frac{0.4}{0.76} \] To simplify the fraction: \[ \frac{0.4}{0.76} = \frac{40}{76} = \frac{10}{19} \] Step 4: Final Answer:
The probability that A hits the target before B hits it is \( \frac{10}{19} \).