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

This problem is a classic example of an infinite geometric series applied to probability. Since they take turns until someone hits the target, the game could theoretically go on forever.

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Identify the Probabilities

For A: $P(A) = 0.4$ and $P(A') = 1 - 0.4 = 0.6$

For B: $P(B) = 0.6$ and $P(B') = 1 - 0.6 = 0.4$

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Determine the Winning Scenarios for A

Since A shoots first and they alternate (A, B, A, B...), A can win on his 1st shot, his 2nd shot, his 3rd shot, and so on.

A wins on 1st shot: A hits immediately.

$P_1 = 0.4$

A wins on 2nd shot: A misses, B misses, then A hits.

$P_2 = 0.6 \times 0.4 \times 0.4 = 0.096$

A wins on 3rd shot: A misses, B misses, A misses, B misses, then A hits.

$P_3 = 0.6 \times 0.4 \times 0.6 \times 0.4 \times 0.4 = 0.02304$

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Use the Infinite Geometric Progression (GP) Formula

The total probability is the sum of these independent events:

$Total\ Probability = 0.4 + 0.096 + 0.02304 + \dots$

This is an infinite GP where:

First term ($a$): $0.4$

Common ratio ($r$): The probability that both miss their respective turns, which is $P(A') \times P(B') = 0.6 \times 0.4 = 0.24$.

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The formula for the sum of an infinite GP is:

$S = \frac{a}{1 - r}$

$S = \frac{0.4}{1 - 0.24}$

$S = \frac{0.4}{0.76}$

$S = \frac{10}{19}$

Result: The probability that A hits the target before B is $10/19$ (or approximately 0.526).