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A and B throw a die alternatively till one of them gets a ′6′ and wins the game. Find their respective probabilities of winning, if A starts first.

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Final answer:

A starts the game and has a probability of 1 of winning, while B has a probability of 0.

Step-by-step explanation:

To find the probabilities of A and B winning the game, we can consider the possible scenarios. Let's assume A starts the game.

Scenario 1: A wins on the first roll by getting a '6'.

Since the die has 6 sides, the probability of A winning on the first roll is 1/6.

Scenario 2: A doesn't win on the first roll, but B wins on the second roll by getting a '6'.

The probability of A not getting a '6' on the first roll is 5/6. And the probability of B getting a '6' on the second roll is 1/6. So, the probability of this scenario is (5/6) * (1/6).

We can continue this pattern for each subsequent roll until either A or B wins.

Therefore, the probability of A winning is the sum of the probabilities of all the scenarios where A wins, which can be written as:

P(A) = 1/6 + (5/6) * (1/6) + (5/6)^2 * (1/6) + ...

Simplifying the equation gives us:

P(A) = 1/6 * (1 + (5/6) + (5/6)^2 + ...)

This is an infinite geometric series where a = 1/6 and r = 5/6.

The formula for the sum of an infinite geometric series is:

Sum = a / (1 - r)

Substituting the values, we get:

P(A) = (1/6) / (1 - (5/6)) = 1/6 * 6/1 = 1

Therefore, the probability of A winning is 1.

Since this is a two-player game, the probability of B winning is 1 - P(A), which is 0.

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