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In a chess tournament, n games are being played, independently. Each game ends in a win for one player with a probability of 0.4 and ends in a draw (tie) with a probability of 0.6. Find the PMFs of the number of games ending in a draw, and of the number of players whose games end in draws.

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Final answer:

The PMF (Probability Mass Function) of the number of games ending in a draw can be calculated using the binomial distribution formula. The PMF of the number of players whose games end in draws can also be calculated using the same formula.

Step-by-step explanation:

The PMF (Probability Mass Function) of the number of games ending in a draw can be calculated using the binomial distribution formula. Let's say there are n games being played, each with a probability of 0.6 of ending in a draw. The PMF of the number of games ending in a draw, k, is given by the formula:

P(k) = C(n, k) * p^k * (1-p)^(n-k)

where C(n, k) is the binomial coefficient of choosing k draws out of n games. For example, if n = 10 and k = 4, the PMF of the number of games ending in a draw is:

P(4) = C(10, 4) * 0.6^4 * 0 .4^6 = 210 * 0.129 6 * 0.046 656 = 0.986 646

To find the PMF of the number of players whose games end in draws, we can use similar logic. If there are n players, each playing one game with a probability of 0.6 of ending in a draw, the PMF of the number of players whose games end in draws, m, is given by the formula:

P(m) = C(n, m) * p^m * (1-p)^(n-m)

For example, if n = 8 and m = 3, the PMF of the number of players whose games end in draws is:

P(3) = C(8, 3) * 0.6^3 * 0.4^5 = 56 * 0.0144 * 0.0819 = 0.0661

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