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What is the probability of having 13 wins and 3 losses in 16 games?
Wiki User
∙ 9y ago
Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is
16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)
However, the assumption of constant and independent probability of a win is rubbish.
Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is
16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)
However, the assumption of constant and independent probability of a win is rubbish.
Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is
16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)
However, the assumption of constant and independent probability of a win is rubbish.
Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is
16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)
However, the assumption of constant and independent probability of a win is rubbish.
Wiki User
∙ 9y ago
The probability of having 13 wins and 3 losses in 16 games depends on the success rate of winning each game. If each game is independent and the probability of winning is constant, you can calculate it using the binomial distribution formula. With the information provided, you can use the formula (P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k}), where n = 16, k = 13, and p is the probability of winning each game.
Wiki User
∙ 9y agoSuppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is
16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)
However, the assumption of constant and independent probability of a win is rubbish.
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