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(C) This is a binomial distribution problem where $n = 7$ matches are played.For each match,there are $3$ possible outcomes (win,draw,defeat),so the p

Vedclass · Accepted Answer

$\frac{280}{3^7}$

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(C) This is a binomial distribution problem where $n = 7$ matches are played.For each match,there are $3$ possible outcomes (win,draw,defeat),so the probability of a correct prediction is $p = \frac{1}{3}$ and the probability of an incorrect prediction is $q = 1 - p = \frac{2}{3}$.We need to find the probability of exactly $k = 4$ correct predictions.The formula for binomial probability is $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$.Substituting the values: $P(X = 4) = {}^7C_4 \cdot (\frac{1}{3})^4 \cdot (\frac{2}{3})^{7-4}$.Calculate the combination: ${}^7C_4 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.Calculate the powers: $(\frac{1}{3})^4 = \frac{1}{81}$ and $(\frac{2}{3})^3 = \frac{8}{27}$.Multiply the values: $P(X = 4) = 35 	imes \frac{1}{81} 	imes \frac{8}{27} = 35 	imes \frac{8}{2187} = \frac{280}{3^7}$.

$A$ contest consists of predicting the results (win,draw,or defeat) of $7$ football matches. $A$ sent his entry by predicting at random. The probability that his entry will contain exactly $4$ correct predictions is:

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