A person fails 4 times in a game when he play | vedclass

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(C) Given that the person fails $4$ times in $9$ games, the number of successes is $9 - 4 = 5$. Thus, the probability of success in a single game is $

Vedclass · Accepted Answer

$\frac{79}{9}\left(\frac{4}{9}\right)^{14}$

Vedclass · Answer

(C) Given that the person fails $4$ times in $9$ games, the number of successes is $9 - 4 = 5$. Thus, the probability of success in a single game is $p = \frac{5}{9}$. Consequently, the probability of failure is $q = 1 - p = 1 - \frac{5}{9} = \frac{4}{9}$. For $n = 15$ trials, we want the probability of at most one success, i.e.,$P(X \leq 1) = P(X = 0) + P(X = 1)$. Using the binomial distribution formula $P(X = r) = {}^{n}C_{r} p^r q^{n-r}$: $P(X = 0) = {}^{15}C_{0} \left(\frac{5}{9}ight)^0 \left(\frac{4}{9}ight)^{15} = \left(\frac{4}{9}ight)^{15}$. $P(X = 1) = {}^{15}C_{1} \left(\frac{5}{9}ight)^1 \left(\frac{4}{9}ight)^{14} = 15 	imes \frac{5}{9} 	imes \left(\frac{4}{9}ight)^{14} = \frac{75}{9} 	imes \left(\frac{4}{9}ight)^{14}$. Adding these probabilities: $P(X \leq 1) = \left(\frac{4}{9}ight)^{15} + \frac{75}{9} 	imes \left(\frac{4}{9}ight)^{14} = \left(\frac{4}{9}ight)^{14} \left[ \frac{4}{9} + \frac{75}{9} ight] = \frac{79}{9} \left(\frac{4}{9}ight)^{14}$.

$A$ person fails $4$ times in a game when he plays $9$ times. If he plays $15$ times, the probability of having success at most one is

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