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(C) Let $X$ be the number of machines out of service. $X$ follows a binomial distribution with $n = 5$ and $p = \frac{1}{4}$.The probability of $r$ ma

Vedclass · Accepted Answer

$\frac{17}{8}$

Vedclass · Answer

(C) Let $X$ be the number of machines out of service. $X$ follows a binomial distribution with $n = 5$ and $p = \frac{1}{4}$.The probability of $r$ machines being out of service is given by $P(X = r) = ^{5}C_{r} (\frac{1}{4})^{r} (\frac{3}{4})^{5-r}$.We need to find the probability that at most $2$ machines are out of service,which is $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.$P(X=0) = ^{5}C_{0} (\frac{1}{4})^{0} (\frac{3}{4})^{5} = (\frac{3}{4})^{5}$.$P(X=1) = ^{5}C_{1} (\frac{1}{4})^{1} (\frac{3}{4})^{4} = 5 	imes \frac{1}{4} 	imes (\frac{3}{4})^{4} = \frac{5}{4} (\frac{3}{4})^{4} = \frac{5 	imes 3}{4 	imes 4} (\frac{3}{4})^{3} = \frac{15}{16} (\frac{3}{4})^{3}$.$P(X=2) = ^{5}C_{2} (\frac{1}{4})^{2} (\frac{3}{4})^{3} = 10 	imes \frac{1}{16} 	imes (\frac{3}{4})^{3} = \frac{10}{16} (\frac{3}{4})^{3} = \frac{5}{8} (\frac{3}{4})^{3}$.Summing these: $P(X \le 2) = (\frac{3}{4})^{2} (\frac{3}{4})^{3} + \frac{15}{16} (\frac{3}{4})^{3} + \frac{10}{16} (\frac{3}{4})^{3} = (\frac{9}{16} + \frac{15}{16} + \frac{10}{16}) (\frac{3}{4})^{3} = \frac{34}{16} (\frac{3}{4})^{3} = \frac{17}{8} (\frac{3}{4})^{3}$.Comparing this with $(\frac{3}{4})^{3} k$,we get $k = \frac{17}{8}$.

In a workshop,there are five machines and the probability of any one of them to be out of service on a day is $\frac{1}{4}$. If the probability that at most two machines will be out of service on the same day is $\left(\frac{3}{4}\right)^{3} k$,then $k$ is equal to

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