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(D) Let $X$ be the number of engineering students in a sample of $n = 8$ people. This follows a binomial distribution with $p = \frac{1}{5}$ and $q =

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$19 	imes \frac{4^7}{5^8}$

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(D) Let $X$ be the number of engineering students in a sample of $n = 8$ people. This follows a binomial distribution with $p = \frac{1}{5}$ and $q = 1 - p = \frac{4}{5}$.We need to find $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.$P(X=0) = \binom{8}{0} (\frac{1}{5})^0 (\frac{4}{5})^8 = (\frac{4}{5})^8 = \frac{4^8}{5^8}$.$P(X=1) = \binom{8}{1} (\frac{1}{5})^1 (\frac{4}{5})^7 = 8 	imes \frac{1}{5} 	imes \frac{4^7}{5^7} = 2 	imes \frac{4 	imes 4^7}{5^8} = 2 	imes \frac{4^8}{5^8}$.$P(X=2) = \binom{8}{2} (\frac{1}{5})^2 (\frac{4}{5})^6 = 28 	imes \frac{1}{25} 	imes \frac{4^6}{5^6} = 28 	imes \frac{4^6}{5^8} = 28 	imes \frac{4^6}{5^8}$.Summing these: $P(X \le 2) = \frac{4^8 + 2 	imes 4^8 + 28 	imes 4^6}{5^8} = \frac{4^6(16 + 32 + 28)}{5^8} = \frac{76 	imes 4^6}{5^8} = \frac{19 	imes 4 	imes 4^6}{5^8} = 19 	imes \frac{4^7}{5^8}$.

In a university campus,the probability that a person chosen at random is an engineering student is $\frac{1}{5}$. The probability of having at most two engineering students in a sample of $8$ people is:

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