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(C) Let $n = 50$ be the total number of problems.Let $p$ be the probability of solving a problem,so $p = \frac{4}{5}$.Let $q$ be the probability of no

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$\frac{54}{5} \left( \frac{4}{5} \right)^{49}$

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(C) Let $n = 50$ be the total number of problems.Let $p$ be the probability of solving a problem,so $p = \frac{4}{5}$.Let $q$ be the probability of not solving a problem,so $q = 1 - p = \frac{1}{5}$.We want to find the probability that the candidate is unable to solve less than two problems,which means the number of problems not solved $(X)$ is $0$ or $1$.Using the binomial distribution formula $P(X = k) = ^{n}C_{k} q^{k} p^{n-k}$:$P(X $P(X = 0) = ^{50}C_{0} \left( \frac{1}{5} \right)^{0} \left( \frac{4}{5} \right)^{50} = 1 \cdot 1 \cdot \left( \frac{4}{5} \right)^{50} = \left( \frac{4}{5} \right)^{50}$$P(X = 1) = ^{50}C_{1} \left( \frac{1}{5} \right)^{1} \left( \frac{4}{5} \right)^{49} = 50 \cdot \frac{1}{5} \cdot \left( \frac{4}{5} \right)^{49} = 10 \cdot \left( \frac{4}{5} \right)^{49}$Summing these probabilities:$P(X $= \left( \frac{4}{5} \right)^{49} \left( \frac{4}{5} + 10 \right)$$= \left( \frac{4}{5} \right)^{49} \left( \frac{4 + 50}{5} \right)$$= \frac{54}{5} \left( \frac{4}{5} \right)^{49}$

For an initial screening of an admission test,a candidate is given $50$ problems to solve. If the probability that the candidate can solve any problem is $\frac{4}{5}$,then the probability that he is unable to solve less than two problems is

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