Question:

Let $Y$ be the number of problems the candidate is unable to solve. The total number of problems is $N=50$. The probability that the candidate can solve any problem is $p = 4/5$. The probability that the candidate is unable to solve any problem is $q = 1 - p = 1 - 4/5 = 1/5$. The number of problems the candidate is unable to solve, $Y$, follows a binomial distribution with parameters $N=50$ and $q=1/5$. Thus, $Y \sim \text{Bin}(50, 1/5)$. The probability mass function for $Y$ is given by $P(Y=k) = \binom{N}{k} q^k p^{N-k}$. We need to find the probability that the candidate is unable to solve less than two problems, which is $P(Y < 2)$. This can be expressed as the sum of probabilities for $Y=0$ and $Y=1$: $$P(Y < 2) = P(Y=0) + P(Y=1)$$ First, calculate $P(Y=0)$: $$P(Y=0) = \binom{50}{0} \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^{50} = 1 \cdot 1 \cdot \left(\frac{4}{5}\right)^{50} = \left(\frac{4}{5}\right)^{50}$$ Next, calculate $P(Y=1)$: $$P(Y=1) = \binom{50}{1} \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^{49} = 50 \cdot \frac{1}{5} \cdot \left(\frac{4}{5}\right)^{49} = 10 \cdot \left(\frac{4}{5}\right)^{49}$$ Now, sum these probabilities: $$P(Y < 2) = \left(\frac{4}{5}\right)^{50} + 10 \cdot \left(\frac{4}{5}\right)^{49}$$ To simplify, factor out $\left(\frac{4}{5}\right)^{48}$: $$P(Y < 2) = \left(\frac{4}{5}\right)^{48} \left[ \left(\frac{4}{5}\right)^2 + 10 \cdot \left(\frac{4}{5}\right)^1 \right]$$ $$P(Y < 2) = \left(\frac{4}{5}\right)^{48} \left[ \frac{16}{25} + \frac{40}{5} \right]$$ $$P(Y < 2) = \left(\frac{4}{5}\right)^{48} \left[ \frac{16}{25} + \frac{200}{25} \right]$$ $$P(Y < 2) = \left(\frac{4}{5}\right)^{48} \left[ \frac{16 + 200}{25} \right]$$ $$P(Y < 2) = \frac{216}{25} \left(\frac{4}{5}\right)^{48}$$