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Question

(C) Let $X$ be the number of problems the candidate is unable to solve. The probability of being unable to solve a problem is $p = 1 - \frac{4}{5} = \

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $X$ be the number of problems the candidate is unable to solve. The probability of being unable to solve a problem is $p = 1 - \frac{4}{5} = \frac{1}{5}$.The probability of being able to solve a problem is $q = \frac{4}{5}$.Given $n = 50$ problems.We need to find the probability that he is unable to solve less than $2$ problems,i.e.,$P(X Using the binomial distribution formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:$P(X = 0) = {}^{50}C_{0} \left(\frac{1}{5}ight)^{0} \left(\frac{4}{5}ight)^{50} = \left(\frac{4}{5}ight)^{50}$.$P(X = 1) = {}^{50}C_{1} \left(\frac{1}{5}ight)^{1} \left(\frac{4}{5}ight)^{49} = 50 	imes \frac{1}{5} 	imes \left(\frac{4}{5}ight)^{49} = 10 	imes \left(\frac{4}{5}ight)^{49}$.$P(X < 2) = \left(\frac{4}{5}ight) \left(\frac{4}{5}ight)^{49} + 10 \left(\frac{4}{5}ight)^{49} = \left(\frac{4}{5} + 10ight) \left(\frac{4}{5}ight)^{49} = \left(\frac{4 + 50}{5}ight) \left(\frac{4}{5}ight)^{49} = \frac{54}{5} \left(\frac{4}{5}ight)^{49}$.

For an initial screening of an entrance exam,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 $2$ problems is:

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