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(B) Consider the Binomial probability distribution where $n = 5$ and $p = 0.2 = \frac{1}{5}$.Probability of success $p = \frac{1}{5}$, so probability

Vedclass · Accepted Answer

$\frac{821}{3125}$

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(B) Consider the Binomial probability distribution where $n = 5$ and $p = 0.2 = \frac{1}{5}$.Probability of success $p = \frac{1}{5}$, so probability of failure $q = 1 - \frac{1}{5} = \frac{4}{5}$.We need to find the probability that at least two subscribers receive the promotion, which is $P(X \geq 2)$.This can be calculated as $P(X \geq 2) = 1 - [P(X = 0) + P(X = 1)]$.$P(X = 0) = { }^5 C_0 \left(\frac{1}{5}ight)^0 \left(\frac{4}{5}ight)^5 = 1 	imes 1 	imes \frac{1024}{3125} = \frac{1024}{3125}$.$P(X = 1) = { }^5 C_1 \left(\frac{1}{5}ight)^1 \left(\frac{4}{5}ight)^4 = 5 	imes \frac{1}{5} 	imes \frac{256}{625} = \frac{256}{625} = \frac{1280}{3125}$.Therefore, $P(X \geq 2) = 1 - \left(\frac{1024 + 1280}{3125}ight) = 1 - \frac{2304}{3125} = \frac{3125 - 2304}{3125} = \frac{821}{3125}$.

As a business strategy, $20 \%$ of the new internet service subscribers selected randomly receive a special promotion. If a group of $5$ such subscribers signs for the service, then the probability that at least two of them get the special promotion is

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