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Q.

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

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a

$\frac{817}{3125}$

b

$\frac{823}{3125}$

c

$\frac{819}{3125}$

d

$\frac{821}{3125}$

answer is B.

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Detailed Solution

Given n=5, P=20%= $\frac{1}{5}$

$q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5}$

Required Probability $= P (r \geq 2)$

$= P (r = 2) + P (r = 3) + P (r = 4) + P (r = 5) = {}^{5}{c_{2}} {(\frac{1}{5})}^{2} {(\frac{4}{5})}^{3} + {}^{5}{c_{3}} {(\frac{1}{5})}^{3} {(\frac{4}{5})}^{2} + {}^{5}{c_{4}} {(\frac{1}{5})}^{4} (\frac{4}{5}) + {}^{5}{c_{5}} {(\frac{1}{5})}^{5} = {(\frac{1}{5})}^{5} [10 \times 64 + 10 \times 16 + 5 \times 4 + 1] = \frac{640 + 160 + 20 + 1}{3125} = \frac{821}{3125}$

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