A machine manufacturing screws is known to produce 5% defectives. In a random sample of 15 screws the probability that there are not more than 3 defective is

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$^{15} C_{3} {(\frac{1}{20})}^{12}$

$\sum_{x = 4}^{15}^{15} C_{x} {(\frac{1}{20})}^{x} {(\frac{19}{20})}^{15 - x}$

$\sum_{x = 0}^{3}^{15} C_{x} {(\frac{1}{20})}^{x} {(\frac{19}{20})}^{15 - x}$

$\sum_{x = 0}^{3}^{15} C_{x} {(\frac{1}{20})}^{x} {(\frac{19}{20})}^{15}$

answer is C.

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Detailed Solution

$n = 15, p = \frac{5}{100} \Rightarrow p = \frac{1}{20}, q = \frac{19}{20}$

$P (x \leq 3) = P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3)$

$\Rightarrow P (x \leq 3) = \sum_{x = 0}^{3}^{15} C_{x} {(\frac{1}{20})}^{x} {(\frac{19}{20})}^{15 - x}$

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