Probability Question 281
Question: The items produced by a firm are supposed to contain 5% defective items. The probability that a sample of 8 items will contain less than 2 defective items, is
[MP PET 1993]
Options:
A) $ \frac{27}{20}{{( \frac{19}{20} )}^{7}} $
B) $ \frac{533}{400}{{( \frac{19}{20} )}^{6}} $
C) $ \frac{153}{20}{{( \frac{1}{20} )}^{7}} $
D) $ \frac{35}{16}{{( \frac{1}{20} )}^{6}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Required probability $ ={}^{8}C_1{{( \frac{1}{20} )}^{1}}{{( \frac{19}{20} )}^{7}}+{}^{8}C_0{{( \frac{1}{20} )}^{0}}{{( \frac{19}{20} )}^{8}}=\frac{27}{20}{{( \frac{19}{20} )}^{7}} $
BETA
