Question

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is(A) 10-1 (B) `(1/2)^5` (C) `(9/(10))^5` (D) `9/(10)` A. `((9)/(10))^5` B. `(9)/(10)` C. `10^-5` D. `((1)/(2))^2` Select the correct answer from above options

Answer 1

Correct Answer - A We have , `p=` Probability that a bulb is defective `=(10)/(100)=(1)/(10)` `therefore q=1-p=(9)/(10)` Let X be the number of defective bulbs in a simple of 5 bulbs. Then, X follows binomial bistribution with `n=5, p=(1)/(10)and q=(9)/(10)` such that `P(X=r)=.^5C_(r)((1)/(10))^r((9)/(10))^(5-r),r=0,1,...,5` `therefore` Required Probability `=P(X=0)` `=.^C_(0)((1)/(10)^0((9)/(10))^5=((9)/(10))^5`.