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`

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`.