A carton contains 20 bulbs ,5 of which are defective. The probability that,if a sample of 3 bulbs in chosen at random from the carton, 2 will be defective, is

Options

A. `(1)/(16)`
B. `(3)/(64)`
C. `(9)/(64)`
D. `(2)/(3)`

Answer

C

Explanation

(c ) We have,
p=Probability that a bulb is defective `=(5)/(20)=(1)/(4)`
`:. " " q==1-p=1-(1)/(4)=(3)/(4)`
Let X denots the number of defective bulbs in a sample of 3 bulbs. Then, X is a binomial variate with parameter n=3 and `p=(1)/(4)` and `q=(3)/(4)` such that
`P(X=r)=""^(3)C_(r )((1)/(4))^(r )((3)/(4))^(3-r)`
Now, for `r=2, P(X=2)=""^(3)C_(r )((1)/(4))^(2)((3)/(4))=(9)/(64)`