Question

A container has ‘a’ defective and 3 non defective items. if a sample of 2 items

are drawn one after another without replacement what is the probability that, the

sample will at most one defective

Answer 1

Let $x=$ the number of defective items among 2 drawn items.

$n=a+3$

$$P(x\leq1)=P(x=0)+P(x=1)$$

$$=\dfrac{\dbinom{a}{0}\dbinom{3}{2}}{\dbinom{a+3}{2}}+\dfrac{\dbinom{a}{1}\dbinom{3}{1}}{\dbinom{a+3}{2}}$$

$$=3\dfrac{2!(a+1)!}{(a+3)!}+3a\dfrac{2!(a+1)!}{(a+3)!}$$

$$=\dfrac{6(a+1)}{(a+2)(a+3)}$$

The probability that, the sample will at most one defective is $\dfrac{6(a+1)}{(a+2)(a+3)} .$