Question

A container has 55 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 in the sample.

$$P(X=0)=\dfrac{\dbinom{3}{2}\dbinom{55}{0}}{\dbinom{55+3}{2}}=\dfrac{3(1)}{1653}=\dfrac{1}{551}$$

$$P(X=1)=\dfrac{\dbinom{3}{1}\dbinom{55}{1}}{\dbinom{55+3}{2}}=\dfrac{3(55)}{1653}=\dfrac{55}{551}$$

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

$$=\dfrac{1}{551}+\dfrac{55}{551}=\dfrac{56}{551}\approx0.101633$$