Answer to Question #225196 in Statistics and Probability for Reyad

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

Total number of items N = 55+3 = 58

n=2

Number of ways to select two items $= \frac{58!}{2!(58-2)!} = \frac{57 \times 58}{2}=1653$

X= number of defective items

P(X≤1) = P(X=0) + P(X=1)

X=0, so there are no defective items

Number of ways to select two non-defective items from 3:

$=\frac{3!}{2!(3-2)!}=3$

$P(X=0) = \frac{3}{1653}=0.001814$

X=1, so one defective item and one non-defective item

Number of ways to select one non-defective item from 3:

$=\frac{3!}{1!(3-1)!}=3$

Number of ways to select one defective item from 55:

$=\frac{55!}{1!(55-1)!}= 55$

$P(X=1) = \frac{3 \times 55}{1653}=0.099818$

P(X≤1) = 0.001814 + 0.099818 = 0.101632

The probability that, the sample will at most one defective is 0.101632.