Question

A statistics Department had purchased 24calculators in which 4 are defective. Calculators are selected one-after-another without replacement and tested. What is the probability that the second calculator found to be defective is the eighth calculator selected.

Answer 1

Use the Hypergeometric distribution

The probability with $n=7, N=24, k=4$

$$P(X=1)=\dfrac{\dbinom{4}{1}\dbinom{24-4}{7-1}}{\dbinom{24}{7}}=\dfrac{340}{759}$$

$$P(second\ defective)=\dfrac{\dbinom{4}{1}\dbinom{24-4}{7-1}}{\dbinom{24}{7}}\cdot\dfrac{4-1}{24-7}$$

$$=\dfrac{340}{759}\cdot\dfrac{3}{17}=\dfrac{20}{253}$$

The probability that the second calculator found to be defective is the eighth calculator selected is $\dfrac{20}{253}.$