Answer to Question #202332 in Statistics and Probability for Aasia Tariq

Question

A bin contains six parts. Two of the parts are defective and four are acceptable. If three of the six parts are selected from the bin, how large is the sample space? Which counting rule did you use, and why? For this sample space, what is the probability that exactly one of the three sampled parts is defective?

Answer 1

3 of 6 parts are selected. Here, order is not important.

Thus, we will use is combinations

Sample space size=$\dbinom{6}{3}=\dfrac{6!}{3!(6-3)!}=20$  

Use the $mn$ counting rule

$$P(one \ D)=\dfrac{\dbinom{2}{1}\dbinom{4}{3-1}}{\dbinom{6}{3}}=\dfrac{2(6)}{20}=0.6$$