Answer to Question #106579 in Statistics and Probability for Junaid

Question

Four defective electric toothbrushes were accidentally shipped to a drugstore by Clean brush Products along with 25 non-defective ones.

a. What is the probability the first two toothbrushes sold will be defective? (Round the final answer to 5 decimal places.)


b. What is the probability the first two electric toothbrushes sold will not be defective? (Round the final answer to 4 decimal places.)

Answer 1

We can choose two toothbrushes among $4+25=29$ toothbrushes in

$$\binom{29}{2}={29! \over 2!(29-2)!}={29\cdot28 \over 1\cdot2}=406\ ways$$

a) We can choose two defective toothbrushes among four defective toothbrushes in

$$\binom{4}{2}={4! \over 2!(4-2)!}={4\cdot3 \over 1\cdot2}=6\ ways$$

We can choose zero non-defective toothbrushes among 25 non-defective toothbrushes in

$$\binom{25}{0}=1\ way$$

Hence the probability the first two toothbrushes sold will be defective is

$$P(2\ first\ defective)={\dbinom{4}{2}\dbinom{25}{0} \over \dbinom{29}{2}}=$$ $$={6\cdot1 \over 406}={3 \over 203}\approx0.01478$$

b) We can choose zero defective toothbrushes among four defective toothbrushes in

$$\binom{4}{0}=1\ way$$

We can choose two non-defective toothbrushes among 25 non-defective toothbrushes in

$$\binom{25}{2}={25! \over 2!(25-2)!}={25\cdot24 \over 1\cdot2}=300\ ways$$

Hence the probability the first two toothbrushes sold will be non-defective is

$$P(2\ first\ non-defective)={\dbinom{4}{0}\dbinom{25}{2} \over \dbinom{29}{2}}=$$ $$={1\cdot300 \over 406}={150 \over 203}\approx0.7389$$