Question

kmf Of the 40 specimens of bacteria in a dish,3 specimens have a certain trait. If 5 specimens are to be selected from the dish at random and without replacement, which of the following represents the probability that only 1 of the 5 specimens selected will have the trait?

Answer 1

Answer:

$\frac{C^{3}_1* C^{47}_4}{C^{40}_5}$

Explanation:

It is given that out of 40, total 5 specimen are selected.

Thus, the number of ways in which 5 specimens are selected from a given sample of 40 is given by

$C^{40}_{5}$

Number of specimen with certain trait is equal to 3.

Now, total 5 specimen are selected out of which one specimen is selected from the set of specimen that has certain trait (i.e out of 3) while the remaining four specimen will be selected from the rest $40-3=37$ specimen.

Thus, the probability of choosing only 1 of the 5 specimens with the trait is equal to

$\frac{C^{3}_1* C^{37}_4}{C^{40}_5}$