Question

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?1) (5/1)/(40/3)
2) (5/1)/(40/5)
3) (40/3)/(40/5)
4) (3/1)(37/4)/(40/3)
5) (3/1)(37/4)/(40/5)

Answer 1

Answer:

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

Step-by-step explanation:

The total number of ways in which 5 specimens can be selected from the  dish at random is given as C(40, 5).

Since only one of the five specimens would have the trait, the number of ways of selecting the one specimen out of the 3 specimens with the trait is C(3, 1).

3 specimens have the trait therefore 37 specimens (40 - 3) do not have the trait. The number of ways in which the remaining 4 specimens out of the 5 spemimens that do not have the trait is C(37, 4).

Therefore, the probability that only 1 of the 5 specimens selected will have the trait = $\frac{C(3,1)*C(37,4)}{C(40,5)} =\frac{(^3_1)*(^{37}_4)}{(^{40}_5)}$