Question

A pharmacist wishes to select three brands of aspirin to sell in his store. He has five major brands to choose from: A, B, C, D and E. If he selects the three brands at random, what is the probability that he will select the following?
(a) brand B
(b) brands B and C
(c) at least one of the two brands B and C

Answer 1

Answer:

(a) 0.6

(b) 0.3

(c) 0.9

Step-by-step explanation:

We are given that a pharmacist wishes to select three brands of aspirin to sell in his store. He has five major brands to choose from: A, B, C, D and E.

Total number of ways to select 3 brands from total of 5 = $^{5}C_3$ = $\frac{5!}{3!*2!}$ = $\frac{20}{2}$ = 10 .

In all the three parts denominators remains at 10

(a) If he selects the three brands at random, the probability that he will select brand B = one brand B will be selected and other 2 brands will be selected from the remaining 4 brands i.e. $^{1}C_1 * ^{4}C_2$

So, required probability = $\frac{^{1}C_1 * ^{4}C_2}{^{5}C_3}$ = $\frac{6}{10}$ = 0.6

(b) The probability that he will select the brands B and C = Two brands will be selected of B and C and one brand will be selected from remaining 3 brands i.e.; $^{2}C_2 * ^{3}C_1$

So, required probability = $\frac{^{2}C_2 * ^{3}C_1}{^{5}C_3}$ = $\frac{3}{10}$ = 0.3 .

(c) The probability that he will select at least one of the two brands B and C    

     = 1 - Probability that neither of the three brands are from B and C

     = 1 - Probability that all three brands are selected from the remaining

              brands of A, D and E

     =  1 - $\frac{^{3}C_3}{^{5}C_3}$ =  $1-\frac{1}{10}$ = $\frac{9}{10}$ = 0.9 .