Answer :
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 = [tex]^{5}C_3[/tex] = [tex]\frac{5!}{3!*2!}[/tex] = [tex]\frac{20}{2}[/tex] = 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. [tex]^{1}C_1 * ^{4}C_2[/tex]
So, required probability = [tex]\frac{^{1}C_1 * ^{4}C_2}{^{5}C_3}[/tex] = [tex]\frac{6}{10}[/tex] = 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.; [tex]^{2}C_2 * ^{3}C_1[/tex]
So, required probability = [tex]\frac{^{2}C_2 * ^{3}C_1}{^{5}C_3}[/tex] = [tex]\frac{3}{10}[/tex] = 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 - [tex]\frac{^{3}C_3}{^{5}C_3}[/tex] = [tex]1-\frac{1}{10}[/tex] = [tex]\frac{9}{10}[/tex] = 0.9 .