Question

A druggist 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 at least one of the two brands B and C

Answer 1

Solve these problems by finding out probability of failure and subtracting it from 1

P(fail) = (3/5)(3/5)(3/5) chose ADE out of the five each time = 0.216

P(success) = 1 - P(fail) = 0.784

Answer 2

hmmm

Answer 3

i think that might not be right

Answer 4

ya i just tried and its wrong :(

Answer 5

there are a couple ways to do it
the idea of computing 1 minus the probability than none are B and C is a good one

Answer 6

that would be
\[\frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\] however

Answer 7

I am assuming there are plenty of each and that he makes three withdrawals, missing B and C each time, thus failing three times in a row. Success is 1 - failure probability, I believe.

Answer 8

then your answer is
\[1-\frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\] whatever that is

Answer 9

Five brands (lots of each) seemed different from five objects, A to E.

Answer 10

thanks!!!
that was right

Answer 11

He has five major brands to choose from
not
he is going to pick bottle at random from an unlimited selection
if that were the problem, he could pick all 5 from brand A

Answer 12

Yes. I misinterpreted the question, had him selecting pills at random from a mixture of them. Makes little sense in retrospect.