Question

A retail store is having a sale. A shopper wanting two medium shirts heads for the sales rack, which is a mess, with sizes jumbled together. Hanging on the rack are 4 medium, 11 large, and 5 extra-large shirts. Find the probability of each event described.

(a) The first two shirts hehe grabs are the wrong size.

(b) The first medium shirt hehe finds is the third one hehe checks.

(c) The first four shirts hehe picks are all extra-large.

(d) At least one of the first four shirts hehe checks is a medium.

Answer 1

Number of ways to select r items from n, nCr = n!/(r! x (n-r)!)

Number of medium = 4

Number of large = 11

Number of extra-large = 5

Total number of shirts = 4 + 11 + 5 = 20

a) P(first 2 shirts are wrong size) = 16C2 / 20C2

= 120/190

= 0.63158

b) P(first medium shirt he finds is the third one) = P(first 2 ones are not medium) x P(next one is medium)

= 16C2/20C2 x 4/18

= 0.14035

c) P(first four shirts he picks are all extra-large) = 5C4/20C4

= 5/4845

= 0.00103

d) P(at least one of the first four shirts he checks is a medium) = 1 - P(all 4 shirts he check are not medium)

= 1 - 16C4/20C4

= 1 - 1820/4845

= 0.62436