Question

5. A box contains two $10 bills, five $5 bills, and eight $1 bills. Two bills are taken at random without replacement from the box.

Answer 1

Answer:-

Given

$ 10 bills ---2
$ 5 bills-----5
$ 1 bills ---- 8
Two bills are drawn randomly without replacement
P($ 10) = 2 / 15
P ($ 5) = 5 / 15
P ( $ 1) = 8 / 15

a) Exactly $ 15

In this case we need $10 and $ 5
P ( $15) = P ($ 10 ) x P ( $5) + P ($ 10 ) x P ( $5)
= 2 /15 x 5 / 15 + 2 /15 x 5 /15
= 10 / 225 + 10 / 225
= 20 /225
P ( $ 15) = 4 / 45

b) probability that the two denominations are of same value

= 2 / 15 x 1 / 14 + 5 / 15 x 4 / 14 + 8 / 15 + 7 / 14 ( Since if $10 is drawn in the first pick there are 14 bills left as one of them is already picked therefore only one is left out of 14. Similar logic for others)
= 2 /210 + 20 / 210 + 56 / 210
= 78 / 210
= 39 / 105