Question

At a fundraiser, a school group charges $6 for tickets for a "grab bag." You choose one bill at random from a bag that contains 39 $1 bills, 19 $5 bills, 3 $10 bills, 7 $20 bills, and 3 $100 bills. Is it likely that you will win enough to pay for your ticket?

Answer 1

no, the probability is only about 18%

Explanation:

there are total of 71 bills of varying denominations in the bag

you need to select a bill that is $10 or more to cover your $6 expense; therefore, P($ ≥ 10) = 3 + 7 + 3 / 71 or 13 /71, about 18%