At a fundraiser, a school group charges $8 for tickets for a "grab bag." You choose one bill at random from a bag that contains 38 $1 bills, 20 $5 bills, 7 $10 bills, 5 $20 bills, and 1 $100 bill. Is it likely that you will win enough to pay for your ticket?

It is that you will win enough to pay for your ticket because the probability of winning enough to pay for your ticket as a simplified fraction is

a. It is not likely that a winning ticket will be selected

b. The probability of winning enough to pay for the ticket expressed as a simplified fraction is 13/71

Step-by-step explanation:

To answer this question, there are some important notes to know.

1. The selection is only once

2. To win, you have to select a note which has a greater value than $8

Now, the total number of notes that we have = 38 + 20 + 7 + 5 + 1 = 71

The number of bills present that are greater than $8 are; 7 $10 bills, 5 $20 bills and 1 $100 bill

This makes a total of 7 + 5 + 1 = 13

Thus, the probability of selecting a bill which would pay for the charges is 13/71

Now, the probability of not selecting is 1 - probability of selecting = 1-13/71 = 58/71

Since the probability of not selecting is greater than the probability of selecting, it means that it is not likely that a winning ticket will be selected