Question

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 36 $1 bills, 19 $5 bills, 6 $10 bills, 7 $20 bills, and 3 $100 bills. Is it likely that you will win enough to pay your ticket?

Answer 1

Answer:

Probability of event = 0.28

Step-by-step explanation:

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 36 $1 bills, 19 $5 bills, 6 $10 bills, 7 $20 bills, and 3 $100 bills.

The total number of bills = 36 + 19 + 6 + 7 +3 =71 bills.

We have to find the probability that i will be able to win enough to pay my ticket that is he must get at least $8.

The price would be sufficient if he gets $10, $20 or $100 bills.

So the favorable cases = 20

Total cases = 71

hence probability = $\frac{20}{71}$ = 0.28

The probability of him getting price greater than his ticket = 0.28