A box contains a one-dollar bill, a five-dollar bill, a ten-dollar bill, and a twenty-dollar bill. Two bills are chosen in succession without replacement.
1. What is the probability that the value of at least one of the bills is even? As a fraction
knoin
To calculate the probability, we first need to determine how many different pairs of bills can be selected. Since there are four bills in the box, the first bill can be chosen in 4 different ways. After the first bill is chosen, there will be 3 bills remaining, so the second bill can be chosen in 3 different ways. Therefore, there are a total of 4 * 3 = 12 possible pairs of bills.
Now, let's determine the number of pairs where the value of at least one bill is even.
Pair 1: Choosing an even bill first and any bill second
- There are 2 even bills in the box (the one-dollar bill and the ten-dollar bill). So, we can choose one of these two even bills as the first bill, leaving 3 remaining bills for the second selection. Therefore, there are 2 * 3 = 6 possible pairs where an even bill is chosen first and any bill is chosen second.
Pair 2: Choosing an odd bill first and an even bill second
- There are 2 odd bills in the box (the five-dollar bill and the twenty-dollar bill) and 2 even bills. So, we can choose one of the 2 odd bills first, leaving 2 even bills for the second selection. Therefore, there are 2 * 2 = 4 possible pairs where an odd bill is chosen first and an even bill is chosen second.
Pair 3: Choosing an even bill first and an even bill second
- There are 2 even bills in the box (the one-dollar bill and the ten-dollar bill). So, we can choose one of these two even bills as the first bill, leaving 1 even bill for the second selection. Therefore, there are 2 * 1 = 2 possible pairs where an even bill is chosen first and an even bill is chosen second.
Therefore, the total number of pairs where the value of at least one bill is even is 6 + 4 + 2 = 12.
Finally, we can calculate the probability as the number of favorable outcomes (pairs where the value of at least one bill is even) divided by the total number of possible outcomes (all pairs of bills):
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 12 / 12
Probability = 1
So, the probability that the value of at least one of the bills is even is 1, which can be written as the fraction 1/1.