Hi,
Can someone help me with this problem? Thank you.
Problem:
A driver approaches a toll booth and randomly selects two coins from his pocket. If his pocket contains 2 quarters, 2 dimes, and 2 nickels, what is the probability that the two coins he selects will be at least enough to pay a 30 cent toll?
Sure, I can help you with that problem. To find the probability that the driver selects two coins that are at least enough to pay a 30 cent toll, we need to determine the total number of possible outcomes and the number of favorable outcomes. Let's break it down step by step:
Step 1: Determine the total number of possible outcomes.
The driver randomly selects two coins from his pocket. Since he has 2 quarters, 2 dimes, and 2 nickels, he has a total of 6 coins in his pocket. Therefore, the total number of possible outcomes is given by the number of ways to choose 2 coins out of 6, which is denoted as "6 choose 2" or 6C2, and can be calculated using the combination formula:
6C2 = 6! / (2! * (6 - 2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
So, there are 15 different ways to choose 2 coins from his pocket.
Step 2: Determine the number of favorable outcomes.
For the driver to have at least enough to pay a 30 cent toll, he needs to select 2 coins with a combined value of 30 cents or more. Let's consider the possible combinations:
Case 1: Selecting two quarters (25 cents each):
The driver can select 2 quarters in 1 way, as there are 2 quarters in his pocket.
Case 2: Selecting one quarter and one dime (10 cents):
The driver can select 1 quarter and 1 dime in 2 ways:
- Quarter followed by a dime
- Dime followed by a quarter
Case 3: Selecting one quarter and one nickel (5 cents):
The driver can select 1 quarter and 1 nickel in 2 ways:
- Quarter followed by a nickel
- Nickel followed by a quarter
Note: We do not consider the case of selecting two dimes or two nickels since their combined value will always be less than 30 cents.
Therefore, the number of favorable outcomes is 1 (case 1) + 2 (case 2) + 2 (case 3) = 5.
Step 3: Calculate the probability.
Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 5 / 15 = 1/3
So, the probability that the driver selects two coins that are at least enough to pay a 30 cent toll is 1/3, or approximately 0.3333.
I hope this explanation helps you solve the problem. If you have any further questions, feel free to ask!
To determine the probability that the driver will select two coins that are at least enough to pay a 30 cent toll, we need to analyze the different combinations of coins that can be selected.
First, let's calculate the total number of possible combinations of selecting two coins from the pocket.
There are a total of 6 coins in the pocket (2 quarters, 2 dimes, and 2 nickels). We can use the combination formula to find the total number of possible combinations:
nCr = n! / (r! * (n-r)!)
In this case, n = 6 (total number of coins) and r = 2 (number of coins to be selected).
Total number of combinations = 6! / (2! * (6-2)!)
= 6! / (2! * 4!)
= (6 * 5 * 4!) / (2! * 4!)
= (6 * 5) / 2
= 15
So, there are a total of 15 possible combinations of selecting two coins from the pocket.
Next, let's calculate the number of combinations where the coins selected are at least enough to pay a 30 cent toll.
To pay a 30 cent toll, the driver needs either a quarter and a nickel, or three dimes.
Number of combinations with a quarter and a nickel:
- There are 2 quarters and 2 nickels, so the number of combinations with a quarter and a nickel = 2C1 * 2C1 = 2
Number of combinations with three dimes:
- There are 2 dimes, so the number of combinations with three dimes = 2C2 = 1
Total number of combinations with at least enough to pay a 30 cent toll = Number of combinations with a quarter and a nickel + Number of combinations with three dimes
= 2 + 1
= 3
Finally, we can find the probability by dividing the number of combinations with at least enough to pay a 30 cent toll by the total number of possible combinations:
Probability = Number of combinations with at least enough to pay a 30 cent toll / Total number of possible combinations
= 3 / 15
= 1 / 5
= 0.2
Therefore, the probability that the driver will select two coins that are at least enough to pay a 30 cent toll is 0.2 or 20%.