all students in peter's class have $1.05, consisting of nickels and/ or dimes. Each student has a different combination of coins, and all possible combinations of coins are included. What is the probability that Peter is unable to give his friend Jenny change for $1.00?
To solve this problem, we need to find the probability that Peter cannot give his friend Jenny change for $1.00. Let's break it down step-by-step.
Step 1: Determine the total possible combinations of coins.
Since each student has a different combination of coins and all possible combinations are included, we need to calculate the total number of possible combinations for $1.05 using nickels and/or dimes.
To find these combinations, we can set up two equations:
Let the number of nickels be represented by 'n', and the number of dimes be represented by 'd'.
Equation 1: 5n + 10d = 105 (in cents, as $1.05 is equal to 105 cents)
Equation 2: n ≠ d (to ensure that each student has a different combination of coins)
Step 2: Solve the equations to find the values of 'n' and 'd'.
To solve these equations, we can use trial and error or any algebraic method. By trying different values of 'n' and 'd', we can find the combinations that satisfy both equations.
After solving the equations, we find that there are 7 possible combinations of coins that satisfy the given conditions:
n = 0, d = 10 (Solution 1: 10 dimes)
n = 5, d = 7 (Solution 2: 5 nickels, 7 dimes)
n = 10, d = 4 (Solution 3: 10 nickels, 4 dimes)
n = 15, d = 1 (Solution 4: 15 nickels, 1 dime)
n = 20, d = 9 (Solution 5: 20 nickels, 9 dimes)
n = 25, d = 6 (Solution 6: 25 nickels, 6 dimes)
n = 30, d = 3 (Solution 7: 30 nickels, 3 dimes)
Step 3: Determine the number of combinations that do not allow Peter to give Jenny change for $1.00.
In order for Peter to be unable to give Jenny change for $1.00, he should not have both nickels and dimes, as a combination of nickels and dimes can make any amount of change.
From the solutions obtained in Step 2, we can see that only Solutions 1 (10 dimes) and 3 (10 nickels, 4 dimes) do not include both nickels and dimes.
So, there are two combinations that do not allow Peter to give Jenny change for $1.00.
Step 4: Calculate the probability.
Now, we can calculate the probability by dividing the number of combinations that do not allow Peter to give Jenny change for $1.00 by the total number of possible combinations.
Probability = (Number of combinations without $1.00 change) ÷ (Total number of possible combinations)
Probability = 2 ÷ 7
Therefore, the probability that Peter is unable to give Jenny change for $1.00 is 2/7 or approximately 0.286.
To find the probability that Peter is unable to give his friend Jenny change for $1.00, we need to determine the number of possible coin combinations that do not allow for Peter to give $1.00 in change.
Let's break down the problem step by step:
1. Determine the possible combinations of nickels and dimes to make $1.05:
- The value of a nickel is $0.05, and the value of a dime is $0.10.
- We need to find how many combinations of nickels and dimes can add up to $1.05.
- We can use a systematic approach to list all the possible combinations.
Here is a table showing the possible combinations:
| Nickels | Dimes |
|---------|--------|
| 21 | 0 |
| 16 | 5 |
| 11 | 10 |
| 6 | 15 |
| 1 | 20 |
| 0 | 25 |
Note that we have exhausted all possible combinations.
2. Determine the total number of possible combinations:
- From the table above, we can see that there are 6 possible coin combinations.
3. Determine the number of combinations where Peter cannot give $1.00 in change:
- To find this number, we need to consider what happens if Peter has one dime.
- We have 5 possible coin combinations when Peter has a dime, all of which are greater than $1.00:
(16 nickels, 5 dimes), (11 nickels, 10 dimes), (6 nickels, 15 dimes), (1 nickel, 20 dimes), (0 nickels, 25 dimes).
Therefore, there are 5 combinations where Peter cannot give $1.00 in change.
4. Calculate the probability:
- The probability is the desired outcome (5 combinations where Peter cannot give $1.00) divided by the total number of outcomes (6 combinations in total).
Thus, the probability that Peter is unable to give his friend Jenny change for $1.00 is 5/6 or approximately 0.8333 (rounded to four decimal places).