All students in Peter's class have $1.05, consisting of nickels and/or dimes and/or quarters. Each student has a different combination of coins, and all possible combination of coins are included. What is the probability that Peter is unable to give his friend Jenny change for $1.00?
To determine the probability that Peter is unable to give his friend Jenny change for $1.00, we first need to find out the possible combinations of nickels, dimes, and quarters that each student can have. Let's denote the number of nickels, dimes, and quarters as n, d, and q, respectively.
We know that each student has a different combination of coins, so we need to find all possible values for n, d, and q that satisfy the given conditions.
Since each student has a total of $1.05, we can write the equation:
0.05n + 0.10d + 0.25q = 1.05
Multiplying both sides of the equation by 20 to eliminate the decimals, we get:
n + 2d + 5q = 21
Now, we need to find all integer solutions for n, d, and q that satisfy this equation.
Using a combination of trial and error, we find that the possible combinations (n, d, q) are:
(0, 0, 4), (0, 2, 3), (0, 4, 2), (0, 6, 1), (1, 1, 3), (1, 3, 2), (1, 5, 1), (2, 0, 3), (2, 2, 2), (2, 4, 1), (2, 6, 0), (3, 1, 2), (3, 3, 1), (3, 5, 0), (4, 0, 1), (4, 2, 0), (5, 1, 0), (6, 0, 0)
There are a total of 18 possible combinations.
To find the probability that Peter is unable to give his friend Jenny change for $1.00, we need to determine the number of combinations where Peter cannot give $1.00.
Looking at the combinations, we can see that Peter cannot give $1.00 if he has 4 quarters and no nickels or dimes (combination (0, 0, 4)), or if he has 6 dimes and no nickels or quarters (combination (0, 6, 1)).
Therefore, there are 2 out of the 18 possible combinations where Peter cannot give $1.00.
Thus, the probability that Peter is unable to give his friend Jenny change for $1.00 is 2/18, which simplifies to 1/9 or approximately 0.1111.