### Instruction: Each of five, standard, six-sided dice is rolled once. What is the probability that there is at least one pair but not a three-of-a-kind (that is, there are two dice showing the same value, but no three dice show the same value)?### Response: To find the probability of an event, I need to divide the number of favorable outcomes by the number of possible outcomes.
The number of possible outcomes is 6^5, since each die has six choices and there are five dice.
The number of favorable outcomes is harder to count, so I will use some combinatorics and the multiplication principle.
First, I need to choose which two dice will form the pair.
There are 5 choose 2 ways to do this, which is 5 times 4 divided by 2, or 10.
Next, I need to choose the value of the pair.
There are 6 choices for this.
Then, I need to choose the values of the other three dice, which must be different from each other and from the pair.
There are 5 choices for the first one, 4 for the second one, and 3 for the third one.
So, the number of favorable outcomes is 10 times 6 times 5 times 4 times 3, which is 3600.
Therefore, the probability is 3600 divided by 6^5, which is 3600 divided by 7776, which is about 0.463.