A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 1 and 2. What is the probability that the third digit is 3?
How many total numbers are available in the combination?
1/(Total-2) = ?
A locker combination has 3 nonzero digits, and the digits cannot be repeated. If the first two digits are even, what is the probability that the third digit is even?
To find the probability that the third digit is 3, we need to know the total number of possible combinations for the three-digit locker combination.
Since the first digit is fixed at 1, there is only one choice for it.
For the second digit, we are told that it is 2. So, again, there is only one choice for it.
For the third digit, we are looking for the probability that it is 3. Since we cannot repeat digits, there is only one choice for the third digit as well.
Therefore, the total number of possible combinations for the three-digit locker combination is 1 choice for the first digit, 1 choice for the second digit, and 1 choice for the third digit. In other words, only one combination is possible.
Since the third digit can only be 3, and there is only one combination in total, the probability that the third digit is 3 is 1 out of 1, which can be simplified to a probability of 1.
So, the probability that the third digit is 3 is 1, or 100%.