1. A locker combination has two nonzero digits, and digits can be used twice. The first number is 8. What is the probability that the second number is 8?
2. My locker combination has three digits. None of the digits are zero. What is the probability that the first digit of my locker combination is less than 4?
I dont expect the answer, i just need help figuring out how to solve it
1. Since you can repeat digits, your second choice will be independent of your first
You can't have a zero, so that leaves 9 to choose from, one of which must be the "8"
prob(of the 8) = ...
2. Well, the first digit could be 1, 2, or 3
How many cases is that?
How many digits were available to you?
To solve these probability problems, we need to understand the concept of counting and calculating the number of favorable outcomes (or desired outcomes) and total possible outcomes.
1. In the first problem, we are given that the locker combination has two nonzero digits, and digits can be used twice. Since the first number is 8, we want to find the probability that the second number is 8.
The total number of possible outcomes can be calculated by considering the number of choices for each digit. Here, each digit can range from 0 to 9 (10 options), but we are excluding zero. So, there are 9 options for the first digit (excluding zero) and 10 options for the second digit (including zero).
The number of favorable outcomes, in this case, is the number of ways the second digit can be 8. As we can use each digit twice, there are two ways to have the 8 in the second digit (8 as the first or second digit).
So, the probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes: Probability = Number of favorable outcomes / Total number of possible outcomes.
2. In the second problem, we have a locker combination with three non-zero digits. We need to find the probability that the first digit is less than 4.
Again, we need to calculate the number of favorable outcomes and the total number of possible outcomes. In this case, the first digit can be any digit from 1 to 9 (excluding zero). But we are only interested in the digits less than 4 (namely 1, 2, and 3).
So, the number of favorable outcomes is 3 (as there are three digits less than 4), and the total number of possible outcomes is 9 (as the first digit can be any of the nine non-zero digits).
We can then calculate the probability as before: Probability = Number of favorable outcomes / Total number of possible outcomes.
By following these steps, we can find the probability for each of these scenarios.