A locker combination has three nonzero digits, and digits cannot be repeated. If the first two digits are even, what is the probability that the third digit is even?
There are 9 possible digits, 4 of which are even and 5 are odd.
Digits cannot be repeated and 1st two are even, so that leaves the possibility of 2 even numbers and 5 odd numbers left.
You should be able to take it from here. :)
thanks for help
To find the probability that the third digit is even, given that the first two digits are even, we need to find the total number of possible combinations where the first two digits are even, as well as the number of those combinations where the third digit is also even.
Let's break down the problem step by step:
Step 1: Count the number of possible combinations where the first digit is even.
Since the digits cannot be repeated and the first digit must be even, we have 4 options for the first digit: 2, 4, 6, or 8.
Step 2: Count the number of possible combinations where the second digit is even.
After choosing one even digit for the first digit, we are left with 3 even digits to choose from for the second digit.
Step 3: Count the number of possible combinations where the third digit is even.
Out of the remaining 8 digits (1, 3, 5, 7, and 9), 4 are even. So, we have 4 options for the third digit.
Step 4: Find the total number of combinations where the first two digits are even.
To find the total number of combinations, we multiply the number of possibilities in each step together. So, the total number of combinations where the first two digits are even is 4 (choices for the first digit) multiplied by 3 (choices for the second digit), which equals 12.
Step 5: Find the number of combinations where the third digit is even.
Since there are 4 options for the third digit, the number of combinations where the third digit is even is also 4.
Step 6: Calculate the probability.
The probability of an event is equal to the number of successful outcomes divided by the total number of possible outcomes. In this case, the successful outcomes are the combinations where both the first two digits are even and the third digit is even. The total number of possible outcomes is the total number of combinations where the first two digits are even.
Therefore, the probability is 4 (number of combinations with even third digit) divided by 12 (total number of combinations with even first two digits), which simplifies to 1/3 or approximately 0.3333.
So, the probability that the third digit is even, given that the first two digits are even, is 1/3 or approximately 0.3333.