A locker combination has three nonzero digits, with no digit repeated. If the first two digits are odd, what is the probability that the third digit is also odd?
hlp
256
its one in 256
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To solve this problem, we need to understand the given conditions and then calculate the probability.
Given:
- The locker combination has three nonzero digits.
- The first two digits are odd, and no digit is repeated.
We know that the first two digits are odd, so there are 5 odd digits to choose from: 1, 3, 5, 7, and 9.
To calculate the probability that the third digit is also odd, we need to find out the total number of possible combinations and the number of combinations where the third digit is odd.
Total number of combinations:
Since no digit is repeated, there are 5 choices for the first digit, 4 choices for the second digit, and 3 choices for the third digit. Therefore, the total number of combinations is 5 * 4 * 3 = 60.
Number of combinations with the third digit odd:
Since the first two digits are odd, we already have used two odd digits. Therefore, for the third digit to be odd, we have 3 odd digits left to choose from. So, the number of combinations with the third digit odd is 3.
Probability:
The probability is the number of favorable outcomes (combinations with the third digit odd) divided by the total number of possible outcomes.
Probability = Number of combinations with the third digit odd / Total number of combinations
Probability = 3 / 60
Simplifying, we get:
Probability = 1 / 20
Therefore, the probability that the third digit is odd is 1/20.