A locker combination has 2 nonzero digits. Digits in a combination are not repeated and range from 2 through 9. Event A = choosi
ng an odd digit for the first number Event B = choosing an odd digit for the second number If all of the digits are equally likely to be chosen, what is P(B|A) expressed in simplest form?.
There are two possibilities; in the first, the first digit is below 5, in the second, it is above. The probability of the first possibility is 3/8, because there are 3 possible digits below 5 and 8 total digits. In this scenario, the chance of the second digit being below 5 is 2/7, because one of the digits is taken. In the other possibility, which has a chance of 5/8, the probability of choosing a number below 5 is 3/7, since all of them are still available. Doing the arithmetic, you find that the total probability is 2/7. Hope this helps!
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