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For the first roll, you expect that you would get an even number. Since there are six sides of a die, then there are 3 ways to get an even number which are numbers 2, 4 and 6. So the probability of an even number is 3/6 or 1/2.
Next for the second roll, you expect that you should not get a number 2. There are 5 ways occurring which are 1, 3, 4, 5, 6. So the probability for this is 5/6.
Multiply the two probabilities because both probabilities will occur for this event.
P(even, then not 2) = (1/2)(5/6) = 5/12
vaughnthompson411

Answer:

For the first roll, you expect that you would get an even number. Since there are six sides of a die, then there are 3 ways to get an even number which are numbers 2, 4 and 6. So the probability of an even number is 3/6 or 1/2.

Next for the second roll, you expect that you should not get a number 2. There are 5 ways occurring which are 1, 3, 4, 5, 6. So the probability for this is 5/6.

Multiply the two probabilities because both probabilities will occur for this event.

P(even, then not 2) = (1/2)(5/6) = 5/12

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