Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
You roll a number cube twice. Find P(even, then not 2). Write the probability as a fraction in simplest form.
There are six possible outcomes when rolling a number cube twice: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).
Of these outcomes, half are even: (1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6).
Of these even outcomes, nine are not 2: (1,1), (1,3), (1,5), (2,4), (2,6), (3,1), (3,3), (3,5), (4,6).
Therefore, the probability of rolling an even number and then not rolling a 2 is:
9/36 = 1/4
Answer: \boxed{\frac{1}{4}}
You roll a number cube twice. Find P(even, then not 2). Write the probability as a fraction in simplest form.
There are six possible outcomes when rolling a number cube: 1, 2, 3, 4, 5, or 6. If we roll twice, there are 6 x 6 = 36 possible outcomes. We want to find the probability of rolling an even number and then not rolling a 2.
There are three even numbers on a number cube: 2, 4, and 6. The probability of rolling an even number on the first roll is 3/6 or 1/2.
If we roll an even number on the first roll, there are only five possible outcomes for the second roll that are not a 2: 1, 3, 4, 5, and 6. The probability of rolling a number that is not a 2 on the second roll, given that we rolled an even number on the first roll, is 5/6.
To find the probability of rolling an even number and then not rolling a 2, we multiply the probabilities of each event.
P(even, then not 2) = (1/2) x (5/6) = 5/12
Therefore, the probability of rolling an even number and then not rolling a 2 is 5/12, in simplest form.
Answer: \boxed{\frac{5}{12}}.