A standard fair dice is rolled twice. What is the probability of getting an even number on the first roll and any number except 5 on the other?

Step-by-step:

Well, the probability for getting an even number on the first roll is 1/2

The probability for getting not 5 would be 5/6

You can make a tree diagram.

      1                     2                   3                    4                5                    6

1  2 3 4 5 6   1 2 3 4 5 6   1 2 3 4 5 6   1 2 3 4 5 6     1 2 3 4 5 6      1 2 3 4 5 6

The numbers on the top represent the first roll. The numbers on the bottom represent the second roll. The first roll should be an even number, so we know that we should be looking at the even numbers. The second one should be anything but 5. So, what I did was look at the numbers that weren't 5. Look at the diagram with the even number at the top. then, look at the numbers that aren't 5 below. Since there are 5 numbers that aren't 5, and 3 even numbers, multiply 3 x 5 and you'll have 15. So add all the numbers at the bottom and that would be 36, so it's 15/36.

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation:

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step-by-step explanation:

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