A fair 6-sided die is rolled twice. The probability that the second roll is strictly less than the first roll can be written as a/b, where a and b are positive, coprime integers. What is the value of a+b?

Assume die is fair.

Rolling twice gives 36 outcomes, of which the second roll is strictly less than the first (successes) can be enumerated as follows:
21,31,32,41,42,43,51,52,53,54,61,62,63,64,65
Count how many successes there are, and divide by the size of the sample space (36).

thanks sir...

You're welcome!

It would be even better if you would not use various aliases.

To find the probability that the second roll is strictly less than the first roll, we can break down the possible outcomes and calculate their probabilities.

Let's consider all possible outcomes for the two dice rolls. Each roll has 6 equally likely outcomes because the die has 6 sides.

The total number of outcomes is given by 6 multiplied by 6, since we have two rolls: 6 x 6 = 36.

Now, let's analyze the possible outcomes where the second roll is strictly less than the first roll. We can list them as follows:

(1, 6), (2, 6), (2, 5), (3, 6), (3, 5), (3, 4), (4, 6), (4, 5), (4, 4), (5, 6), (5, 5), (5, 4), (5, 3), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1)

There are 19 outcomes where the second roll is strictly less than the first roll. Therefore, the probability is 19/36.

Since 19 and 36 are already coprime integers, we have a = 19 and b = 36.

The sum of a and b is a + b = 19 + 36 = 55.

Therefore, the value of a + b is 55.