A die is thrown twice. What is the probability that the number in the second throw is higher than the number in the first throw ?

$\frac{5}{12}$

The total number of outcomes of two throw of dice =  6×6

[ 6 outcomes for the first throw and 6 outcomes for the second throw]

To find the probability that the number in the second throw is higher than the number in the first throw, we can consider the possible outcomes:

1. If the first is = 1, then possible outcomes for the second throw (2, 3, 4, 5, 6) = 5

2. If the firs is =  2, then possible outcomes for the second throw (3, 4, 5, 6) = 4

3. If the first throw is= 3, then possible outcomes for the second throw (4, 5, 6) = 3

4. If the first throw is = 4, then possible outcomes for the second throw (5, 6) = 2

5. If the first throw is = 5, then possible outcome for the second throw (6) = 1

6. If the first throw is = 6, there are no possible outcomes for the second throw .

So, Total possible outcomes = 5 + 4 + 3 + 2 + 1 + 0

= 15

Probability that the number in the second throw is higher than the number in the first throw

= \(\frac{15}{36}\)

= \(\frac{5}{12}\)