Question: A dice is rolled twice. What is the probability that the number in the second roll will be higher than that in the first?
Explanation:
Total combinations when a die is rolled twice = 6 × 6 = 36
Case 1: The first roll = 1
The second roll = (2, 3, 4, 5, 6)
∴ Favourable outcomes = 5
Case 2 : The first roll = 2
The second roll = (3, 4, 5, 6)
∴ Favourable outcomes = 4
Case 3 : The first roll = 3
The second roll = (4, 5, 6)
∴ Favourable outcomes = 3
Case 4 : The first roll = 4
The second roll = (5, 6)
∴ Favourable outcomes = 2
Case 5 : The first roll = 5
The second roll = (6)
∴ Favourable outcomes = 1
Hence, the number of favourable outcomes = 5 + 4 + 3 + 2 + 1 = 15
Therefore, the probability = 15/36
Alternately,
Let us calculate the number of ways a different number comes up both the die = 6 × 5 = 30.
Out of these 30 ways, in half of these first die will have higher number and in other half second die will have higher number.
∴ Probability that second die has higher number = 15/36.
Hence, option (c).