Two fair dice are thrown. The number of cases where the number appearing on the upper face of the first die is not less than that on the lower face of the second die is

Correct option is D

Given:
Two fair dice are thrown.
We must count the number of favorable outcomes such that:

​​Number on the **first die** ≥ Number on the **second die**

Each die can show values from 1 to 6, so total possible outcomes:

$$\text{Total outcomes} = 6 \times 6 = 36$$

Formula Used:
Count the number of ordered pairs
(a, b) such that:

​$$a \geq b \quad \text{where } a, b \in \{1, 2, 3, 4, 5, 6\}$$

Solution:

We will count the number of outcomes where first die ≥ second die:

List all such combinations:

• First die = 1 → (1,1) → 1 case

• First die = 2 → (2,1), (2,2) → 2 cases

• First die = 3 → (3,1), (3,2), (3,3) → 3 cases

• First die = 4 → (4,1), (4,2), (4,3), (4,4) → 4 cases

• First die = 5 → (5,1), (5,2), (5,3), (5,4), (5,5) → 5 cases

• First die = 6 → (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) → 6 cases

Now sum them up:

​$$1 + 2 + 3 + 4 + 5 + 6 = 21 \text{ favorable outcomes}$$

Final Answer: (D) 21​​