Two fair cubical dice are thrown. What is the probability that at least one of them shows 4?

Correct option is B

Given :- Two fair dices are thrown. We have to find the probability that at least one of them shows 4

Concept Used : We will write all the favourable outcomes.

Solution: 

​$$\begin{aligned}&\text{Total outcomes when two dice are thrown: } 6 \times 6 = 36 \\[6pt]&\text{Favorable outcomes (at least one 4):} \\[3pt]&\quad \text{Use complement rule: } P(\text{at least one 4}) = 1 - P(\text{no 4}) \\[6pt]&\text{No 4 on either die: choices = } 5 \times 5 = 25 \\[6pt]&\Rightarrow P(\text{at least one 4}) = 1 - \frac{25}{36} = \frac{11}{36} \\[6pt]\end{aligned}$$​​

outcomes which would be

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

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

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

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

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

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

Here total outcomes which contains at least 4 are 11

hence probability is $$\boxed{\text{Answer: } \frac{11}{36}}$$​​