Question

When the two dice are thrown, the probability of getting a number always greater than 4 on the second dice is
(A). $\dfrac{1}{6}$
(B). $\dfrac{1}{3}$
(C). $\dfrac{1}{{36}}$
(D). $\dfrac{5}{{36}}$

Answer 1

Hint: In these types of questions which are based on the concept of probability which helps in finding every possibility of the outcomes of the event as in this question 2 dice are thrown to have a better approach in this question find the total outcomes of the event and the number of outcomes in which number in the second dice is greater than 4.
We know that dice as a number from 1 to 6. So, the possible outcome will be $6 \times 6 = 36$.

Complete step-by-step solution -
Let us first write down all the favorable outcome when in the question is asked the number always greater than 4 (the numbers are only 5, 6 which are greater than 4) on the second dice

DIce 1	Dice 2
1	5
1	6
2	5
2	6
3	5
3	6
4	5
4	6
5	5
5	6
6	5
6	6

Here, we can clearly see that there are 12 outcomes. Thus, the probability will be $\dfrac{\text{(Favorable outcome)}}{\text{(Possible outcome)}}$ which is $\dfrac{12}{36}$= $\dfrac{1}{3}$.
Hence, option B is correct.

Note: Probability is the number of ways to achieve success. A full number of potential tests. In other words, the chance something is going to happen. How probable it is to occur with any case. For example, the chance of flipping a coin and being head is $\dfrac{1}{2}$, since there is 1 way to get ahead and 2(had a tail) is the total number of potential outcomes. Thus, p (heads) =$\dfrac{1}{2}$.