Question

Two dice are thrown simultaneously. Find the probability of getting:
(i) the sum as 8.
(ii) a multiple of 2 on one dice and a multiple of 3 on other dice.
(iii) a total of at least 10.

Answer 1

In this question, we are given that two dice are thrown simultaneously and we have to find various probabilities of numbers shown on both dice. For this, we will first make sample space and then use that to find favorable outcomes for finding each probability. Total outcomes will be given as the number of elements in sample space. Probability of any event is given as $\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{Total outcomes}}$.

Complete step by step answer:
Here, we are given that two dice are thrown simultaneously. As we know, a dice has 6 possibilities, therefore for two dice, the number of possibilities will be $6\times 6=36$. Let us draw sample space for the given event.
(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)
Hence, the total number of outcomes are 36.
With the help of this sample space we will find required elements for every part.
(i) Here we have to find the probability of getting the sum as 8. Therefore, let us analyze the sample space and count the numbers whose sum is 8. As we can see, following are required cases:
(2,6), (3,5), (4,4), (5,3), (6,2)
Hence, the number of favorable outcomes is 5. So,
$\text{Probability}=\dfrac{5}{36}$.

Hence, the probability of getting the sum as 8 is $\dfrac{5}{36}$.

(ii) Let us analyze the sample space and count numbers whose one number is multiple of 2 i.e. 2, 4, 6 and another number is multiple of 3 i.e. 3, 6. As we can see, following are required cases:
(2,3), (2,6), (4,3), (4,6), (6,3), (6,6), (3,2), (3,4), (3,6), (6,2), (6,4).
Hence, the number of favorable outcomes is 11. So,
$\text{Probability}=\dfrac{11}{36}$.

Hence, probability of getting a multiple of 2 on one dice and multiple of 3 on other dice is $\dfrac{11}{36}$.

(iii) Now, let us count numbers whose sum is at least 10, therefore, we have to count numbers whose sum is 10, 11 or 12. As we can see, following are the required cases:
(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)
Hence, the number of favorable outcomes are 6. So,
$\text{Probability}=\dfrac{6}{36}=\dfrac{1}{6}$.

Hence, the probability of getting a sum at least 10 is $\dfrac{1}{6}$.

Note: Students should carefully count all the possibilities while calculating probability. In (ii) part, students should note that multiple of 2 or 3 can be on any of the two dices. For example, (2,3) and (3,2) both are favorable cases. In (iii) part, students should note that sum should be at least 10, so, they have to consider sum as 10 or higher. Try to avoid mistakes while making sample space.