If two dice are thrown simultaneously then what will be the probability of getting sum of 8?

Question

Accepted Answer

The correct option is 2:
To solve this problem, we need to understand the concept of probability and how to determine the possible outcomes when two dice are thrown simultaneously.

First, let's recall the basic formula for probability:
$$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$

When two dice are thrown simultaneously, each die has 6 possible outcomes (1, 2, 3, 4, 5, 6). The total number of possible outcomes when throwing two dice is the product of the outcomes for each die:
Total number of possible outcomes = $6 \times 6 = 36$.

Next, we need to find the number of favorable outcomes, which are the combinations where the sum of the numbers on the two dice is 8. Let's list them:  * (2, 6) - The first die shows 2, and the second die shows 6.
  * (3, 5) - The first die shows 3, and the second die shows 5.
  * (4, 4) - The first die shows 4, and the second die shows 4.
  * (5, 3) - The first die shows 5, and the second die shows 3.
  * (6, 2) - The first die shows 6, and the second die shows 2.There are 5 favorable outcomes.

Now, we can calculate the probability of getting a sum of 8:
$$P(\text{sum of 8}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{5}{36}$$
Thus, the correct option is 2, which states $\frac{5}{36}$.

The incorrect option is 1:
$\frac{1}{36}$ would be the probability of a single specific outcome, for example, getting a sum of 2 (only (1,1)) or a sum of 12 (only (6,6)). This does not account for all combinations that sum to 8.

The incorrect option is 3:
$\frac{6}{36}$ (or $\frac{1}{6}$) represents the probability of getting a sum of 7 (combinations: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)) or the probability of a specific number appearing on one die (e.g., rolling a 1 on the first die). It is not the probability of getting a sum of 8.

The incorrect option is 4:
$\frac{4}{36}$ would be the probability of getting a sum of 5 (combinations: (1,4), (2,3), (3,2), (4,1)) or a sum of 9 (combinations: (3,6), (4,5), (5,4), (6,3)). This is incorrect for a sum of 8.