Question

In a single throw of two dice, find the probability of getting a sum different from $8.$

Answer 1

Step 1: Find the possible outcomes:

The probability of occurrence of an event = $\frac{Thetotalnumberoffavourableoutcomes}{Totalnumberofoutcomes}$

Thus, the possible outcomes are as follows:

$$\begin{gathered} (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) \\ (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) \\ (3,1)(3,2)(3,3)(3,4)(3,5)(3,6) \\ (4,1)(4,2)(4,3)(4,4)(4,5)(4,6) \\ (5,1)(5,2)(5,3)(5,4)(5,5)(5,6) \\ (6,1)(6,2)(6,3)(6,4)(6,5)(6,6) \end{gathered}$$

Step 2: Find the probability of getting a sum different from $8.$

Therefore, the total number of outcomes =$36$

There are $5$ possibilities $\left\{\right(2,6\left)\right(3,5\left)\right(4,4\left)\right(5,3\left)\right(6,2\left)\right\}$ for getting a sum as $8$

Thus, the probability is,

$$\begin{gathered} =\frac{36-5}{36} \\ =\frac{31}{36} \end{gathered}$$

Hence, the probability of getting a sum different from $8$ is $\frac{31}{36}$