Question

Find the probability of getting a total of $7$ or $11$ when a pair of dice is tossed.

• A. $\frac{1}{9}$
• B. $\frac{7}{9}$
• C. $\frac{2}{9}$
• D. $\frac{5}{9}$

Answer 1

The correct option is C $\frac{2}{9}$

Total number of all possible outcomes, $n\left(S\right)=6\times 6=36$

Let A: be the event that the sum is $7$.

Hence the favourable outcomes are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$, i.e., $n\left(A\right)=6$

Hence the probability of getting a total of $7$. $P\left(E\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{6}{36}$

Similarly, let B: be the event that the sum is $11$

Hence the favourable outcomes are {(5, 6), (6, 5)}, i.e., $n\left(E\right)=2$

Hence the probability of getting a total of 11 $P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{2}{36}$

Now, the probability of getting a total of $7$ or $11$ is

$P\left(A\right)+P\left(B\right)⟹=\frac{6}{36}+\frac{2}{36}=\frac{8}{36}=\frac{2}{9}$