Question

Find the probability of getting neither total of $7$ nor $11$ when a pair of dice is tossed.

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

Answer 1

Answer: (C)

$\frac{7}{9}$

Given: A pair of dice is thrown

To find: the probability of getting neither total of 7 nor 11

Sol:

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 $P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{2}{36}$

Since the events getting sum7 and getting sum 11 are independent of each other and mutually exclusive. Hence, $P(A\cap B)=0$

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

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

Therefore, the probability that the sum is neither 7 nor 11 is

$P(A\cup B{)}^{\prime }=1-P(A\cup B)=1-\frac{8}{36}=\frac{28}{36}=\frac{7}{9}$