Question

If two unbiased six-faced dice are thrown simultaneously until a sum of either $7$ or $11$ occurs, then the probability that $7$ comes before $11$ is

• A. $\frac{1}{4}$
• B. $\frac{3}{4}$
• C. $\frac{5}{9}$
• D. $\frac{5}{18}$

Answer 1

Answer: (B)

Let $A$ be the event of obtained sum of $7$ and $B$ be the event of obtained sum of $11$.
$\therefore n(A)=\{(2,5),(5,2),(3,4),(4,3),(1,1),(6,1)\}=6$
Now, $P(A)=\frac{6}{36}=\frac{1}{6}$
and $n(B)=\{(5,6),(6,5)\}=2$
$\therefore P(B)=\frac{2}{36}=\frac{1}{18}$
$C=$ Neither a sum of $11$ nor a sum of $7$ shows of
$\therefore P(C)=\frac{36-(6+2)}{36}$
$=\frac{36-8}{36}=\frac{28}{36}=\frac{7}{9}$
Required probability $(p)$
$=\frac{6}{36}+\frac{6}{36}\times \frac{28}{36}+{\left(\frac{28}{36}\right)}^2\left(\frac{6}{36}\right)+{\left(\frac{28}{36}\right)}^3\left(\frac{6}{36}\right)\dots$
$=\frac{1}{6}\left[\frac{1}{1-\frac{7}{9}}\right]$
$=\frac{1}{6}\times \frac{9}{2}=\frac{3}{4}$