Question

On a toss of two dice, A throws a total of $5$, then the probability that he will throw another $5$ before he throws $7$, is:

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{2}{5}$
• D. $\frac{1}{9}$

Answer 1

The correct option is C $\frac{2}{5}$

5 can be thrown in 4 ways and 7 in 6 ways in a single throw with a pair of dice. Hence number of ways of throwing neither 5 nor 7 is 36 - ( 4 + 6 ) = 26

Hence probability of throwing a 5 in a single throw with a pair of dice is $\frac{4}{36}=\frac{1}{9}$

And probabilty of throwing neither 5 nor 7 is $\left(\frac{26}{36}\right)=\frac{13}{18}$

Hence the required probability

$=\frac{1}{9}+\frac{13}{18}\times \frac{1}{9}+{\left(\frac{13}{18}\right)}^2\times \frac{1}{9}+{\left(\frac{13}{18}\right)}^3\times \frac{1}{9}+.....$

$=\frac{\frac{1}{9}}{1-\frac{13}{18}}=\frac{2}{5}$