Question

A pair of dice is rolled until a sum of either 6 or 9 appears. Find the probability that a sum of 6occurs first.

Answer 1

Answer:

$P(\text{Sum of 6 occurs first})=\frac{10}{81}$

Step-by-step explanation:

We have been given that a pair of dice is rolled until a sum of either 6 or 9 appears. We are asked to find the probability that a sum of 6 occurs first.

We know that there are 5 ways, when sum of 6 can occur that are: (1,5), (2,4), (3,3), (4,2), (5,1).

There are 4 ways, when sum of 9 can occur that are: (3,6), (4,5), (5,4), (6,3).

Total possible outcomes of two dice rolled at a time are 36.

$P(\text{Sum of 6})=\frac{5}{36}$

$P(\text{Sum of 9})=\frac{4}{36}$

Let us find the probability that sum is not 9 as:

$P(\text{Sum not 9})=1-\frac{4}{36}$

$P(\text{Sum not 9})=\frac{36}{36}-\frac{4}{36}$

$P(\text{Sum not 9})=\frac{32}{36}$

To find the probability that a sum of 6 occurs first, we will multiply the probability of getting sum of six with probability of not getting sum of 9 as:

$P(\text{Sum of 6 occurs first})=\frac{5}{36}\times \frac{32}{36}$

$P(\text{Sum of 6 occurs first})=\frac{5}{36}\times \frac{8}{9}$

$P(\text{Sum of 6 occurs first})=\frac{5}{9}\times \frac{2}{9}$

$P(\text{Sum of 6 occurs first})=\frac{5*2}{9*9}$

$P(\text{Sum of 6 occurs first})=\frac{10}{81}$

Therefore, the probability that sum of 6 occurs first would be 10/81.