a pair of dice is rolled together till a sum of 6 or 8 is obtained.find the probability that 6 comes before 8.
Since 6 and 8 are equally likely on any one roll, the probability that "6 comes before an 8" (which means that the sequence of rolls terminate on a roll of 6, not 8) is exactly the same as "8 comes before a 6". Since one of those two must happen by the rules of this problem, their sum must be 1: each is equal to 1/2.
Hello, psparmar!
There is an "intuitive" solution for this.
We find that: .$\displaystyle \begin{array}{ccc} P(\text{sum of 6}) &=& \frac{5}{36} \\ \\[-3mm] P(\text{sum of 8}) &=& \frac{5}{36} \end{array}$A pair of dice is rolled together till a sum of 6 or 8 is obtained.
Find the probability that 6 comes before 8.
Their probabilities are equal.
Either event is equally likely to happen.
Therefore, the probability that a sum of 6 precedes a sum of 8 is: .$\displaystyle \frac{1}{2}$