A pair of dice is rolled and the sum is determined. The probability that a sum of 5 is rolled before a sum of 8 is rolled in a sequence of rolls of the dice is ____.
The given answer is 4/9.
Probability sum of 5 before sum of 7
I have found this link as well but I want to know where I am going wrong.
My approach:
Here 2 events are independent. Getting a sum of 5 won't be dependent on getting sum of 8 in next roll. so the problem reduces to probability of getting a sum equal to 5. Hence the probability is equal to 4/36 = 1/9. Even if you assume P(A/B) where A is the the event where sum is 5 and B is the event sum of 8. then P(A/B) = P(A) —- events are independent.
why to apply bayes theorem? what mistake I am doing, aren't this 2 events independent?
Best Answer
It certainly isn't true that the probability of getting a $5$ first is just the probability of getting a $5$. To see that, suppose you were asked for the probability of getting a $5$ before a $9$. You'd still get $\frac 19$ but since a roll of $5$ is just as likely as a roll of $9$ the answer must be $\frac 12$ by symmetry.
You can modify your argument to make it work. Ignore every roll other than an $8$ or a $5$. Then there are $9$ rolls which have meaning, and $4$ of those yield a $5$ so the answer is $\frac 49$.