Question

A die is rolled repeatedly and each roll is independent of each other. Then, what's the probability that "2" appears before "5" in the sequence?

Answer 1

we need to sum a geometric series for this one
do you know how to do it?

Answer 2

we compute the probability of getting a 2 before a 5
how can we do it?
first rolls is a 2, probability
\(\frac{1}{6}\)
first roll is not a 2 or a 5, second roll is a 2 probability
\(\frac{2}{3}\times \frac{1}{6}\)
first roll not a 2 or 5, second roll not a 2 or 5, third roll 2 probability
\[(\frac{2}{3})^2\times \frac{1}{6}\]

etc etc you have to add
\[\sum_{k=0}^{\infty}\left(\frac{2}{3}\right)^k\frac{1}{6}\]

Answer 3

Hm I kind of get it? I know how to compute the sum, but this looks like it's considering just when is the first "2" going to show. But what about the "5", it has to appear after the 2... I'm not sure how the above takes this into account

Answer 4

Oh Wait.. is it because we "don't care" about the 5 because as long as the 2 shows up first (without the 5 showing up), then it doesn't matter if subsequent rolls have the 5, because we are guaranteeing that 2 will show up before 5 regardless??