Question

two fair dice are rolled repeatedly until the total number of spots that comes up is divisible by 3. what is the probability that this first happens on a roll number divisble by 3

Answer 1

probably have to sum a geometric series

Answer 2

we can do this for sure
i think that you have to sum a series, but it is not hard

Answer 3

P(sum divisible by 3) = 1/3
do you mean i have to do this?
(1/3)^3+(1/3)^6+......

Answer 4

if it first happens on the third roll, probability is
\[\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}=\frac{4}{27}\]

Answer 5

since the probability it is divisible by 3 is \(\frac{1}{3}\) and the probability it is not divisible by 3 is \(\frac{2}{3}\) so that particular sequence represents
"not , not, is"

Answer 6

then the sequence for getting it for the first time on the sixth roll is
\[\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}\times\frac{1}{3}\]

Answer 7

and so on
the series will look like
\[\frac{4}{27}+\frac{4}{27}\times \frac{8}{27}+\frac{4}{27}\times \left(\frac{8}{27}\right)^2+...\]

Answer 8

you can add this up via
\[\frac{a}{1-r}\] with \(a=\frac{4}{27}\) and \(r=\frac{8}{27}\)

Answer 9

ok

Answer 10

i get \(\frac{4}{19}\)
does that seem right?

Answer 11

yes. that's what I got too

Answer 12

ok