If it is a good day (G) there are 60% chances tomorrow will be G and 40% chances tomorrow will be bad (B). If it is a B day, there 30% chances tomorrow will be G and 70% chances tomorrow will be B. If today is B, what is the expected number of days before seeing another G?
I solved it but I'm not sure if the solution is correct. Please let me know how to go about it if it's wrong. $E = P(G) + P(B) * (E+1)$
$E = 0.3 + 0.7 * (E+1)$
$E = \frac{10}{3}$
but this solution seems to be missing something so I followed another approach I've seen on a similar question.
$E(X)=1 + 0.7 × (1 + 0.7 × (1 + 0.7 × (1+...)+...)$
$E(X) = 1 + 0.7 × \frac{a}{1-r} = 1 + 0.7 × \frac{1}{1-0.7} = \frac{10}{3}$
(using sum of infinite geometric series $= \frac{a}{1−r}$, a = 1,r = 0.7)
Can someone please help me understand the second solution and why both gives the same result?