Answer:
Part a:
Assume that each test is independent.
You want P(Y1<1, Y2<1, Y3<1, Y4<1, Y5>1) = (P(Y<1))4 P(Y>1) by independence
Part b:
Here we have failures on 4,5,6, and 7 and success on 8. So 4 failures and a success. This is exactly the same as part a
Part c:
P(odd) = P(Y>1) + P(Y>1)(P(Y<1))2 + P(Y>1)(P(Y<1))4 . . .
= P(Y>1) Σ (P(Y<1))2n for n in the natural numbers.
P(even) = 1 - P(odd)