A fair coin is tossed 10 times. What is the probability that on the last toss you get heads, given that in total there were 8 heads or more?
Usage of conditional probability:
Pr(Head on tenth flip|8 or more heads in the first 10 flips)
What I don't understand is the transition in the solutions sheet.
P(H in the 10th flip | 8 or more H in 10 flips)=
=P(H in the 10th flip AND 7 or more H in flips 1–9)/P(8 or more H in 10 flips)
=P(H in the 10th flip) · P(7 or more H in flips 1–9)/P(8 or more H in 10 flips)
Mainly the second step. We need to find the intersection of the groups so we leave out the tenth flip, but why is it 7 heads in flips 1-9 and not 8 ?