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?

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 ?

• Because one of the 8'th H is the 10'th flip so only 7 among flip 1-9 – H. H. Rugh Aug 4 '16 at 14:18
• Ohh it's based on the assumption that the 10th will be an head. Thanks for the clarification. – RonaldB Aug 4 '16 at 14:19
• "H in 10th AND $\geq8$ times H in the first 10" is the same as "H in 10th AND $\geq7$ times H in the first 9" – drhab Aug 4 '16 at 14:32

Let A denote the event in which the last out of $10$ tosses is head.
Let B denote the event in which $8$ or more out of $10$ tosses are head.
Then $P(A|B)=\frac{P(A\cap B)}{P(B)}=$
$$\frac{\sum\limits_{n=7}^{9}\binom{9}{n}\cdot\left(\frac12\right)^{10}}{\sum\limits_{n=8}^{10}\binom{10}{n}\cdot\left(\frac12\right)^{10}}=\frac{23}{28}$$