0
$\begingroup$

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 ?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Because one of the 8'th H is the 10'th flip so only 7 among flip 1-9 $\endgroup$ – H. H. Rugh Aug 4 '16 at 14:18
  • $\begingroup$ Ohh it's based on the assumption that the 10th will be an head. Thanks for the clarification. $\endgroup$ – RonaldB Aug 4 '16 at 14:19
  • $\begingroup$ "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" $\endgroup$ – drhab Aug 4 '16 at 14:32
1
$\begingroup$

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 $P(A|B)=\frac{P(A\cap B)}{P(B)}=$

$$\frac{\sum\limits_{n=7}^{9}\binom{9}{n}\cdot\left(\frac12\right)^{10}}{\sum\limits_{n=8}^{10}\binom{10}{n}\cdot\left(\frac12\right)^{10}}=\frac{23}{28}$$

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.