Question

A coin is tossed 8 times. Calculate the probability that the first 4 tosses and the last 4 tosses result in the same number of heads.

Answer 1

i hate these coin questions :(

Answer 2

The answer is 0.273... so not 15/64... but I'm not sure how to arrive at the answer.

Answer 3

0 heads: 0.5^8
1 head in each: 0.5^8*16
2 heads in each: 0.5^8*36
3 heads in each: 0.5^8*16
4 heads: 0.5 ^8

Answer 4

which totals to 0.273, indeed

Answer 5

okay, what i punched in was the number of ways, you can get heads in the first four tosses, my bad

Answer 6

and i got the sample space wrong too, pellete

Answer 7

0 1 2 3 4 5 6 7 8
1 8 28 56 70 56 28 8 1

if my pascal is right ...

70 * .5^4 * .5^4 = 35/128 = .2734

Answer 8

henkjan, That doesn't add up to the answer... But for all those who want to know how to get it, I finally solved it on my own. First, by getting the binomial distributions and then timesing each distribution by 2, and then adding them together :)

Also, Amistre is right.

Answer 9

thanks :)

Answer 10

it does add up to the answer... I think you pressed the wrong button on your calculator..
total = 0.5^8 *(1+16+36+16+1) = 0.5^8*70 = 0.273

Answer 11

just to practice the latex :)
\[{n \choose r}\ p^r\ q^r\]
\[{8 \choose 4}\ .5^4\ .5^4\]

Answer 12

Hmmm. I'll have to double check that! Sorry..

Answer 13

It still doesn't add up...

Answer 14

ok, that's strange, it works here... I don't know how to help you further than..

Answer 15

It's Ok, I already have the answer :) Amistre has the correct answer, and I eventually found it myself. But thanks for trying anyway...