Here's the question you clicked on:
order
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.
i hate these coin questions :(
The answer is 0.273... so not 15/64... but I'm not sure how to arrive at the answer.
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
which totals to 0.273, indeed
okay, what i punched in was the number of ways, you can get heads in the first four tosses, my bad
and i got the sample space wrong too, pellete
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
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.
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
just to practice the latex :) \[{n \choose r}\ p^r\ q^r\] \[{8 \choose 4}\ .5^4\ .5^4\]
Hmmm. I'll have to double check that! Sorry..
It still doesn't add up...
ok, that's strange, it works here... I don't know how to help you further than..
It's Ok, I already have the answer :) Amistre has the correct answer, and I eventually found it myself. But thanks for trying anyway...