Identity for the probability of a coin having an even number of heads

[koab1mjr]

Homework Statement

Given that n independent tosses having probability of p of coming up heads are made, show that an even number of heads results is 0.5(1+(q-p)^n) where q is 1-p, by proving the identity
Sigma from i=0 to n/2 of (n choose 2i) (p^2i)(q^(n-2i))=0.5(((p+q)^n)+(q-p)^n)

Homework Equations

The Attempt at a Solution

I expanded it, (n choose 0)p^0 q^n + (n choose 2)p^2 q^(n-2)+...letting i=n/2 we get (n choose n)p^n q (n-n) ends up being just p^n.

 

[tiny-tim]

Hi koab1mjr!

(have a sigma: ∑ and try using the X² and X₂ tags just above the Reply box )

Homework Statement

Given that n independent tosses having probability of p of coming up heads are made, show that an even number of heads results is 0.5(1+(q-p)^n) where q is 1-p, by proving the identity
Sigma from i=0 to n/2 of (n choose 2i) (p^2i)(q^(n-2i))=0.5(((p+q)^n)+(q-p)^n)

Hint: start at the end …

what is the coefficient of p^jq^n-j of 0.5(((p+q)ⁿ)+(q-p)ⁿ) for j even? and for j odd?

 

[koab1mjr]

thanks, but...
I tried it, but i am still not getting it, because starting from the end for even, i got p^2q^(n-2), p^4q^(n-4) and so on, plus i cannot get the factorials to work out...

 

[tiny-tim]

(just got up :zzz: …)

…for even, i got p^2q^(n-2), p^4q^(n-4) and so on …

(please use the X² tag just above the Reply box)

p²q^n-2 time what? p⁴q^n-4 times what? … p^2jq^n-2j times what?

 

[Mark44]

Couldn't resist: The probability of a coin having an even number of heads should be zero.

 

[koab1mjr]

timing each one by (n choose 2), (n choose 4), and so on until i hits n/2, and at that point all becomes just p^n
?

 

[tiny-tim]

Hi koab1mjr! :smile

timing each one by (n choose 2), (n choose 4), and so on until i hits n/2, and at that point all becomes just p^n
?

(you can write that ⁿC₂ etc )

Yup!

And that's exactly the left-hand-side of the original identity, isn't it?

 

[koab1mjr]

yeah, when you plug in i=n/2 on the he left side you get p^n, but when you expand the right side and at the end you do the exact same thing, but the problem is that the identity of (1/2)[(p+q)^n + (q-p)^n) doesn't show up at all??

 

[tiny-tim]

Given that n independent tosses having probability of p of coming up heads are made, show that an even number of heads results is 0.5(1+(q-p)^n) where q is 1-p, by proving the identity
Sigma from i=0 to n/2 of (n choose 2i) (p^2i)(q^(n-2i))=0.5(((p+q)^n)+(q-p)^n)

… the problem is that the identity of (1/2)[(p+q)^n + (q-p)^n) doesn't show up at all??

Sorry, I'm not following you at all.

 

[koab1mjr]

Hi Tiny Tim

I finally got it. I applied Newton's bionomial expansion to prove the identity. Thank you so much for your help. IT help get me thinking in the right direction. You are the best.