# proof of the identity for a coin having even number of heads

• Nov 24th 2009, 08:25 AM
libragirl79
proof of the identity for a coin having even number of heads
Suppose that n independent tosses of a coin having probability p of coming up heads are made. Show that the probability that an even number of heads results is (1/2)(1+(q-p)^n) where q=1-p. Do this by utilizing the identity

sigma from i=0 to n/2 of (n choose 2i) p^2i q^(n-2i)=(1/2)[(p=q)^n+(q-p)^n]
where n/2 is the largest integer less than or equal to n/2.

I tried expanding it, but every time i end up with p^n only.

Thanks so much!!
• Nov 26th 2009, 08:23 AM
awkward
Quote:

Originally Posted by libragirl79
Suppose that n independent tosses of a coin having probability p of coming up heads are made. Show that the probability that an even number of heads results is (1/2)(1+(q-p)^n) where q=1-p. Do this by utilizing the identity

sigma from i=0 to n/2 of (n choose 2i) p^2i q^(n-2i)=(1/2)[(p=q)^n+(q-p)^n]
where n/2 is the largest integer less than or equal to n/2.

I tried expanding it, but every time i end up with p^n only.

Thanks so much!!

You want to show
$\sum_{i=0}^{\lfloor n/2 \rfloor} \binom{n}{2i} p^{2i} q^{n-2i} = (1/2) \; [(q+p)^n + (q-p)^n]$.

Expand $(p+q)^n$ and $(p-q)^n$ by the Binomial Theorem.

Add the two equations and observe that half the terms in the expansions cancel.