need help with this nice math problem

[jota]

A coin with p{h}=p-1 is tossed n times. Show that the probability that the number of heads is even equals 0.5[1+(q-p)^n]

I appreciatte your any of your valuable contributions.
Thanks

 

[the_jackal]

Does p{h} mean p{Heads}? What is q?

 

[jota]

p-heads
q-probability of not happening hats-(tails)

thanks

Added after 4 hours 55 minutes:

corection to the problem

A coin with p{h}=p=1-q is tossed n times. Show that the probability that the number of heads is even equals 0.5[1+(q-p)^n].

 

[steve10]

another thinking ...

 

[jota]

I found a hint for this problem. The hint is saying that we should write down the binomial theorem expansions of (q + p)^n and (q – p)^n, then add them together.
This would eventually result in expected solution-- 0.5[1+(q - p)^n]. The only problem is, I can not get to this expected solution. I need you help.

 

[steve10]

That's brilliant! Yes, the sum of all the even terms is just ((q+p)^n + (q-p)^n)/2, which is exactly your answer.

 

[jota]

where do you get this 1 from? 0.5[1+(q-p)^n]. [/b]

 

[steve10]

p+q=1. Isn't it?

 

[jota]

ooops!!!

 

[freewilly30]

p+q sure equals one as it conveys all the probabilities