Probability of even number of successes in a series of independent trials

Question

Consider a series of independent trials at each of which there is a success of a failure with probabilities $p$ and $1-p$ respectively. I am finding it difficult to derive the probability of even number of successes occurring at the nth trial. Any assistance will be much appreciated. Thanks

Answer 1

Use induction. If $P_n$ is the probability of even number of successes in n trials, then

$$ P_n = p(1-P_{n-1}) + (1-p)P_{n-1}$$

This results in

$$ P_n = \frac{1+(1-2p)^n}{2}$$

Answer 2

Since this follows a binomial distribution, your chance of a particular number of successes $s$ is $$\binom{n}{s}p^s(1-p)^{n-s}$$

So you could take $$\sum_{s=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2s}p^{2s}(1-p)^{n-2s}$$

Answer 3

The generating function for the probability of success in $n$ trials is $$f(x) = (1-p+px)^n$$ The generating function for the cases where the number of successes is even is $$\frac{f(x) + f(-x)}{2}$$ To find the total probability of success in an even number of trials, just set $x=1$ in the generating function: $$\frac{f(1) + f(-1)}{2} = \frac{1 + (1-2p)^n}{2}$$