Calculate Probability of Getting Even Number of Heads with Biased and Fair Coins Let's say that we have $5$ coins. Three of the coins are biased, with a $60$% chance of getting heads, and the other $2$ coins are normal, fair coins. What is the probability of getting an even number of heads when you flip all five coins.
My current understanding of the problem is that we have to calculate the probability of getting $0$ heads, getting $2$ heads, getting $4$ heads, and then adding those probabilities together.
I think I have the right solution to getting the probability of 0 heads, by multiplying $(\frac{1}{2})^2*(\frac{3}{5})^3$. I don't know how to get the probability for getting $2$ and $4$ heads however.
 A: The answer is $\frac{1}{2}$ because it is entirely determined by the last (fair) coin.
Consider $p$ the probability of an even number of heads in your first $4$ tosses. Hence $1-p$ is the probability you have an odd number in the first $4$ tosses.
To get an even number in all five tosses you either have an even number in the first $4$ and then toss a tail, $p \cdot \frac{1}{2}$ or you have an odd number and then toss a head, $ (1-p) \cdot \frac{1}{2}$
Hence the probability of an even number of heads in the five tosses is just $p \cdot \frac{1}{2} + (1-p) \cdot \frac{1}{2} = \frac{1}{2} $
A: You are correct about your general approach: calculate the sum of the probabilities $Pr(H=0)+Pr(H=2)+Pr(H=4)$
$Pr(H=0)=(\frac{2}{5})^3(\frac{1}{2})^2$
$Pr(H=2)=3[(\frac{3}{5})^2(\frac{2}{5})(\frac{1}{2})^2]+(\frac{2}{5})^3(\frac{1}{2})^2 + 6[(\frac{3}{5})(\frac{2}{5})^2(\frac{1}{2})^2]$
$Pr(H=4)=2[(\frac{3}{5})^3(\frac{1}{2})^2]+2[(\frac{3}{5})^2(\frac{2}{5})(\frac{1}{2})^2]$
Edit: I can't comment because I am new and I need over 50 points to do so. But, I want to point out that the answer from Oskar is incorrect.  The flipping of the coins is not ordered as far as the problem states, so the last coin flipped does not have to be unbiased.  If the biased coin is flipped last, the logic breaks down.  Sorry to be a debby downer because I love elegant arguments too.  You can use that argument with modification by considering the average of all 5 coins holding the last position.  Recognize the p's won't cancel for biased coins.
