Finding probability using Var

Question

I'm stuck on the following question:

Suppose you toss three coins: two fair coins and a positively biased coin (more likely to show Heads than Tails). The variance of the total number of Heads in the three tosses is 0.55. What is the probability that the biased coin shows head?

To start off, I made the following expression: Var(2F+B) = 0.55

I then seperated and simplified to 2Var(F) + Var(B) = 0.55

I used the equation of Var for the binomial distribution (which is np(1-p)) to get

2(2*0.5)(1-0.5) + Var(B) = 0.55

so Var(B) = -0.45. I then used the same equation of Var again to get

p(1-p) = -0.45 but after expanding and solving the quadratic equation, I get 2 values for the probabilities that are less than 0 and greater than 1.

Can anyone tell me where I went wrong.

Answer 1

The main problem here is that you assume Var to be liniear, which it is not.

$Var(2F+B) = 2^2Var(F) + Var(B)$

Also note that the Variance can never be negative. Otherwise you are on a good path.

Answer 2

The problem here is that you are mistreating the distribution $F$. If $F$ is the distribution for the number of Heads resulting from tossing two fair coins then you want to write $$\text{Var}(F+B)=\text{Var}(F)+\text{Var}(B)$$ We get $\text{Var}(F)=.5$ and all is good. We get $$.5+\text{Var}(B)=.55$$ which makes perfect sense.

But in your first expression, you appear to want $F$ to be the distribution for a single coin. In that case you should have written $$\text{Var}(F_1+F_2+B)=2\text{Var}(F)+\text{Var}(B)$$

Where the $F_i$ are i.i.d. and both equal $F$. But in this case $\text{Var}(F)=.25$ and again we get $.5+\text{Var}(B)=.55$

In either case, you should not have written $2F$ as no variable is being multiplied by $2$.