There is a red coin for which P(Heads) = 0.4, a green coin for which P(Heads) = 0.5 and a yellow coin for which P(Heads) = 0.6.
Let N be the number of heads in 300 tosses. Before each toss, I choose 1 of the 3 coins at random (each coin is equally likely to be selected).
What is the mean and variance of N?
I model N as follows: $$N = T_1 + T_2 + \cdots + T_{300}$$
$$\mathbb{E}[T_i] = \frac{0.4 + 0.5 + 0.6}{3} = 0.5$$
$$\begin{eqnarray}
Var(T_i) &=& \frac{1}{3}(0.4-\mathbb{E}[T_i])^2 + \frac{1}{3}(0.5-0.5)^2 + \frac{1}{3}(0.6-0.5)^2
\\ &=& \frac{(0.4-0.5)^2 + (0.5-0.5)^2 + (0.6-0.5)^2}{3}
\\ &=& \frac{0.02}{3}
\\ &=& \frac{1}{150}
\end{eqnarray}$$
By linearity of expectations and making use of the fact that all the Ts are iid,
$$\begin{eqnarray}
\mathbb{E}[N] &=& \mathbb{E}[T_1] + \mathbb{E}[T_2] + \cdots + \mathbb{E}[T_{300}]
\\ &=& 300 \cdot \mathbb{E}[T_1]
\\ &=& 150
\end{eqnarray}$$
Similarly, since the Ts are independent,
$$\begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + \cdots + T_{300})
\\ &=& Var(T_1) + Var(T_2) + \cdots + Var(T_{300})
\\ &=& 300 \cdot Var(T_1)
\\ &=& 300 \cdot \frac{1}{150}
\\ &=& 2
\end{eqnarray}$$
I know that my calculation of the mean is correct. The variance is incorrect, but I haven't been able to figure out why.
Best Answer
The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.
By the law of total variance, $$\begin{eqnarray} Var(T_i) &=& \mathbb{E}[Var(T_i|coin)] + Var(\mathbb{E}[T_i|coin]) \\ &=& \mathbb{E}[P(1-P)] + Var(P) \\ &=& \mathbb{E}[P] - \mathbb{E}[P^2] + \mathbb{E}[P^2] - (\mathbb{E}[P])^2 \\ &=& 0.5 - 0.5^2 \\ &=& 0.25 \end{eqnarray}$$
$$\begin{eqnarray} Var(N) &=& Var(T_1 + T_2 + \cdots + T_{300}) \\ &=& 300 \cdot Var(T_i) \\ &=& 300 \cdot 0.25 \\ &=& 75 \end{eqnarray}$$
Still don't understand why my first approach is wrong though.