Suppose one coin with probability $0.4$ for heads, one with probability $0.6$ for heads, and one that is a fair coin are each tossed once. Find the pmf of the total number of heads obtained. Is it a Binomial distribution?

Here's what I am thinking:

$X=$ # of heads

When $X=0$

No head $(TTT)$ which means $0.6$ for the first one, $0.4$ for the second one and $0.5$ for the third one. Therefore, $P(X=0) = (0.6)(0.4)(0.5) = 0.12$

Similarly,it will be $P(X=3) = (0.6)(0.4)(0.5) = 0.12$

I am stuck when $X=1$ and $X=2$

When $X=1$ which means it will be $\left \{ HTT,THT,TTH \right \}$

Let's say the first coin always is the one with Heads. So, we have $3\*(0.4)(0.4)(0.5)$

And when the Head is from the second coin, we have $3\*(0.6)(0.6)(0.5)$

And when it's coin three, we have $3\*(0.6)(0.4)(0.5)$

So, $P(X=1)$ will be sum of all these. Same thing for $P(X=2)$

I am not seeing a binomial distribution because $P(X=1)$ is giving me after adding all the terms above $3\*(0.5)\left \{(0.4)^2 + (0.6)^2 + (0.6)(0.4) \right \}$

Am I doing it right?