Confusion in probability (Bernoulli trials and binomial distribution)

Question

I am having a problem solving the following question-

"A coin is biased so that it has 40% chances of landing on heads. If it is thrown 3 times find the probability of getting 2 heads and a tail."

My first attempt was as follows-

I chose p=probability of getting head=2/5

& q=probability of getting tails=3/5

now for 2 H and 1 T probability=p^2*q=12/125 and percentage comes out to be around 9.6%....

BUT

I got a second thought of solving it using the Binomial distribution formula which is nCr (p)^r (q)^(n-r)

so it this case n=3, r=2 and we get an extra 3 multiplied in the numerator giving us the probability to be 36/125 !!

Now this has got me confused....Please Guide me on where i am wrong and when i should use these 2 methods?

Answer 1

You could get HHT, HTH, or THH.

For HHT, the probability is exactly how you calculated it. For the other two you get exactly the same probability.

This is actually where the extra factor of 3 comes from.