If you flip three fair coins, what is the probability that you'll get two tails and one head in any order?

Question

I can't find a solution that doesn't involve listing out all the possible combinations. Is there a way that I can use combinations/permutations to easily calculate this by hand?

For example, when I'm solving "If you flip three fair coins, what is the probability that you'll get exactly two tails?" I can use combinations to find how many ways of getting exactly $2$ tails: $(\frac{3!}{2!\times1!})$. Then I can divide the answer by $8$, which is the total number of possible combinations $(2^3)$.

Answer 1

It's just $$\binom{3}{2}*\dfrac{1}{2^3}=\dfrac{3}{8}$$

Think of it this way.

You have 3 slots.

$\text{_ _ _}$

I want two of them to be tail out of 3 (hence the $\binom{3}{2}$), and the probability of $2$ tails and $1$ head is basically $\dfrac{1}{2^3}$, as there are $1/2$ probability of head/tail for each toss, and there are 3 tosses.

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Similarly I can choose 1 head out of 3, so that would make it $\binom{3}{1}$, and nothing else would change.

Answer 2

When you flip $3$ coins, you either get a head or a tail. So getting 2 tails and 1 heads is exactly the same event as getting exactly two tails. Thus, as you noted in the last part of your question, the answer is $\frac{3}{8}$.