Find the Probability Distribution of 12 Coin Tosses (Conditional Probability)

in few words... given that "exactly" 2 H occurred, your 12-sized vector is this, for example

$$\{1,1,0,0,0,0,0,0,0,0,0,0\}$$

and all the possible 12-tuple with 10 "zero's" and 2 "one's"

any of these 66 possible tuples are equiprobable with probability

$$\frac{1}{\binom{12}{2}}=\frac{1}{66}$$

If you want you can better formalize this sketch using binomial conditional distribution...

$$\mathbb{P}[\mathbf{X}=\mathbf{x}|A]=\frac{\left(\frac{1}{2}\right)^{12}}{\binom{12}{2}\cdot\left(\frac{1}{2}\right)^{12}}=\frac{1}{\binom{12}{2}}=\frac{1}{66}$$

This because at the numerator you have the probability of one single specified combination you can observe

Examples

$$\mathbb{P}[\{1,1,0,0,0,0,0,0,0,0,0,0\}|A]=\frac{\left(\frac{1}{2}\right)^{12}}{\binom{12}{2}\cdot\left(\frac{1}{2}\right)^{12}}$$

$$\mathbb{P}[\{1,0,0,0,0,0,0,0,1,0,0,0\}|A]=\frac{\left(\frac{1}{2}\right)^{12}}{\binom{12}{2}\cdot\left(\frac{1}{2}\right)^{12}}$$

and so on...

...thus the conditional distribution is uniform all over its support