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

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Question

A fair coin is tossed 12 times. Let A be the event - "exactly 2 tosses are head".

Find the probability distribution of the coin tossing vector $\left(X_{1},\ldots,X_{12}\right) $, given the even A has happened.

My Take

I'm looking to find the probability distribution of the following expression: $$ P\left(X_{1},\ldots,X_{12}=\left(k_{1},\ldots,k_{12}\right)\left|A\right.\right) $$ Which is: $$ \frac{P\left(X_{1},\ldots,X_{12}=\left(k_{1},\ldots,k_{12}\right)\cap A\right)}{P\left(A\right)} $$ I have used binomial distribution to find that: $$ P\left(A\right)={12 \choose 2}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{10}=66\cdot\left(\frac{1}{2}\right)^{12} $$ but I'm confused with finding $P\left(X_{1},\ldots,X_{12}=\left(k_{1},\ldots,k_{12}\right)\cap A\right)$.

The Book's Answer

The book's answer is $\frac{1}{66}$.


Judging by the book's answer it holds that: $$ P\left(X_{1},\ldots,X_{12}=\left(k_{1},\ldots,k_{12}\right)\cap A\right)=\left(\frac{1}{2}\right)^{12} $$

but I can't tell why.

This probability and $P\left(A\right)$ seems to me like the same one.

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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