# What is the probability of a symmetric coin fall

The prompt: We have 10 coins and 1 of them is non-symmetric (with probability of head equal $\frac{1}{3}$). We toss a randomly selected coin 6 times, and obtain 3 tails. What is the probability that we tossed a symmetrical coin?

The way I went on about solving the problem was first assuming it to be a win or success if we obtain a Tails, assuming that NS = Non symmetric coin, S = Symmetric coin, H = heads, T = tails $$P(S) = \frac{9}{10}; \text{ } P(NS) = \frac{1}{10}; \text{ } P(T|S) = \frac{1}{2}; \text{ }P(T|NS) = \frac{1}{3}$$

Now Probability of getting a Tail can be obtained using total probability theorem, $$P(T) = P(T|S)\cdot P(S) + P(T|NS) \cdot P(NS)$$ $$P(T) = \frac{1}{2}\cdot \frac{9}{10} + \frac{1}{3} \cdot \frac{1}{10} = \frac{29}{60}$$

Using Bernoulli's trials, where winning/success(p) is defined as getting a tails, given by $P(T) = \frac{29}{60}$ and failure(q) is defined by getting a head, $P(H) = \frac{31}{60}$ and n = number of tosses.

$$P(X = x) = {n \choose x} (p)^n \cdot (q)^{n - x}$$ $$P(X = 3) = {6 \choose 3} (\frac{29}{60})^6 \cdot (\frac{31}{60})^3$$ Assuming this to be an event E, we now have $P(E) = {6 \choose 3} (\frac{29}{60})^6 \cdot (\frac{31}{60})^3$, The desired result is $P(S|E)$, I'm not sure how to go on about finding that, any hint would be much appreciated.

The probability of getting exactly 3 tails out of 6 tosses is:

a) ${6\choose 3}\left(\frac{1}{2}\right)^6$ in the case of symmetric coin.

b) ${6\choose 3}\left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^3$ in the case of the asymmetric coin.

Then, if you picked your initial coin uniformly at random (probability $1/10$ for each coin), the probability that you picked a symmentric coin is:

$\frac{9\cdot\frac{1}{10}\cdot {6\choose 3}\left(\frac{1}{2}\right)^6} {9\cdot\frac{1}{10}\cdot {6\choose 3}\left(\frac{1}{2}\right)^6 + \frac{1}{10}{6\choose 3}\left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^3} = \frac{9\cdot\left(\frac{1}{2}\right)^6}{9\cdot\left(\frac{1}{2}\right)^6 + \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^3} =\frac{6561}{6561+512}=\frac{6561}{7073}\approx 92.76\%$

Sanity check: it does make sense that it's slightly better than $90\%$ (the a priori probability) since $3$ heads and $3$ tails is somewhat more in line with what you'd expect from a symmetric coin than from an asymmetric one.

Borrowing your notation, the value we want is: $P(S \mid E) = \frac{P(E \mid S)~P(S)}{P(E)}$, which is Bayes' rule.

Note that: $$P(E \mid S) = {6 \choose 3} \left( \frac{1}{2} \right)^6 = 0.3125$$ $$P(E \mid NS) = {6 \choose 3} \left( \frac{1}{3} \right)^3 \left(\frac{2}{3} \right)^3=0.219$$

Thus, $P(E) = \sum_i P(E \mid C_i)P(C_i)=\frac{1}{10} \cdot 0.3125 + \frac{9}{10} \cdot 0.219 = 0.0566$. Plug these numbers into the quotient above to get the answer.