How many coins are there in total?

• $F$ be no. of fair coins
• $U$ be no. of unfair coins
• $N = F+U$

$\begin{aligned} Pr( i\text{s fair coin | 1 H and 1 T }) &= \cfrac{ Pr\text{(get fair coin and flip 1 H and flip 1 T)} }{ Pr \text{(get a coin at random and flip 1 H and flip 1 T) } }= \cfrac{1}{2} \\ \\ Pr\text{(get fair coin and flip 1 H and flip 1 T)} &= \cfrac{F}{N} \cdot \cfrac{2}{4} = \cfrac{1}{2} \cdot \cfrac{F}{N} \\ \\ Pr\text{(get a coin at random and flip 1 H and flip 1 T) } &= \cfrac{F}{N} \cdot \cfrac{2}{4} + \cfrac{U}{N} \cdot 2 \cdot \cfrac{2}{3} \cfrac{1}{3} = \cfrac{1}{2} \cdot \cfrac{F}{N}+ \cfrac{4}{9}\cdot \cfrac{U}{N} \\ \\ \implies \cfrac{F}{N} &= \cfrac{1}{2} \cdot \cfrac{F}{N} + \cfrac{U}{N} \cdot \cfrac{4}{9} \\ \implies \cfrac{1}{2} \cdot \cfrac{F}{N} &= \cfrac{4}{9} \cdot \cfrac{U}{N} \\ \implies 9F &=8U \end{aligned}$

$\therefore \text{ min. \boxed{ N = 17 } when F=8, U=9}$

Say there are "m" fair coins and "n" biased coins. Two events could have happened:

• drawing a fair coin and then flipping it twice to get heads and tails

• drawing a biased coin and then flipping it twice to get heads and tails

Both the two events have equal probability of $\frac{1}{2}$ : a fair coin of $\frac{1}{2}$ and a biased coin 1- $\frac{1}{2}$ = $\frac{1}{2}$ .

The probability of the event of drawing a fair coin and flipping heads and tails is $\frac{m}{n+m}$ $\times$ $\frac{1}{2}$ $\times$ $\frac{1}{2}$ .

The probability of the event of drawing a biased coin and flipping heads and tails is $\frac{n}{n+m}$ $\times$ $\frac{2}{3}$ $\times$ $\frac{1}{3}$ .

Since both the event probabilities are equal:

$\frac{1}{4}$ $\times$ $\frac{m}{n+m}$ = $\frac{2}{9}$ $\times$ $\frac{n}{n+m}$

so

$\frac{1}{4}$ $\times$ m = $\frac{2}{9}$ $\times$ n

or

$\frac{n}{m}$ = $\frac{9}{8}$

and the total number of coins is n + m, or 8 + 9 = $\boxed{17}$

We are given that there is some number of fair coins and some number of unfair coins in a box and that randomly drawing a coin and flipping it twice yields a T and a H. In addition, the probability that we selected a fair coin is $\frac{1}{2}$ . This problem is a perfect application of Bayes's Theorem.

We know that $P(\text{fair} | \text{heads and tails}) = \frac{1}{2}$ and, by application of Bayes's Theorem, $P(\text{fair} | \text{heads and tails}) = \frac{P(\text{fair and heads and tails})}{P(\text{heads and tails})}$ . $P(\text{fair and heads and tails}) = 2p(\frac{1}{2})^2$ where $p$ is the probability of selecting a fair coin and $P(\text{heads and tails}) = 2p(\frac{1}{2})^2 + 2(1 - p)(\frac{2}{3})(\frac{1}{3})$ . Thus, $\displaystyle \frac{1}{2}=\frac{2p\left(\frac{1}{2}\right)^{2}}{2p\left(\frac{1}{2}\right)^{2}+2\left(1-p\right)\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)}$ $\displaystyle \frac{1}{4}p+\frac{2}{9}\left(1-p\right)=\frac{1}{2}p$ $\displaystyle \frac{2}{9}=\frac{17}{36}p$ $p=\frac{8}{17}$

The value of $p$ we have found is the reduced ratio of fair coins to the total number of coins. The total number of coins could be any multiple of 17. But since the question asks what the minimum number of coins in the box is, we can simply conclude that the minimum is $\boxed{17}$