Classic Probability

There are $n$ fair coins in the bag along with the one two-headed coin, so the probability of choosing a fair coin is $\frac{n}{n+1}$ and the probability of choosing the two-sided coin is $\frac{1}{n+1}$ .

If a fair coin is chosen then the probability that the toss results in a head is $\frac{1}{2}$ , and if the two-sided coin is chosen then the probability that the toss results in a head is $1$ . So given that the probability that the toss of a randomly chosen coin results in a head is $\frac{7}{12}$ , we have that

$(\frac{n}{n+1})*(\frac{1}{2}) + (\frac{1}{n+1})*1 = \frac{7}{12}$

$\Longrightarrow \dfrac{n + 2}{2(n + 1)} = \dfrac{7}{12}$

$\Longrightarrow 12n + 24 = 14n + 14 \Longrightarrow 10 = 2n \Longrightarrow n = \boxed{5}$ .

Mine is pure intuition and I guessed it correctly. The P(head) is 7/12, then the P(tail) is 5/12. So there must be 5 fair coins in order for 5 tail to occur.