Question

A bag contains n+1 coins. It is known that one of these coins has heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that the toss results in heads is 7/12, find n

Answer 1

Let $E_1:$ a coin with two heads is selected
$E_2:$ a fair coin is selected
$A:$ the toss results in heads
Then, $P\left(E_1\right)=\frac{1}{n+1},P\left(E_2\right)=\frac{n}{n+1},P\left(\frac{A}{E_1}\right)=1$ and $P\left(\frac{A}{E_2}\right)=\frac{1}{2}$
$\therefore P\left(A\right)=P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)$
$\Rightarrow \frac{7}{12}=\frac{1}{n+1}.1+\frac{n}{n+1}.\frac{1}{2}\Rightarrow 12+6n=7n+7\Rightarrow n=5$