Question

A box contains three coins: two regular coins and one fake two-headed coin (P(H)=1)
You pick a coin at random and toss it, and get heads. The probability that it is the two-headed coin = ___

Answer 1

Let’s first name our events so that we can use them easily.
Let A be the event coin 1 is selected
Let B be the event coin 2 is selected.
Let C be the event coin 3, that is the coin with both the faces head is selected.
Let H be the event that we get a head.
You pick a coin at random and toss it, and get heads. We want to know the probability that the coin picked up is C or the biased coin.
It can be represented as $P\left(\frac{C}{H}\right)$
Using the formula for conditional probability, we get this as
$P\left(\frac{C}{H}\right)=\frac{P(H\cap C)}{P\left(H\right)}$
P(H) is the probability of getting heads
We can get head in three different ways. We select A and get head or we select B and get head or we select C and get a head
$\Rightarrow P\left(H\right)=P\left(\frac{H}{A}\right)P\left(A\right)+P\left(\frac{H}{B}\right)P\left(B\right)+P\left(\frac{H}{C}\right)P\left(C\right)$
$=\frac{1}{2}\times \frac{1}{3}+\frac{1}{2}\times \frac{1}{3}+1\times \frac{1}{3}$
$=\frac{2}{3}----------\left(1\right)$
$P(H\cap C)$ can be calculated using the expression for finding
$P\left(\frac{H}{C}\right)P\left(\frac{H}{C}\right)=\frac{P(H\cap C)}{P\left(C\right)}$
$\Rightarrow P(H\cap C)=P\left(C\right)\times P\left(\frac{H}{C}\right)$
$P\left(\frac{H}{C}\right)$ is the probability of getting head given that we chose coin C. Since coin C is biased, we get
$P\left(\frac{H}{C}\right)=1$
$\Rightarrow P(H\cap C)=\frac{1}{3}\times 1$
$=\frac{1}{3}-----------\left(2\right)$
Using (1)and (2)we get $P\left(\frac{C}{H}\right)=\frac{P(H\cap C)}{P\left(H\right)}$
$=\frac{\frac{1}{3}}{\frac{2}{3}}$
$=\frac{1}{2}$