Question

Consider four coins labelled as $1,2,3$ and $4.$ Suppose that the probability of obtaining a 'head' in a single toss of the $i^{th}$ coin is $\frac{i}{4},i=1,2,3,4.$ A coin is chosen uniformly at random and flipped. If it is given that the flip resulted in a 'head', and the conditional probability that the coin was labelled either $1$ or $2,$ is $p,$ then the value of $20p$ is

Answer 1

Let $C_i$ be the event that coin $C_i$ is chosen.
Let $H$ be the event that the flip shows head.
Given $P\left(H|C_i\right)=\frac{i}{4}$
We have to find $P\left(\right(C_1\cup C_2\left)|H\right)$.
By Bayes' theorem,
$P\left(\right(C_1\cup C_2\left)|H\right)$
$=\frac{P\left(\right(C_1\cup C_2)\cap H)}{P\left(H\right)}$

$P((C_1\cup C_2)\cap H)$
$=P((C_1\cap H)\cup (C_2\cap H))$
$=P(C_1\cap H)+P(C_2\cap H)$
$=P\left(C_1\right)\cdot P\left(H|C_1\right)+P\left(C_2\right)\cdot P\left(H|C_2\right)$
$=\frac{1}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{2}{4}=\frac{3}{16}$

$P\left(H\right)=\sum _{i=1}^{4}P\left(C_i\right)\cdot P\left(H|C_i\right)$
$=\frac{1}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{2}{4}+\frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{4}{4}=\frac{10}{16}$

$\therefore p=\frac{3/16}{10/16}=\frac{3}{10}$
Hence, $20p=6$