Question

A letter is known to have come either from $LONDON$ or $CLIFTON$. On the envelop just two consecutive letters $ON$ are visible. What is the probability that the letter has come from
$\left(i\right)LONDON$
$\left(ii\right)CLIFTON$ ?

Answer 1

Let $E_1$ and $E_2$ be two events that the letter came from LONDON and CLIFTON respectively.

$P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$

Let $A$ be the event that two consecutive letters on the envelope are $ON$.

If $E_1$ occurs then the letter came from LONDON and if $E_2$ occurs then the letter came from CLIFTON.

Considering one of ON's as $1$.

$P\left(A/E_1\right)=\frac{2}{5}$

$P\left(A/E_2\right)=\frac{1}{6}$

Now,

Using Bay's theorem

$P\left(E_1/A\right)=\frac{P\left(E_1\right)\cdot P\left(A/E_1\right)}{P\left(E_1\right)\cdot P\left(A/E_1\right)+P\left(E_2\right)\cdot P\left(A/E_2\right)}$

$\Rightarrow P\left(E_1/A\right)=\frac{\frac{1}{2}\times \frac{2}{5}}{\frac{1}{2}\times \frac{2}{5}+\frac{1}{2}\times \frac{1}{6}}$

$\Rightarrow P\left(E_1/A\right)=\frac{12}{17}$

$P\left(E_2/A\right)=\frac{P\left(E_1\right)\cdot P\left(A/E_1\right)}{P\left(E_2\right)\cdot P\left(A/E_2\right)+P\left(E_2\right)\cdot P\left(A/E_2\right)}$

$\Rightarrow P\left(E_2/A\right)=\frac{\frac{1}{2}\times \frac{1}{6}}{\frac{1}{2}\times \frac{2}{5}+\frac{1}{2}\times \frac{1}{6}}=\frac{5}{17}$