Question

A letter is known to have come either from $TATANAGAR$ or $KOLKATA$. On the envelope, only the two consecutive letters $TA$ are visible. What is the probability that the letter has come from
(i) $KOLKATA$, (ii) $TATANAGAR$

	(i)	(ii)
(a)	$\frac{4}{5}\,\,\,\,$	$\frac{1}{5}$
(b)	$\frac{1}{5}$	$\frac{4}{5}$
(c)	$\frac{3}{5}$	$\frac{2}{5}$
(d)	$\frac{2}{5}$	$\frac{3}{5}$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (D)

Let $E_1$, $E_2$ and $A$ be the events defined as follows :
$E_1 =$ letter has come from $KOLKATA$,
$E_2 =$ letter has come from $TATANAGAR$ and
$A =$ two consecutive visible letters are $TA$.
Letter can come either from $KOLKATA$ or $TATANAGAR$, so
$P\left(E_{1}\right) = \frac{1}{2} = P\left(E_{2}\right)$
The word $KOLKATA$ has $7$ letters, so there are $6$ groups of two consecutive letters $— KO$, $OL$, $LK$, $KA$, $AT$, $TA$.
Only one of these is $'TA'$.
$\therefore P(A|E_1) = $ probability of event $A$ when $E_1$ has occurred i.e. when letter has come from $KOLKATA$
$= \frac{1}{6}$
The word $TATANAGAR$ has $9$ letters, so there are $8$ groups of two consecutive letters $— TA$, $AT$, $TA$, $AN$, $NA$, $AG$, $GA$, $AR$. Two out of these are $TA$'.
$\therefore P(A|E_2) = $ probability of event $A$ when $E_2$ has occurred i.e. when the letter has come from $TATANAGAR$
$= \frac{2}{8} = \frac{1}{4}$.
(i) We want to find $P(E_1|A)$.
By Bayes' theorem, we have

$P\left(E_{1}|A\right) = \frac{P\left(E_{1}\right)P\left(A |E_{1}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)}$
$=\frac{\frac{1}{2}\cdot\frac{1}{6}}{\frac{1}{2}\cdot\frac{1}{6}+\frac{1}{2}\cdot\frac{1}{4}} = \frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{4}}$
$= \frac{1}{6}\times\frac{12}{5}= \frac{2}{5}$
(ii) We want to find $P(E_2|A)$.
By Bayes' theorem, we have

$P\left(E_{2}|A\right) = \frac{P\left(E_{2}\right)P\left(A |E_{2}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)}$
$=\frac{\frac{1}{2}\cdot\frac{1}{4}}{\frac{1}{2}\cdot\frac{1}{6}+\frac{1}{2}\cdot\frac{1}{4}} = \frac{\frac{1}{4}}{\frac{1}{6}+\frac{1}{4}}$
$= \frac{1}{4}\times\frac{12}{5}= \frac{3}{5}$