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) $TATANAGAR$

• A. a
• B. b
• C. c
• D. d

Answer 1

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