Question

A letter is known to have come either from TATANAGAR or CALCUTTA. On the envelop just two consecutive letter TA are visible. What is the probability that the letter has come from CALCUTTA?

Answer 1

Let $E_1$ be the event that the letter has come from $TATANAGAR$ and $E_2$ be the event that it has come from $CALCUTTA$

Let $A$ denote the event that the two consecutive letters on the envelope are $TA$

$\therefore E_1$ and $E_2$ are mutually and exhaustive events.

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

$P\left(A/E_1\right)=$Probability that the consecutive letters $TA$ visible on the envelope belong to $TATANAGAR$

$=\frac{2}{8}=\frac{1}{4}$ (There are $8$ consecutive letters namely $TA,AT,RA,AN,NA,AG,GA,AR$ out of which $2$ cases are favourable.)

Similarly,$P\left(A/E_2\right)=$Probability that the consecutive letters $TA$ visible on the envelope belong to $CALCUTTA$

$=\frac{1}{7}$ (There are $7$ consecutive letters namely $CA,AL,LC,CU,UT,TT,TA$ out of which $1$ case is favourable.)

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

$\frac{\frac{1}{2}\times \frac{1}{7}}{\frac{1}{2}\times \frac{1}{4}+\frac{1}{2}\times \frac{1}{7}}=\frac{1}{7}\times \frac{28}{11}=\frac{4}{11}$