Question

A letter is known to have come from “TATA NAGAR” or from “CALCUTTA”. On the envelope, just two consecutive letters TA are visible. What is the probability that the letter came from TATA NAGAR?

Answer 1

Hint: To solve this question, we need to use the Bayes theorem. We have to find the conditional probability of the occurrence of consecutive words “TA” separately for the cases that the letter came from the first and the second place.

Complete step-by-step answer:
Let ${E_1}$ and ${E_2}$ be the respective events corresponding to the event that the letter came from TATA NAGAR, and the event that the letter came from CALCUTTA.
Also, let us assume that $A$ denotes the event that the consecutive letters which appear on the envelope are TA.
Now, since ${E_1}$ and ${E_2}$ are the mutually exclusive events, so we have
\[P\left( {{E_1}} \right) + P\left( {{E_2}} \right) = 1\]
Also, since the letter is came from either of the two places TATANAGAR or CALCUTTA, so we have
\[P\left( {{E_1}} \right) = \dfrac{1}{2}\] (1)
\[P\left( {{E_2}} \right) = \dfrac{1}{2}\] (2)
Now, let us assume that the event ${E_1}$ has occurred, that is, the letter has come from TATA NAGAR. In this word, a total of $10$ words are there (including the space between A and N). The list of two consecutive letters which can be present in this word is
$\left\{ {TA,AT,TA,A{\text{ }},{\text{ }}N,NA,AG,GA,AR} \right\}$
So we can see that there are a total of $9$ combinations there, out of which $2$ combinations correspond to the combination TA.
So the probability of occurrence of the event $A$ is
\[P\left( {A/{E_1}} \right) = \dfrac{2}{9}\] (3)
Now, we assume that the event ${E_2}$ has occurred, that is, the letter has come from CALCUTTA. In this word, a total of $8$ words are there. The list of two consecutive letters which can be present in this word is
$\left\{ {CA,AL,LC,CU,UT,TT,TA} \right\}$
So we can see that there are a total of $7$ combinations there, out of which $1$ combinations correspond to the combination TA.
So the probability of occurrence of the event $A$ is
\[P\left( {A/{E_2}} \right) = \dfrac{1}{7}\] (4)
Now, the probability that the letter came from Calcutta, provided that TA letters are visible is given by $P\left( {{E_1}/A} \right)$.
From the base theorem we can write it as
$P\left( {{E_1}/A} \right) = \dfrac{{P\left( {A/{E_1}} \right)P\left( {{E_1}} \right)}}{{P\left( {A/{E_1}} \right)P\left( {{E_1}} \right) + P\left( {A/{E_2}} \right)P\left( {{E_2}} \right)}}$
From (1), (2), (3) and (4) we have
$P\left( {{E_1}/A} \right) = \dfrac{{2/9 \times 1/2}}{{2/9 \times 1/2 + 1/7 \times 1/2}}$
Dividing the numerator and denominator by $1/2$ we get
\[P\left( {{E_1}/A} \right) = \dfrac{{2/9}}{{2/9 + 1/7}}\]
 \[ \Rightarrow P\left( {{E_1}/A} \right) = \dfrac{{2/9}}{{23/63}}\]
On solving we get
\[P\left( {{E_1}/A} \right) = \dfrac{{14}}{{23}}\]
Hence, the required probability is equal to \[\dfrac{{14}}{{23}}\].

Note:
We should not ignore the space present in the word “TATA NAGAR”. Ignoring it will give us a different answer.