Answer

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**Hint:**Before solving this question, we should know about concept of conditional probability which is as follows

If A and B are any events in S then the conditional probability of B relative to A is given by

\[P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P(B)}\]

Also, it is important to know the concept of Bayer’s theorem as follows

Let \[{{A}_{1}},{{A}_{2}},{{A}_{3}},....{{A}_{n}}\] be n mutually exclusive and exhaustive events of the sample space S and A is the event which can occur with any of the events then

\[P\left( \dfrac{{{A}_{i}}}{A} \right)=\dfrac{P\left( {{A}_{i}} \right)\cdot P\left( \dfrac{A}{{{A}_{i}}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{A}_{i}} \right)\cdot P\left( \dfrac{A}{{{A}_{i}}} \right)}}\]

This formula gives the required probability.

Now, using the above mentioned Bayes theorem, we will solve this question as well.

**Complete step by step answer:**

> Let \[({{E}_{1}})\] denote the event that the letter came from TATANAGAR and \[({{E}_{2}})\] denote the event that the letter came from CALCUTTA.

> Let A be the event that the two consecutive alphabets visible on the envelope are TA. We have to find the probability that the letter came from TATANAGAR given the letter has TA in it, that is, \[P\left( \dfrac{{{E}_{1}}}{A} \right)\].

> As we know that the letter has come from either Calcutta or Tatanagar, we can write the following,

\[~P\left( {{E}_{1}} \right)=\dfrac{1}{2},P\left( {{E}_{2}} \right)=\dfrac{1}{2}\ \ \ \ \ ...(a)\]

Now, if \[({{E}_{1}})\] has taken place then the letter has come from TATANAGAR.

Now, in the word TATANAGAR, there are 8 pairs of consecutive alphabets, that is, {TA, AT, TA, AN, NA, AG, GA, AR}. TA occurs two times in the above mentioned cases.

We know that, if A and B are any events in S then the conditional probability of B given A is given by,

\[P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P(B)}\]

\[~P\left( {{E}_{1}} \right)=\dfrac{1}{2},P\left( {{E}_{2}} \right)=\dfrac{1}{2}\ \ \ \ \ ...(a)\]

Now, if \[({{E}_{1}})\] has taken place then the letter has come from TATANAGAR.

Now, in the word TATANAGAR, there are 8 pairs of consecutive alphabets, that is, {TA, AT, TA, AN, NA, AG, GA, AR}. TA occurs two times in the above mentioned cases.

We know that, if A and B are any events in S then the conditional probability of B given A is given by,

\[P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P(B)}\]

Hence, the probability of A given \[({{E}_{1}})\] is,

\[\begin{align}

& P\left( A/(E_1) \right)=\dfrac{P(A\bigcap (E_1))}{P(E_1)} \\

&\Rightarrow P\left( A/(E_1) \right)=\dfrac{2}{8}\ =\ \dfrac{1}{4}\ \ \ \ \ ...(b) \\

\end{align}\]

Now, if \[({{E}_{2}})\] has occurred then the letter has come from CALCUTTA.

Now, in the word CALCUTTA,

There are 7 pairs of consecutive alphabets, that is, {CA, AL, LC, CU, UT, TT, TA}

And TA occurs one time in the above mentioned cases.

We know that, if A and B are any events in S then the conditional probability of B given A is given by,

\[P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P(B)}\]

\[\begin{align}

& P\left( A/(E_1) \right)=\dfrac{P(A\bigcap (E_1))}{P(E_1)} \\

&\Rightarrow P\left( A/(E_1) \right)=\dfrac{2}{8}\ =\ \dfrac{1}{4}\ \ \ \ \ ...(b) \\

\end{align}\]

Now, if \[({{E}_{2}})\] has occurred then the letter has come from CALCUTTA.

Now, in the word CALCUTTA,

There are 7 pairs of consecutive alphabets, that is, {CA, AL, LC, CU, UT, TT, TA}

And TA occurs one time in the above mentioned cases.

We know that, if A and B are any events in S then the conditional probability of B given A is given by,

\[P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P(B)}\]

Hence, the probability of A given \[({{E}_{2}})\] is,

\[\begin{align}

&\Rightarrow P\left( A/(E2) \right)=\dfrac{P(A\bigcap (E2))}{P(E2)} \\

&\Rightarrow P\left( A/(E2) \right)=\dfrac{1}{7}\ \ \ \ \ ...(c) \\

\end{align}\]

We have to find the probability when the letter came from TATANAGAR given the letter has TA in it, that is, \[P\left( \dfrac{{{E}_{1}}}{A} \right)\].

Using Bayes theorem,

\[\Rightarrow P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{P\left( E_1 \right)P\left( \dfrac{A}{{{E}_{1}}} \right)}{P\left( E_1 \right)P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( E2 \right)P\left( \dfrac{A}{{{E}_{2}}} \right)}\]

\[\begin{align}

&\Rightarrow P\left( A/(E2) \right)=\dfrac{P(A\bigcap (E2))}{P(E2)} \\

&\Rightarrow P\left( A/(E2) \right)=\dfrac{1}{7}\ \ \ \ \ ...(c) \\

\end{align}\]

We have to find the probability when the letter came from TATANAGAR given the letter has TA in it, that is, \[P\left( \dfrac{{{E}_{1}}}{A} \right)\].

Using Bayes theorem,

\[\Rightarrow P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{P\left( E_1 \right)P\left( \dfrac{A}{{{E}_{1}}} \right)}{P\left( E_1 \right)P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( E2 \right)P\left( \dfrac{A}{{{E}_{2}}} \right)}\]

We have,

\[\Rightarrow P\left( \dfrac{A}{{{E}_{1}}} \right)=\dfrac{1}{4}\]

\[\Rightarrow P\left( \dfrac{A}{{{E}_{2}}} \right)=\dfrac{1}{7}\]

\[\Rightarrow [~P\left( {{E}_{1}} \right)=\dfrac{1}{2},P\left( {{E}_{2}} \right)=\dfrac{1}{2}\]

\[\therefore P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{\dfrac{1}{2}\times \dfrac{1}{4}}{\dfrac{1}{2}\times \dfrac{1}{4}+\dfrac{1}{2}\times \dfrac{1}{7}}\]

\[=\dfrac{\dfrac{1}{8}}{\dfrac{1}{8}+\dfrac{1}{14}}\]

\[\therefore P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{7}{11}\]

**So, the correct answer is “Option A”.**

**Note:**

> If A and B are any events, then \[P\left( {}^{A}/{}_{B} \right)\] is read as ‘Probability of A given B’. This represents the probability of A given that B has already occurred.

> The same question can be asked for any number of events. Another place with ‘TA’ in it could have been introduced, in which case this becomes ${{E}_{3}}$. Now all the three events ${{E}_{1}}$,${{E}_{2}}$ and ${{E}_{3}}$ will have equal probability $\left( =\dfrac{1}{3} \right)$ . Nevertheless, the method remains the same.

> The same question can be asked for any number of events. Another place with ‘TA’ in it could have been introduced, in which case this becomes ${{E}_{3}}$. Now all the three events ${{E}_{1}}$,${{E}_{2}}$ and ${{E}_{3}}$ will have equal probability $\left( =\dfrac{1}{3} \right)$ . Nevertheless, the method remains the same.

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