Answer
418.5k+ views
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, \[P\left( \dfrac{{{E}_{1}}}{A} \right)\].
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 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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)