Question

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

Answer 1

Hint: We will first assume $E_1$ as the event that letter comes from TATANAGAR and $E_2$ as the event that the letter comes from CALCUTTA. We have only two events, so $P\left(E_1\right)=P\left(E_2\right)={\displaystyle \frac{1}{2}}$ . We will then find $P\left(E_2/A\right)$ by applying the formula of Bayes Theorem, that is, $$P\left(E_2/A\right)={\displaystyle \frac{P\left(E_2\right)P\left(A/E_2\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)}}$$.

Formula used: $$P\left(E_2/A\right)={\displaystyle \frac{P\left(E_2\right)P\left(A/E_2\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)}}$$.

Complete step by step answer:
It is given in the question that a letter is known to have come either from TATANAGAR or CALCUTTA. On the envelope just two consecutive letters TA are visible. And we have been asked to find the probability that the letter has come CALCUTTA.
We will first assume that $E_1$ is the event that letter comes from TATANAGAR and $E_2$ is the event that the letter comes from CALCUTTA.
So, there are only two possible events here, that is, the letter either comes from TATANAGAR or CALCUTTA. So, we will get the probability of occurrence of the event $E_1$ , that is, $$P\left(E_1\right)$$ as ${\displaystyle \frac{1}{2}}$ and also the probability of occurrence of the event $E_2$ , that is, $$P\left(E_2\right)$$ as ${\displaystyle \frac{1}{2}}$ . So, we can write,
 $P\left(E_1\right)=P\left(E_2\right)={\displaystyle \frac{1}{2}}$
Let us assume A as the event that the two consecutive letters visible on the envelope are TA.
Now, we will consider the case that the letter is from TATANAGAR, that is, the event $E_1$ . So, the possible consecutive letters from the word TATANAGAR are,
{TA, AT, TA, AN, NA, AG, GA, AR}
So, out of 8 possible consecutive letters from the word TATANAGAR, ‘TA’ is repeated twice. So, the probability that ‘TA’ is from TATANAGAR will be,
 $P\left(A/E_1\right)={\displaystyle \frac{2}{8}}={\displaystyle \frac{1}{4}}$
Where $P\left(A/E_1\right)$ denotes probability of A given $E_1$ , that is the probability of occurrence of the consecutive letters ‘TA’ on the envelope, given the letter is from TATANAGAR.
Now, we will assume that the letter is from CALCUTTA. So, the possible consecutive letters from the word CALCUTTA are,
{CA, AL, LC, CU, UT, TT, TA}
So, out of 7 possible consecutive letters from the word CALCUTTA, ‘TA’ is repeating once. So, the probability that ‘TA’ is from CALCUTTA will be,
 $P\left(A/E_2\right)={\displaystyle \frac{1}{7}}$
Where $P\left(A/E_2\right)$ denotes probability of A given $E_2$ , that is the probability of occurrence of the consecutive letters ‘TA’ on the envelope, given the letter is from CALCUTTA.
As we have to find the probability that the letter has come from CALCUTTA, that is, $P\left(E_2/A\right)$ we will use the formula of Bayes Theorem, that is, $P\left(A/B\right)={\displaystyle \frac{P\left(A\right)P\left(B/A\right)}{P\left(B\right)}}$ , where $P\left(A\right),P\left(B\right)$ are the probabilities of occurrence of only A and only B respectively and $P\left(A/B\right),P\left(B/A\right)$ are the probability of A given B and the probability of B given A respectively.
On applying the Bayes theorem, we will get,
$$P\left(E_2/A\right)={\displaystyle \frac{P\left(E_2\right)P\left(A/E_2\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)}}$$
So, we have, $P\left(E_1\right)={\displaystyle \frac{1}{2}},P\left(E_2\right)={\displaystyle \frac{1}{2}},P\left(A/E_1\right)={\displaystyle \frac{1}{4}},P\left(A/E_2\right)={\displaystyle \frac{1}{7}}$ . On substituting these values in the above equation, we will get,
$$P\left(E_2/A\right)={\displaystyle \frac{{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{7}}}{{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{7}}}}$$
On taking ${\displaystyle \frac{1}{2}}$ common from the numerator and the denominator and on cancelling it, we get,
$$P\left(E_2/A\right)={\displaystyle \frac{{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{7}}\right)}{{\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{7}})}}={\displaystyle \frac{{\displaystyle \frac{1}{7}}}{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{7}}}}$$
On taking the LCM of the denominator, we get,
$$P\left(E_2/A\right)={\displaystyle \frac{{\displaystyle \frac{1}{7}}}{{\displaystyle \frac{7+4}{28}}}}$$
On further simplifying, we get,
$$\begin{array}{rl}& P\left(E_2/A\right)={\displaystyle \frac{{\displaystyle \frac{1}{7}}}{{\displaystyle \frac{11}{28}}}}\\ & P\left(E_2/A\right)={\displaystyle \frac{1}{7}}\times {\displaystyle \frac{28}{11}}={\displaystyle \frac{4}{11}}\end{array}$$
Therefore, the probability that the letter has come from CALCUTTA is ${\displaystyle \frac{4}{11}}$ .

Note: It is observed that many students make mistakes while writing the consecutive letters of TATANAGAR and CALCUTTA. They may write the consecutive letters of TATANAGAR as {TA, TA, NA, GA, R}, but this is wrong and this will result in getting the wrong answers. Also, the students must remember the Bayes theorem formula, that is, $P\left(A/B\right)={\displaystyle \frac{P\left(A\right)P\left(B/A\right)}{P\left(B\right)}}$ in order to solve these types of questions.