Question

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

• A. $\frac{7}{11}$
  
• B. $\frac{24}{29}$
• C. $\frac{4}{11}$
• D. $\frac{1}{11}$

Answer 1

Answer: (A)

$\frac{7}{11}$

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.

Since the letters have come from either Calcutta or Tatanagar.

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

If $E_1$ has occurred then the letter has come from TATANAGAR. In the word TATANAGAR there are $8$ consecutive alphabets $i.e.,\{TA,AT,TA,AN,NA,AG,GA,AR\}$ and TA occurs two times.

$\therefore P\left(A/E_1\right)=\frac{2}{8}=\frac{1}{4}$

If $E_2$ has occurred then the letter has come from CALCUTTA. In the word CALCUTTA there are $7$ consecutive alphabets $i.e.,\{CA,AL,LC,CU,UT,TT,TA\}$ and TA occurs once.

$\therefore P\left(A/E_2\right)=\frac{1}{7}$

The probability that the letter has come from TATANAGAR is $P\left(E_1/A\right)$

By Bayes theorem, $P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)}$

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

$=\frac{\frac{1}{8}}{\frac{1}{8}+\frac{1}{14}}$

$\therefore P\left(E_1/A\right)=\frac{7}{11}$

Hence the probability that the letter has come from TATANAGAR is $\frac{7}{11}$.