Question

A letter is known to have come for 'RAVINDRA' or 'DEVINDRAN'. On the envelop just two consecutive letters 'RA' are visible. The probability that the letter has come for 'RAVINDRA' is:

• A. \(\frac{16}{23}\)
• B. \(\frac{7}{11}\)
• C. \(\frac{4}{11}\)
• D. \(\frac{7}{23}\)

Hint:

Use Bayes' theorem. \(P(RA|RAVINDRA)=\frac{2}{7}\) and \(P(RA|DEVINDRAN)=\frac{1}{8}\).

Answer 1

The Correct Option is A

Step 1: Name the events:
Let \(E_1\) be the event that the letter is for 'RAVINDRA' and \(E_2\) the event that it is for 'DEVINDRAN'. The letter came for one of the two names with no other information, so \(P(E_1)=P(E_2)=\frac{1}{2}\).
Let \(E\) be the event that two consecutive visible letters on the envelope are 'RA'. We need \(P(E_1|E)\), which is a job for Bayes' theorem.

Step 2: Count the pairs of consecutive letters:
'RAVINDRA' has 8 letters R-A-V-I-N-D-R-A, so it has 7 pairs of consecutive letters: RA, AV, VI, IN, ND, DR, RA. Two of these pairs are 'RA'. So \(P(E|E_1)=\frac{2}{7}\).
'DEVINDRAN' has 9 letters D-E-V-I-N-D-R-A-N, so it has 8 pairs: DE, EV, VI, IN, ND, DR, RA, AN. Only one pair is 'RA'. So \(P(E|E_2)=\frac{1}{8}\).

Step 3: Apply Bayes' theorem:
\[ P(E_1|E)=\frac{P(E_1)P(E|E_1)}{P(E_1)P(E|E_1)+P(E_2)P(E|E_2)} \]
\[ =\frac{\frac{1}{2}\cdot\frac{2}{7}}{\frac{1}{2}\cdot\frac{2}{7}+\frac{1}{2}\cdot\frac{1}{8}} \]

Step 4: Simplify:
The common factor \(\frac{1}{2}\) cancels: \[ \frac{\frac{2}{7}}{\frac{2}{7}+\frac{1}{8}}=\frac{\frac{16}{56}}{\frac{16}{56}+\frac{7}{56}}=\frac{16}{23} \]

Step 5: Why the other options fail:
\(\frac{7}{23}\) is the probability of the other name, 'DEVINDRAN'. It is the complement of our answer. Options \(\frac{7}{11}\) and \(\frac{4}{11}\) come from wrongly counting pairs. Only option 1 fits.

Final Answer:
The probability is \(\frac{16}{23}\), which is option 1. \[ \boxed{\frac{16}{23}} \]