Question

A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:

• A. $\frac{7}{10}$
• B. $\frac{10}{17}$
• C. $\frac{12}{19}$
• D. $\frac{7}{19}$

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This problem is an application of conditional probability, specifically Bayes' Theorem. We have two possible events (the letter is from KANPUR or ANANTPUR), and we are given new evidence (the visible letters are 'AN'). We need to update our belief about the origin of the letter based on this evidence.
Step 2: Key Formula or Approach:
We will use Bayes' Theorem. Let's define the events:
$E_K$: The letter is from KANPUR.
$E_A$: The letter is from ANANTPUR.
$V$: The event that the two visible consecutive letters are 'AN'.
We want to calculate $P(E_A | V)$.
Bayes' Theorem is given by:
\[ P(E_A | V) = \frac{P(V | E_A) P(E_A)}{P(V | E_K) P(E_K) + P(V | E_A) P(E_A)} \] Step 3: Detailed Explanation:
Prior Probabilities: Since the letter is from one of two places with no other information, we assume each origin is equally likely.
\[ P(E_K) = P(E_A) = \frac{1}{2} \] Likelihoods: Now we calculate the probability of the evidence occurring given each possible origin.
{If the letter is from KANPUR ($E_K$):}
The word is KANPUR. The possible pairs of two consecutive letters are: KA, AN, NP, PU, UR.
Total number of such pairs = 5.
The pair 'AN' appears 1 time.
So, the probability of seeing 'AN' given it's from KANPUR is:
\[ P(V | E_K) = \frac{1}{5} \] {If the letter is from ANANTPUR ($E_A$):}
The word is ANANTPUR. The possible pairs of two consecutive letters are: AN, NA, AN, NT, TP, PU, UR.
Total number of such pairs = 7.
The pair 'AN' appears 2 times.
So, the probability of seeing 'AN' given it's from ANANTPUR is:
\[ P(V | E_A) = \frac{2}{7} \] Applying Bayes' Theorem:
Now we substitute these values into the formula for $P(E_A | V)$:
\[ P(E_A | V) = \frac{\left(\frac{2}{7}\right) \left(\frac{1}{2}\right)}{\left(\frac{1}{5}\right) \left(\frac{1}{2}\right) + \left(\frac{2}{7}\right) \left(\frac{1}{2}\right)} \] The common factor of $\frac{1}{2}$ cancels from the numerator and denominator:
\[ P(E_A | V) = \frac{\frac{2}{7}}{\frac{1}{5} + \frac{2}{7}} \] To simplify the denominator, find a common base:
\[ \frac{1}{5} + \frac{2}{7} = \frac{7+10}{35} = \frac{17}{35} \] Now, complete the calculation:
\[ P(E_A | V) = \frac{2/7}{17/35} = \frac{2}{7} \times \frac{35}{17} = \frac{2 \times 5}{17} = \frac{10}{17} \] Step 4: Final Answer:
The probability that the letter came from ANANTPUR is $\frac{10}{17}$.