Question

On a postcard one of the two words either KANPUR or ANANTPUR is written. If only two consecutive letters AN are visible on the postcard, then the probability that the written word is ANANTPUR, is:

• A. \( 3/17 \)
• B. \( 10/17 \)
• C. \( 2/17 \)
• D. \( 4/17 \)

Hint:

When counting consecutive pairs in a word of length \( n \), there are always \( n-1 \) such pairs. This is the denominator for your conditional probability.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a problem of conditional probability (Bayes' Theorem). We need to find the probability of a word given an observed event (visible letters "AN").

Step 2: Key Formula or Approach:
Let \( E_1 \) be the event that the word is KANPUR, and \( E_2 \) be the event that it is ANANTPUR. Let \( A \) be the event that "AN" is visible. \[ P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \]

Step 3: Detailed Explanation:
1. Assume \( P(E_1) = P(E_2) = 1/2 \). 2. In KANPUR (6 letters), consecutive pairs are: KA, AN, NP, PU, UR (Total 5). "AN" appears once. So, \( P(A|E_1) = 1/5 \). 3. In ANANTPUR (8 letters), consecutive pairs are: AN, NA, AN, NT, TP, PU, UR (Total 7). "AN" appears twice. So, \( P(A|E_2) = 2/7 \). 4. Apply Bayes' Theorem: \[ P(E_2|A) = \frac{\frac{1}{2} \cdot \frac{2}{7}}{\frac{1}{2} \cdot \frac{1}{5} + \frac{1}{2} \cdot \frac{2}{7}} = \frac{2/7}{1/5 + 2/7} = \frac{2/7}{17/35} = \frac{2}{7} \cdot \frac{35}{17} = \frac{10}{17} \].

Step 4: Final Answer:
The probability is 10/17.