Problem 17 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters. TA are visible. The probability that the letters have come from CALCUTTA is (a) \(1 / 3\) (b) \(4 / 11\) (c) \(5 / 12\) (d) None of these (b) Let \(A\) : 'the event that letters came from TATANAGAR' B: 'the event that letters came from CALCUTTA' \(C\) : 'the event that two consecutive letters visible by \(T A^{\prime}\) Then \(P(A)=1 / 2, P(B)=1 / 2, P(C / B)=1 / 7\) Hence, by Bayes' theorem $$ P(B / C)=\frac{P(B) \times P(C / B)}{P(A) P(C / A)+P(B) P(C / B)}=\frac{4}{11} $$

Step-by-Step Solution

Answer

The probability that the letters came from CALCUTTA is \(\frac{4}{11}\).

Step 1: Identify Known Variables and Events

First, identify the events given in the problem: event \(A\) represents the letter coming from TATANAGAR, and event \(B\) represents the letter coming from CALCUTTA. We use \(C\) to denote the event that 'TA' is seen on the envelope.

Step 2: Determine Basic Probabilities

Find basic probabilities: \(P(A) = 1/2\) because the letter can come from either city with equal likelihood, and \(P(B) = 1/2\) for the same reason.

Step 3: Calculate Conditional Probability of Seeing 'TA' Given CALCUTTA

The probability of seeing 'TA' from CALCUTTA, \(P(C|B)\), is \(1/7\). There are 7 possible consecutive pairs in CALCUTTA, and only one of them shows 'TA'.

Step 4: Calculate Conditional Probability of Seeing 'TA' Given TATANAGAR

Calculate \(P(C|A)\): In TATANAGAR, 'TA' appears multiple times. Count each instance of 'TA': it appears 3 times: 'TA', 'TA', and within 'NTA'. Since there are 8 possible pairs, \(P(C|A) = 3/8\).

Step 5: Apply Bayes' Theorem

Use Bayes' theorem to find \(P(B|C)\), the probability that the letter is from CALCUTTA given that 'TA' is seen:\[P(B|C) = \frac{P(B) \cdot P(C|B)}{P(A) \cdot P(C|A) + P(B) \cdot P(C|B)} = \frac{\frac{1}{2} \cdot \frac{1}{7}}{\frac{1}{2} \cdot \frac{3}{8} + \frac{1}{2} \cdot \frac{1}{7}}\]Simplify the expression:\[P(B|C) = \frac{\frac{1}{14}}{\frac{3}{16} + \frac{1}{14}}\]

Step 6: Simplify and Solve

Calculate the final probability:\[P(B|C) = \frac{\frac{1}{14}}{\frac{21}{112} + \frac{8}{112}} = \frac{\frac{1}{14}}{\frac{29}{112}} = \frac{1}{14} \times \frac{112}{29} = \frac{8}{29} = \frac{4}{11}\]Thus, \(P(B|C)\) is \(\frac{4}{11}\).

Key Concepts

Conditional Probability

Conditional probability is a fundamental concept in probability theory. It describes the probability of an event occurring given that another event has already happened. In the given exercise, we are interested in finding the probability that the letter came from CALCUTTA (event \(B\)) given that we know the visible letters are "TA" (event \(C\)). This is expressed as \(P(B|C)\). Conditional probability can be computed using Bayes' Theorem, which helps in revising probabilities in light of new information.

Bayes' Theorem states:

• \(P(B|C) = \frac{P(B) \times P(C|B)}{P(A) \times P(C|A) + P(B) \times P(C|B)}\)

In our solution, we use the theorem to combine the probabilities of CALCUTTA and TATANAGAR together with the visibility of \("TA"\), to determine \(P(B|C)\). This theorem works on the principle of updating and revising previously known probabilities when new data is available.

Event Probability

To solve probability exercises like this one, first determine the probability of basic events.

• \(P(A)\): Probability that the letters came from TATANAGAR is \(\frac{1}{2}\). This is because there are two cities (TATANAGAR and CALCUTTA), giving each an equal chance initially.
• \(P(B)\): Probability that the letters came from CALCUTTA is also \(\frac{1}{2}\), for the same reason.

These simple probabilities serve as the foundation for calculating more complex conditional probabilities later in the problem. It is essential to accurately determine these base probabilities because they are fundamental inputs for applying formulas like Bayes’ Theorem. Keep in mind that each of the basic events is mutually exclusive – meaning if one happens, the other does not in this context.

Combinatorics

Combinatorics deals with counting possibilities and is a key component in calculating probabilities. In this exercise, we have to use combinatorial thinking to determine how likely the combination of letters "TA" is, in different contexts.

Let's analyze the cities:

• **CALCUTTA**: There are 7 consecutive letter pairs possible (CA, AL, LC, CU, UT, TT, TA). Only one, "TA," matches what we see. Therefore, \(P(C|B) = \frac{1}{7}\).
• **TATANAGAR**: Here, "TA" can form three different ways (TA, TA, T in the middle of NT). With 8 possible pairs total, \(P(C|A) = \frac{3}{8}\).

Counting these sequences accurately is crucial for determining the probabilities of events given conditions. Combinatorics offers structured methods for systematically tallying such scenarios, laying the groundwork for further probabilistic calculations.