Problem 143 - A Complete Resource Book in Mathematics for JEE Main

Question

Assertion: A bag contains \(n+1\) coins. It is known that one of these coins has a head on both sides while the other coins are fair. One coin is selected at random and tossed. If head turns up, the probability that the selected coin was fair, is \(\frac{n}{n+2}\) Reason: If an event \(A\) occurs with two mutually exclusive and exhanstive events \(E_{1}\) and \(E_{2}\), then \(P(E / A)\) \(=\frac{P\left(E_{i}\right) P\left(A / E_{i}\right)}{P\left(E_{1}\right) P\left(A / E_{1}\right)+P\left(E_{2}\right) P\left(A / E_{2}\right)}, i=1,2 .\)

Step-by-Step Solution

Answer

The probability that the selected coin was fair is \( \frac{n}{n+2} \).

Step 1: Define the Events

Let's call the event of selecting the fair coin as \( E_1 \) and the event of selecting the two-headed coin as \( E_2 \). The event of getting a head when tossing the coin is denoted by \( A \). We need to find the probability \( P(E_1 / A) \) that the selected coin was fair, given that a head is tossed.

Step 2: Determine Individual Probabilities

Calculate \( P(E_1) \) and \( P(E_2) \). Since one coin is selected at random from \( n+1 \) coins, \( P(E_1) = \frac{n}{n+1} \) and \( P(E_2) = \frac{1}{n+1} \).

Step 3: Determine Conditional Probabilities

Calculate \( P(A/E_1) \) and \( P(A/E_2) \). If the selected coin is fair (\( E_1 \)), the probability of getting a head \( A \) is \( \frac{1}{2} \). If the selected coin is two-headed (\( E_2 \)), then \( P(A/E_2) = 1 \) because it always shows heads.

Step 4: Calculate Total Probability of a Head

Using the law of total probability, find \( P(A) \), the probability of getting a head, by computing:\( P(A) = P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) = \frac{n}{n+1} \cdot \frac{1}{2} + \frac{1}{n+1} \cdot 1 \).

Step 5: Apply Bayes' Theorem

Using Bayes' Theorem, calculate \( P(E_1 / A) \):\[P(E_1 / A) = \frac{P(E_1) \cdot P(A / E_1)}{P(A)} = \frac{\left(\frac{n}{n+1}\right) \cdot \frac{1}{2}}{\left(\frac{n}{n+1} \cdot \frac{1}{2} + \frac{1}{n+1}\right)}.\]

Step 6: Simplify the Expression

Simplify the expression obtained in Step 5:\[P(E_1 / A) = \frac{\frac{n}{2(n+1)}}{\frac{n+2}{2(n+1)}} = \frac{n}{n+2}.\]Hence, the probability that the selected coin was fair is indeed \( \frac{n}{n+2} \).

Key Concepts

Bayes' Theorem

Bayes' Theorem is a powerful mathematical formula used to determine the probability of an event based on prior knowledge of conditions that might affect the event. In simpler terms, it allows us to update the probability estimates for events by incorporating new evidence or information.
Bayes' Theorem can be written as:

• \( P(E_i / A) = \frac{P(E_i) \cdot P(A / E_i)}{P(A)} \)

This means that to find the probability of an event \( E_i \) given that another event \( A \) has occurred, you multiply the prior probability of \( E_i \) by the probability of \( A \) given \( E_i \), and then divide by the total probability of \( A \).
In the exercise, Bayes' Theorem is used to find the likelihood that a randomly selected coin is fair, given it lands on heads after being tossed. The known probabilities and conditional probabilities inform the calculation, helping clarify which condition is more likely to occur based on the given evidence.

Law of Total Probability

The Law of Total Probability is a fundamental rule relating marginal probabilities to conditional probabilities. It helps in breaking down complex probability scenarios into simpler parts, enabling the calculation of the total probability of an event based on the probabilities of different ways the event can occur.
The law can be expressed as:

• \( P(A) = P(E_1) \cdot P(A / E_1) + P(E_2) \cdot P(A / E_2) + \ldots + P(E_n) \cdot P(A / E_n) \)

Here, event \( A \) can happen through one of several ways, each represented by the mutually exclusive events \( E_1, E_2, \ldots, E_n \).
In the bag of coins problem, the event \( A \) is getting a head when a coin is tossed. The Law of Total Probability is applied to compute \( P(A) \) by considering the chances of picking either a fair coin or a two-headed coin and then getting heads accordingly. This setup provides the necessary groundwork for applying Bayes' Theorem to find the desired probability.

Conditional Probability

Conditional Probability signifies the likelihood of an event occurring, given that another event has already occurred. It is a critical concept in probability, allowing for more precise probability calculations by considering how the probability of one event might be affected by the knowledge of another event occurring.
The notation \( P(A / B) \) represents the probability of event \( A \) happening given that event \( B \) has happened. It can be explained using the formula:

• \( P(A / B) = \frac{P(A \cap B)}{P(B)} \)

This formula demonstrates how the probability is adjusted when considering another event. In the context of the coin exercise, conditional probabilities like \( P(A / E_1) \) and \( P(A / E_2) \) denote the probabilities of flipping a head given the selection of a regular or double-headed coin, respectively.
Understanding these probabilities is crucial for applying the Law of Total Probability and Bayes' Theorem correctly, as they serve as the foundation for these calculations.