JEE Advanced 2002

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Animated Solution for Mathematics - Probability: A box contains $N$ coins, $m$ of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is $1/2$ , while it is $2/3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. What is the probability that the coin drawn is fair?

Q: Why does the total number of coins $N$ cancel out in the final step?

It cancels out because $N$ appears in the denominator of every term in our Bayes' equation. When you have a common factor in every term of a fraction's numerator and denominator, it effectively becomes a multiplier of $1$, allowing us to simplify the expression without losing any information. It's a beautiful moment of algebraic elegance!

Progressive Hints

Hint 1

The Spark: This problem is a classic application of Bayes' Theorem. You are given the outcome (a specific sequence of tosses) and asked to infer the cause (the type of coin). Think of the box as a system with two distinct paths: the 'Fair' path and the 'Biased' path. Your goal is to determine the probability of the 'Fair' path given the evidence.

Hint 2

The Arsenal: Remember that the two tosses are independent events. For any coin, the probability of the sequence 'Head then Tail' is simply the product of the probability of getting a Head and the probability of getting a Tail. Calculate this probability separately for the fair coin and the biased coin.

Hint 3

The Execution: Start by setting up the Bayes' Theorem formula:

P (F ∣ E) = \frac{P ( E ∣ F ) P ( F )}{P ( E ∣ F ) P ( F ) + P ( E ∣ B ) P ( B )}

, where

F

is the event of picking a fair coin,

B

is the event of picking a biased coin, and

E

is the observed event of getting a Head followed by a Tail. Substitute your calculated probabilities and simplify the resulting algebraic expression.

Visualized Solution

Initial Setup: The Box of Coins

Total coins in the box = $N$
Number of fair coins = $m$
Number of biased coins = $N - m$

Prior Probabilities

Probability of picking a fair coin, $P (F) = \frac{m}{N}$
Probability of picking a biased coin, $P (B) = \frac{N - m}{N}$

Defining the Event $E$

A coin is drawn and tossed twice.
Event $E$ : The first toss shows Head ( $H$ ) and the second shows Tail ( $T$ ).

Conditional Probability: Fair Coin

For a fair coin ( $F$ ): $P (H) = \frac{1}{2}$ , $P (T) = \frac{1}{2}$
Conditional probability $P (E ∣ F) = P (H) \times P (T)$
$P (E ∣ F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Conditional Probability: Biased Coin

For a biased coin ( $B$ ): $P (H) = \frac{2}{3}$ , $P (T) = 1 - \frac{2}{3} = \frac{1}{3}$
Conditional probability $P (E ∣ B) = P (H) \times P (T)$
$P (E ∣ B) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$

Applying Bayes' Theorem

We need to find $P (F ∣ E)$ : the probability the coin is fair given event $E$ occurred.
Bayes' Theorem: $P (F ∣ E) = \frac{P ( E ∣ F ) P ( F )}{P ( E ∣ F ) P ( F ) + P ( E ∣ B ) P ( B )}$
The numerator represents the 'Fair' path, and the denominator is the total probability of Event $E$ .

Substituting the Values

$P (F ∣ E) = \frac{\frac{1}{4} \cdot \frac{m}{N}}{\frac{1}{4} \cdot \frac{m}{N} + \frac{2}{9} \cdot \frac{N - m}{N}}$

Simplifying the Expression

Notice that $\frac{1}{N}$ is a common factor in both the numerator and the denominator.
Cancelling $N$ : $P (F ∣ E) = \frac{\frac{m}{4}}{\frac{m}{4} + \frac{2 ( N - m )}{9}}$

Clearing the Fractions

Multiply the numerator and denominator by $36$ (the LCM of $4$ and $9$ ).
Numerator: $36 \times \frac{m}{4} = 9 m$
Denominator: $36 \times (\frac{m}{4} + \frac{2 ( N - m )}{9}) = 9 m + 8 (N - m)$
$P (F ∣ E) = \frac{9 m}{9 m + 8 ( N - m )}$

Final Simplification

Expand the denominator: $9 m + 8 N - 8 m$
Combine like terms: $m + 8 N$
Final Probability: $P (F ∣ E) = \frac{9 m}{m + 8 N}$

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The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Setup

Imagine you are standing in front of a mysterious box containing

N

coins. Some are fair, and some are biased. You reach in, pull out a coin, and toss it twice, observing a Head followed by a Tail.

We are tasked with finding the probability that the coin is fair. This is a classic application of Bayesian inference, where we use observed evidence to update our prior beliefs.

Defining the Evidence

We have

N

total coins, with

m

of them being fair. Consequently, the number of biased coins is

N - m

If we pick a coin at random, the prior probabilities are:

P (F) = \frac{m}{N}

P (B) = \frac{N - m}{N}

The Two Paths

Let

E

be the event of observing a Head followed by a Tail. If the coin is fair, the probability of this sequence is:

P (E ∣ F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

If the coin is biased, the probability of a Head is

\frac{2}{3}

, which implies the probability of a Tail is

1 - \frac{2}{3} = \frac{1}{3}

. Thus, the probability of the sequence given the coin is biased is:

P (E ∣ B) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}

Bayes' Theorem as a Detective Tool

We use Bayes' Theorem to calculate the posterior probability

P (F ∣ E)

, which represents the probability the coin is fair given the evidence

E

P (F ∣ E) = \frac{P ( E ∣ F ) P ( F )}{P ( E ∣ F ) P ( F ) + P ( E ∣ B ) P ( B )}

This formula acts as a bridge between the 'cause' (the coin type) and the 'effect' (the observed tosses). The numerator represents the 'Fair' path, while the denominator represents the total probability of observing

E

across all possible coin types.

The Algebraic Cleanup

Substituting our known values into the theorem, we get:

P (F ∣ E) = \frac{\frac{1}{4} \cdot \frac{m}{N}}{\frac{1}{4} \cdot \frac{m}{N} + \frac{2}{9} \cdot \frac{N - m}{N}}

Since

\frac{1}{N}

is a common factor in every term, we cancel it out to simplify the expression:

P (F ∣ E) = \frac{\frac{m}{4}}{\frac{m}{4} + \frac{2 ( N - m )}{9}}

To eliminate the fractions, we multiply the numerator and the denominator by

36

, the least common multiple of

4

and

9

P (F ∣ E) = \frac{9 m}{9 m + 8 ( N - m )}

The Final Revelation

Expanding the denominator, we obtain

9 m + 8 N - 8 m

, which simplifies to

m + 8 N

The final probability that the coin is fair is: $P (F ∣ E) = \frac{9 m}{m + 8 N}$

This result is clean, logical, and perfectly derived from the evidence. You have successfully navigated the uncertainty of the box using the power of Bayesian inference.

Common Traps & FAQs

Wait, why do we multiply the probabilities for the two tosses? Shouldn't we add them?

Great question! We multiply because the two tosses are independent events. In probability, when we want the probability of two independent events both happening (Event A AND Event B), we multiply their individual probabilities. Adding them would imply these are mutually exclusive outcomes, which is not the case here.

Why can't we just use the ratio of fair coins to biased coins to find the answer?

That would only work if both types of coins were equally likely to produce the observed result (Head then Tail). However, the fair coin and the biased coin have different probabilities of producing that specific sequence. Bayes' Theorem allows us to 'weight' the prior probability of picking a coin by how likely that coin is to produce the evidence we actually saw.

Why does the total number of coins

N

cancel out in the final step?

It cancels out because

N

appears in the denominator of every term in our Bayes' equation. When you have a common factor in every term of a fraction's numerator and denominator, it effectively becomes a multiplier of

1

, allowing us to simplify the expression without losing any information. It's a beautiful moment of algebraic elegance!

Visualized Solution

Initial Setup: The Box of Coins

Prior Probabilities

Defining the Event $E$

Conditional Probability: Fair Coin

Conditional Probability: Biased Coin

Applying Bayes' Theorem

Substituting the Values

Simplifying the Expression

Clearing the Fractions

Final Simplification

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Defining the Evidence

The Two Paths

Bayes' Theorem as a Detective Tool

The Algebraic Cleanup

The Final Revelation

Similar Questions

A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is $\frac{m}{n}$ , $g cd (m, n) = 1$ , then $n^{2} - m^{2}$ is equal to :

Comprehension Passage

The probability of the drawn ball from $U_{2}$ being white is

Given that the drawn ball from $U_{2}$ is white, the probability that head appeared on the coin is

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

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Comprehension Passage

If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$ , the probability that all 3 drawn balls are of the same colour is

If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box $B_{2}$ is

Comprehension Passage

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct option(s) with the possible values of $n_{1}$ and $n_{2}$ is(are)

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and $n$ white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $\frac{6}{11}$ , then $n$ is equal to ____.

Comprehension Passage

If $P (u_{i}) \propto i$ , where $i = 1, 2, 3, \dots, n$ , then $lim_{n \to \infty} P (w) =$

If $P (u_{i}) = c$ , (a constant) then $P (u_{n} / w) =$

Let $P (u_{i}) = \frac{1}{n}$ , if $n$ is even and $E$ denotes the event of choosing even numbered urn, then the value of $P (w / E)$ is

Visualized Solution

Initial Setup: The Box of Coins

Prior Probabilities

Defining the Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Conditional Probability: Fair Coin

Conditional Probability: Biased Coin

Applying Bayes' Theorem

Substituting the Values

Simplifying the Expression

Clearing the Fractions

Final Simplification

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Defining the Evidence

The Two Paths

Bayes' Theorem as a Detective Tool

The Algebraic Cleanup

The Final Revelation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability of the drawn ball from U2​ being white is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given that the drawn ball from U2​ is white, the probability that head appeared on the coin is

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1 ball is drawn from each of the boxes B1​,B2​ and B3​, the probability that all 3 drawn balls are of the same colour is

Comprehension Passage

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P(ui​)∝i, where i=1,2,3,…,n, then limn→∞​P(w)=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P(ui​)=c, (a constant) then P(un​/w)=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P(ui​)=n1​, if n is even and E denotes the event of choosing even numbered urn, then the value of P(w/E) is

Defining the Event $E$

The probability of the drawn ball from $U_{2}$ being white is

Given that the drawn ball from $U_{2}$ is white, the probability that head appeared on the coin is

If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$ , the probability that all 3 drawn balls are of the same colour is

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If $P (u_{i}) = c$ , (a constant) then $P (u_{n} / w) =$

Let $P (u_{i}) = \frac{1}{n}$ , if $n$ is even and $E$ denotes the event of choosing even numbered urn, then the value of $P (w / E)$ is