JEE Main 2025 April

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Animated Solution for Mathematics - Probability: 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 :

Q: Wait, why can't we just say the probability is $19/20$ because there are 19 unbiased coins?

That is a very common trap! The $19/20$ is the *prior* probability—the chance of picking an unbiased coin *before* you toss it. But once you toss the coin and see a Head, you have new information. The biased coin is 'more likely' to produce a Head than the unbiased one. Bayes' Theorem helps us adjust our initial guess based on this new evidence.

Q: Is there a faster way to calculate $21^2 - 19^2$ without doing the squares?

Absolutely! Whenever you see $n^2 - m^2$, your brain should immediately jump to the algebraic identity $a^2 - b^2 = (a - b)(a + b)$. Here, $21^2 - 19^2 = (21 - 19)(21 + 19) = 2 \times 40 = 80$. It saves time and prevents calculation errors!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. You are dealing with a classic 'inverse probability' scenario. We know the outcome (a Head appeared), but we are uncertain about the cause (was it the fair coin or the biased one?). This is the perfect setup for Bayes' Theorem, which allows us to update our initial beliefs based on new evidence.

Hint 2

Hint 2: The Arsenal. You need to define your events clearly. Let

U

be the event of choosing an unbiased coin and

B

be the event of choosing the biased coin. The probability of picking each is based on the total count of 20 coins. Then, consider the conditional probability of getting a Head given each coin:

P (H ∣ U)

1/2

, while

P (H ∣ B)

1

because it is double-headed.

Hint 3

Hint 3: The Execution. Apply Bayes' Theorem:

P (U ∣ H) = \frac{P ( U ) P ( H ∣ U )}{P ( U ) P ( H ∣ U ) + P ( B ) P ( H ∣ B )}

. Substitute the values:

P (U) = \frac{19}{20}

P (B) = \frac{1}{20}

P (H ∣ U) = \frac{1}{2}

, and

P (H ∣ B) = 1

. Once you find the fraction

\frac{m}{n}

, calculate

n^{2} - m^{2}

using the difference of squares identity

(n - m) (n + m)

to simplify your arithmetic.

Visualized Solution

Visualizing the Bag of Coins

Total coins = $20$
Unbiased coins = $19$
Biased coins = $1$

Defining the First Draw

Let $U$ be the event of drawing an unbiased coin.
Let $B$ be the event of drawing the biased coin.
$P (U) = \frac{19}{20}$
$P (B) = \frac{1}{20}$

The Toss and Conditional Probabilities

Let $H$ be the event of getting a Head.
$P (H ∣ U) = \frac{1}{2}$ (Fair coin)
$P (H ∣ B) = 1$ (Double-headed coin)

Bayes' Theorem Setup

We need to find $P (U ∣ H)$ .
Using Bayes' Theorem:
$P (U ∣ H) = \frac{P ( U ) P ( H ∣ U )}{P ( U ) P ( H ∣ U ) + P ( B ) P ( H ∣ B )}$

Substituting the Values

Substitute the known probabilities:
$P (U ∣ H) = \frac{\frac{19}{20} \cdot \frac{1}{2}}{\frac{19}{20} \cdot \frac{1}{2} + \frac{1}{20} \cdot 1}$

Simplifying the Probabilities

Numerator $= \frac{19}{40}$
Denominator $= \frac{19}{40} + \frac{2}{40} = \frac{21}{40}$
$P (U ∣ H) = \frac{19}{21}$

Equating to $\frac{m}{n}$

Given $P (U ∣ H) = \frac{m}{n}$
Since $g cd (19, 21) = 1$ , we can directly compare.
$m = 19$
$n = 21$

Calculating $n^{2} - m^{2}$

We need to calculate $n^{2} - m^{2}$ .
$n^{2} - m^{2} = 2 1^{2} - 1 9^{2}$
Use the identity $a^{2} - b^{2} = (a - b) (a + b)$

Final Answer

$(21 - 19) (21 + 19)$
$= 2 \cdot 40$
$= 80$
Final Answer: $80$

00:00 / 00:00

The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Setup

We begin with a total of

20

coins. The probability of selecting an unbiased coin, denoted as event

U

, is:

P (U) = \frac{19}{20}

The probability of selecting the biased coin, denoted as event

B

, is:

P (B) = \frac{1}{20}

These values represent our 'prior' knowledge before observing the outcome of the coin toss.

The Evidence

Next, we consider the evidence: the coin toss resulted in a Head (

H

). We must determine the conditional probability of this outcome based on the coin selected.

If we selected an unbiased coin, the probability of obtaining a Head is:

P (H ∣ U) = \frac{1}{2}

If we selected the biased coin, the probability of obtaining a Head is:

P (H ∣ B) = 1

This is because the biased coin is double-headed, making a Head the only possible outcome.

The Logic of Bayes

We aim to find

P (U ∣ H)

, the probability that the coin is unbiased given that we observed a Head. Bayes' Theorem provides the framework to update our belief based on the observed evidence:

P (U ∣ H) = \frac{P ( U ) P ( H ∣ U )}{P ( U ) P ( H ∣ U ) + P ( B ) P ( H ∣ B )}

The numerator represents the probability of the path where we picked an unbiased coin AND obtained a Head. The denominator represents the total probability of obtaining a Head across all possible scenarios.

The Calculation

Substituting our values into the theorem, the numerator is:

\frac{19}{20} \cdot \frac{1}{2} = \frac{19}{40}

The denominator is the sum of the unbiased path and the biased path:

\frac{19}{40} + (\frac{1}{20} \cdot 1) = \frac{19}{40} + \frac{2}{40} = \frac{21}{40}

Dividing the numerator by the denominator, the

40

s cancel out, yielding:

P (U ∣ H) = \frac{19}{21}

Here,

m = 19

and

n = 21

Final Calculation

The problem requires the value of

n^{2} - m^{2}

. Using the difference of squares identity,

a^{2} - b^{2} = (a - b) (a + b)

, we calculate:

n^{2} - m^{2} = (21 - 19) (21 + 19)

n^{2} - m^{2} = 2 \cdot 40 = 80

The final result is 80.

Common Traps & FAQs

Wait, why can't we just say the probability is

19/20

because there are 19 unbiased coins?

That is a very common trap! The

19/20

is the *prior* probability—the chance of picking an unbiased coin *before* you toss it. But once you toss the coin and see a Head, you have new information. The biased coin is 'more likely' to produce a Head than the unbiased one. Bayes' Theorem helps us adjust our initial guess based on this new evidence.

Why is

P (H ∣ B) = 1

? Shouldn't it be

1/2

Remember the problem description: the biased coin has a Head on *both* sides. That means every single time you toss it, it *must* result in a Head. There is no chance of getting a Tail. Therefore, the probability of getting a Head given that you have the biased coin is

100%

, or

1

Is there a faster way to calculate

2 1^{2} - 1 9^{2}

without doing the squares?

Absolutely! Whenever you see

n^{2} - m^{2}

, your brain should immediately jump to the algebraic identity

a^{2} - b^{2} = (a - b) (a + b)

. Here,

2 1^{2} - 1 9^{2} = (21 - 19) (21 + 19) = 2 \times 40 = 80

. It saves time and prevents calculation errors!

Animated Solution for Mathematics - Probability: 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 :

Select Answer:

Visualized Solution

Visualizing the Bag of Coins

Defining the First Draw

The Toss and Conditional Probabilities

Bayes' Theorem Setup

Substituting the Values

Simplifying the Probabilities

Equating to $\frac{m}{n}$

Calculating $n^{2} - m^{2}$

Final Answer

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

The Evidence

The Logic of Bayes

The Calculation

Final Calculation

Similar Questions

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

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 ____.

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:

A card from a pack of 52 cards is lost. From the remaining 51 cards, $n$ cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$ , then $n$ is equal to

A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :

A bag contains 10 balls out of which $k$ are red and $(10 - k)$ are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 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 :

Select Answer:

Visualized Solution

Visualizing the Bag of Coins

Defining the First Draw

The Toss and Conditional Probabilities

Bayes' Theorem Setup

Substituting the Values

Simplifying the Probabilities

Equating to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nm​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2−m2

Final Answer

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

The Evidence

The Logic of Bayes

The Calculation

Final Calculation

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:

A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, 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 :

Equating to $\frac{m}{n}$

Calculating $n^{2} - m^{2}$

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