Maths Q10 – JEE-MAIN 2025

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}

gcd (m, n) = 1

, then

n^{2} - m^{2}

is equal to :

Clue 1 of 3: Understand the problem setup

Identify the different types of coins and the events involved: drawing a specific type of coin and getting a head.

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Step 1: Identify Probabilities of Events✦ Active

There are 19 unbiased coins and 1 double-headed coin, making a total of 20 coins. Let $U$ be the event of drawing an unbiased coin, and $D$ be the event of drawing a double-headed coin. Let $H$ be the event of getting a head.

P (U) = \frac{19}{20}, P (D) = \frac{1}{20}

P (H | U) = \frac{1}{2} (for an unbiased coin)

P (H | D) = 1 (for a double-headed coin)

Step 2: Calculate the Conditional Probability○ Expand

Using Bayes' Theorem, the probability that the drawn coin was unbiased given that a head turned up is $P (U | H) = \frac{P (H | U) P (U)}{P (H)}$ . First, calculate $P (H)$ using the law of total probability.

P (H) = P (H | U) P (U) + P (H | D) P (D) = (\frac{1}{2}) (\frac{19}{20}) + (1) (\frac{1}{20}) = \frac{19}{40} + \frac{2}{40} = \frac{21}{40}

P (U | H) = \frac{\frac{1}{2} \times \frac{19}{20}}{\frac{21}{40}} = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21}

💡 Teacher's Secret Hint

Ensure to correctly apply the law of total probability for the denominator.

Step 3: Calculate the Final Expression○ Expand

The probability is given as $\frac{m}{n} = \frac{19}{21}$ , where $gcd (m, n) = 1$ . Thus, $m = 19$ and $n = 21$ . We need to find $n^{2} - m^{2}$ .

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

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

n^{2} - m^{2} = (2) (40) = 80

💡 Teacher's Secret Hint

Remember the difference of squares factorization $a^{2} - b^{2} = (a - b) (a + b)$ to simplify calculations.

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Maths Question 10 – JEE-MAIN 2025

Clue 1 of 3: Understand the problem setup

🧠 Full Solution Path