Key Concepts and Formulas

• Bayes' Theorem: This theorem is fundamental for calculating conditional probabilities, especially when we want to find the probability of a "cause" given an "effect." If $A$ is an event and $H$ is the observed evidence, then: $P(A|H) = \frac{P(H|A) \cdot P(A)}{P(H)}$ Where $P(A|H)$ is the posterior probability, $P(H|A)$ is the likelihood, $P(A)$ is the prior probability, and $P(H)$ is the total probability of the evidence.

• Law of Total Probability: Used to calculate the overall probability of an event $H$ by considering all mutually exclusive and exhaustive ways it can occur. If events $A_1, A_2, \dots, A_k$ form a partition of the sample space, then: $P(H) = \sum_{i=1}^{k} P(H|A_i) \cdot P(A_i)$ In this problem, the two types of coins (unbiased and two-headed) form such a partition.

• Difference of Squares Identity: A useful algebraic identity for simplifying calculations: $a^2 - b^2 = (a - b)(a + b)$

Step-by-Step Solution

Step 1: Define Events and Determine Prior Probabilities To begin, we clearly define the events involved in the problem. This helps in setting up the probability expressions correctly.

• Let $U$ be the event that the drawn coin is unbiased.
• Let $B$ be the event that the drawn coin is the two-headed (biased) coin.
• Let $H$ be the event that a head turns up when the drawn coin is tossed.

Next, we calculate the initial (prior) probabilities of drawing each type of coin from the bag, based on the composition of the bag before any coin is tossed.

• Number of unbiased coins = 19
• Number of two-headed coins = 1
• Total number of coins in the bag = $19 + 1 = 20$

The prior probabilities are:

• Prior probability of drawing an unbiased coin: $P(U) = \frac{\text{Number of unbiased coins}}{\text{Total number of coins}} = \frac{19}{20}$
• Prior probability of drawing the two-headed coin: $P(B) = \frac{\text{Number of two-headed coins}}{\text{Total number of coins}} = \frac{1}{20}$ These events ($U$ and $B$) are mutually exclusive and exhaustive, meaning $P(U) + P(B) = 1$.

Step 2: Determine Conditional Probabilities (Likelihoods) of Observing a Head Now, we determine the probability of observing a head, given the type of coin that was drawn. These are the "likelihoods" required for Bayes' Theorem.

• Probability of getting a head given the coin is unbiased ($U$): An unbiased coin has an equal chance of landing heads or tails. $P(H|U) = \frac{1}{2}$

• Probability of getting a head given the coin is two-headed ($B$): A two-headed coin will always show a head when tossed. $P(H|B) = 1$

Step 3: Calculate the Total Probability of Observing a Head, $P(H)$ Before applying Bayes' Theorem, we need to calculate the overall probability of observing a head, $P(H)$. This is done using the Law of Total Probability, which states that a head can turn up either by drawing an unbiased coin and getting a head, OR by drawing the two-headed coin and getting a head. $P(H) = P(H|U) \cdot P(U) + P(H|B) \cdot P(B)$ Substitute the values from Step 1 and Step 2: $P(H) = \left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right) + (1) \cdot \left(\frac{1}{20}\right)$ $P(H) = \frac{19}{40} + \frac{1}{20}$ To add these fractions, we find a common denominator (40): $P(H) = \frac{19}{40} + \frac{1 \times 2}{20 \times 2} = \frac{19}{40} + \frac{2}{40}$ $P(H) = \frac{19 + 2}{40} = \frac{21}{40}$ So, the total probability of observing a head is $\frac{21}{40}$.

Step 4: Apply Bayes' Theorem to Find the Posterior Probability $P(U|H)$ We are asked to find the probability that the drawn coin was unbiased, given that a head turned up. This is $P(U|H)$. Using Bayes' Theorem: $P(U|H) = \frac{P(H|U) \cdot P(U)}{P(H)}$ Substitute the values calculated in the previous steps:

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

$P(U|H) = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right)}{\frac{21}{40}}$ First, calculate the numerator: $\text{Numerator} = \frac{1 \cdot 19}{2 \cdot 20} = \frac{19}{40}$ Now, substitute this back into the Bayes' formula: $P(U|H) = \frac{\frac{19}{40}}{\frac{21}{40}}$ To simplify this complex fraction, we multiply the numerator by the reciprocal of the denominator: $P(U|H) = \frac{19}{40} \cdot \frac{40}{21}$ The 40 in the numerator and denominator cancel out: $P(U|H) = \frac{19}{21}$

Step 5: Determine $m$ and $n$ and Calculate $n^2 - m^2$ The problem states that the probability is $\frac{m}{n}$, where $\gcd(m, n) = 1$. From our calculation, $P(U|H) = \frac{19}{21}$. Therefore, $m = 19$ and $n = 21$. We verify that $\gcd(19, 21) = 1$ (19 is prime, 21 is $3 \times 7$, so they share no common factors other than 1).

Finally, we calculate $n^2 - m^2$: $n^2 - m^2 = 21^2 - 19^2$ Using the difference of squares identity, $a^2 - b^2 = (a - b)(a + b)$: $n^2 - m^2 = (21 - 19)(21 + 19)$ $n^2 - m^2 = (2)(40)$ $n^2 - m^2 = 80$

Common Mistakes & Tips

• Confusing Conditional Probabilities: A common error is mixing up $P(A|H)$ and $P(H|A)$. Always be clear about which event is conditional on which.
• Incorrect Denominator Calculation: The most frequent mistake in Bayes' Theorem problems is an incorrect calculation of $P(H)$ using the Law of Total Probability. Ensure all possible pathways to event $H$ are included.
• Arithmetic Errors: Double-check fraction additions and multiplications, especially when dealing with multiple fractions.
• Algebraic Simplification: Remember to utilize algebraic identities like the difference of squares $(a^2 - b^2)$ to simplify calculations, particularly in competitive exams.

Summary

This problem is a classic application of Bayes' Theorem, which allows us to update the probability of an event (drawing an unbiased coin) given new evidence (observing a head). We first defined the events and determined their prior probabilities. Then, we calculated the likelihoods of observing a head for each type of coin. Using the Law of Total Probability, we found the overall probability of observing a head. Finally, we applied Bayes' Theorem to compute the posterior probability that the drawn coin was unbiased, given the observed head. This probability was found to be $\frac{19}{21}$, leading to $m=19$ and $n=21$. The required value $n^2 - m^2$ was then calculated using the difference of squares formula.

The final answer is \boxed{80}, which corresponds to option (B).