Probability Problem: Unbiased Coin Scenario

This problem involves calculating a conditional probability using Bayes' Theorem.

Defining Events and Probabilities

Let U be the event that the drawn coin is unbiased.

Let B be the event that the drawn coin is the biased (double-headed) coin.

Let H be the event that the coin toss results in a head.

There are 20 coins in total (19 unbiased + 1 biased).

• Probability of drawing an unbiased coin: $P(U) = \frac{19}{20}$
• Probability of drawing the biased coin: $P(B) = \frac{1}{20}$
• Probability of getting a head if the coin is unbiased: $P(H|U) = \frac{1}{2}$
• Probability of getting a head if the coin is biased (double-headed): $P(H|B) = 1$

Applying Bayes' Theorem

We need to find the probability that the coin was unbiased given that a head turned up, which is $P(U|H)$.

Bayes' Theorem states: $P(U|H) = \frac{P(H|U) P(U)}{P(H)}$

First, calculate the total probability of getting a head, $P(H)$:

$P(H) = P(H|U) P(U) + P(H|B) P(B)$

$P(H) = (\frac{1}{2} \times \frac{19}{20}) + (1 \times \frac{1}{20})$

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

Now, apply Bayes' Theorem:

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

$P(U|H) = \frac{19}{21}$

Calculating $n^2 - m^2$

The probability that the drawn coin was unbiased is given as $\frac{m}{n}$, where gcd(m, n) = 1.

From our calculation, $P(U|H) = \frac{19}{21}$.

Therefore, $m = 19$ and $n = 21$.

We check the greatest common divisor: gcd(19, 21) = 1, which satisfies the condition.

We need to calculate $n^2 - m^2$:

$n^2 - m^2 = 21^2 - 19^2$

Using the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$):

$n^2 - m^2 = (21 - 19)(21 + 19)$

$n^2 - m^2 = (2)(40)$

$n^2 - m^2 = 80$