Question

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

• A. 80
• B. 60
• C. 72
• D. 64

Hint:

In probability problems involving conditional probability, Bayes' Theorem is a powerful tool to compute the probability of an event given some evidence.

Answer 1

The Correct Option is A

Let's solve the problem utilizing probability concepts and ensure a step-by-step approach with clear explanations. The task is to find \( n^2 - m^2 \) where \( \frac{m}{n} \) is the probability that a coin drawn from the bag is unbiased, given that heads turn up.

The bag contains a total of 20 coins: 19 unbiased coins and one coin with two heads.

When a coin is drawn at random, the probability of drawing the unbiased coin is \( \frac{19}{20} \) and the two-headed coin is \( \frac{1}{20} \).

For obtaining heads:

• If the unbiased coin is drawn, the probability of getting heads is \( \frac{1}{2} \).
• If the two-headed coin is drawn, heads is certain, so the probability is 1.

Using Bayes' theorem, calculate the probability that the drawn coin is unbiased given that heads appear:

\[P(\text{Unbiased} \mid \text{Heads}) = \frac{P(\text{Heads} \mid \text{Unbiased}) \cdot P(\text{Unbiased})}{P(\text{Heads})}\]

• \( P(\text{Heads} \mid \text{Unbiased}) = \frac{1}{2} \)
• \( P(\text{Unbiased}) = \frac{19}{20} \)
• \( P(\text{Heads}) = P(\text{Heads} \mid \text{Unbiased}) \cdot P(\text{Unbiased}) + P(\text{Heads} \mid \text{Two-headed}) \cdot P(\text{Two-headed}) \)
• \( P(\text{Heads}) = \frac{1}{2} \cdot \frac{19}{20} + 1 \cdot \frac{1}{20} = \frac{19}{40} + \frac{1}{20} = \frac{21}{40} \)

Substitute these into the Bayes' theorem formula:

\[P(\text{Unbiased} \mid \text{Heads}) = \frac{\frac{1}{2} \cdot \frac{19}{20}}{\frac{21}{40}} = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21}\]

Thus, \( \frac{m}{n} = \frac{19}{21} \) where \( m = 19 \) and \( n = 21 \).

Calculate \( n^2 - m^2 \):

\[n^2 - m^2 = 21^2 - 19^2 = (21 + 19)(21 - 19) = 40 \times 2 = 80\]

Therefore, the answer is 80.

Answer 2

Step 1: Understanding the problem.
We are given that there are 19 unbiased coins and 1 biased coin in the bag. The problem asks for the probability that the drawn coin was unbiased given that heads turned up. This is a conditional probability problem, and we can apply Bayes' Theorem to solve it. 
Let: 
- \( A \) be the event that the coin drawn is unbiased. 
- \( B \) be the event that heads turns up. We are required to find \( P(A | B) \), the probability that the coin is unbiased given that heads turned up.
According to Bayes' Theorem: \[ P(A | B) = \frac{P(B | A) P(A)}{P(B)} \] 
Step 2: Finding individual probabilities.
\( P(A) = \frac{19}{20} \), the probability of drawing an unbiased coin. \( P(B | A) = \frac{1}{2} \), the probability of getting heads when the coin is unbiased. \( P(B | A^c) = 1 \), the probability of getting heads when the coin is biased (since both sides are heads). \( P(A^c) = \frac{1}{20} \), the probability of drawing the biased coin. Now, calculate \( P(B) \), the total probability of getting heads: \[ P(B) = P(B | A) P(A) + P(B | A^c) P(A^c) \] \[ P(B) = \left(\frac{1}{2}\right) \left(\frac{19}{20}\right) + 1 \left(\frac{1}{20}\right) \] \[ P(B) = \frac{19}{40} + \frac{1}{20} = \frac{19}{40} + \frac{2}{40} = \frac{21}{40} \] 
Step 3: Using Bayes' Theorem.
Now, applying Bayes' Theorem: \[ P(A | B) = \frac{P(B | A) P(A)}{P(B)} = \frac{\left(\frac{1}{2}\right) \left(\frac{19}{20}\right)}{\frac{21}{40}} \] \[ P(A | B) = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21} \] This gives the probability that the coin was unbiased as \( \frac{19}{21} \). So, \( m = 19 \) and \( n = 21 \). 
Step 4: Finding \( n^2 - m^2 \).
Now we calculate \( n^2 - m^2 \): \[ n^2 - m^2 = 21^2 - 19^2 = (21 + 19)(21 - 19) = 40 \times 2 = 80 \]