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

We will determine \( n^2 - m^2 \) by applying probability principles and a systematic, well-explained methodology. The objective is to compute \( n^2 - m^2 \), given that \( \frac{m}{n} \) represents the probability of selecting an unbiased coin, conditional on observing heads.

The collection comprises 20 coins in total: 19 are unbiased, and one possesses two heads.

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

Regarding the outcome of heads:

• The probability of obtaining heads from an unbiased coin is \( \frac{1}{2} \).
• For the two-headed coin, heads is a certainty, hence the probability is 1.

Employing Bayes' theorem, we compute the probability that the selected coin is unbiased, given that heads occurred:

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

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

We substitute these values into Bayes' theorem:

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

Consequently, \( \frac{m}{n} = \frac{19}{21} \), which implies \( m = 19 \) and \( n = 21 \).

We now compute \( n^2 - m^2 \):

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

The final result is 80.