A bag contains 19 unbiased coins and one coin with head on both sides.

Question

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 ...

Accepted Answer

Define the Events:

Let $A$ be the event that an unbiased coin is drawn.

Let $B$ be the event that the coin with heads on both sides is drawn.

Let $H$ be the event that a head turns up when the coin is tossed.

State the Given Probabilities:

$P(A) = \frac{19}{20}$ (since there are 19 unbiased coins out of 20 total coins)

$P(B) = \frac{1}{20}$ (since there is 1 coin with two heads out of 20 total coins)

$P(H|A) = \frac{1}{2}$ (probability of getting a head given an unbiased coin is drawn)

$P(H|B) = 1$ (probability of getting a head given the coin with two heads is drawn)

Use Bayes' Theorem:

We want to find $P(A|H)$, which is the probability that the drawn coin was unbiased given that a head turned up. Bayes' Theorem states:

$$P(A|H) = \frac{P(H|A) \cdot P(A)}{P(H)}$$

Calculate $P(H)$:

We can find $P(H)$ using the law of total probability:

$$P(H) = P(H|A) \cdot P(A) + P(H|B) \cdot P(B)$$

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

Apply Bayes' Theorem to find $P(A|H)$:

$$P(A|H) = \frac{P(H|A) \cdot P(A)}{P(H)} = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right)}{\frac{21}{40}} = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21}$$

Determine $m$ and $n$:

Since $P(A|H) = \frac{m}{n}$ and $\gcd(m, n) = 1$, we have $m = 19$ and $n = 21$.

Calculate $n^2 - m^2$:

$$n^2 - m^2 = 21^2 - 19^2 = (21 + 19)(21 - 19) = (40)(2) = 80$$

Therefore, $n^2 - m^2 = 80$.

So, the correct answer is:

Option B: 80

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