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 to :

Let

    $$U$$ : the drawn coin is unbiased (ordinary coin)
    $$D$$ : the drawn coin is double-headed
    $$H$$ : head turns up on the toss

Bayes’ theorem states

$$P(U\,|\,H)=\dfrac{P(U)\,P(H\,|\,U)}{P(U)\,P(H\,|\,U)+P(D)\,P(H\,|\,D)}$$   $$-(1)$$

Step 1: Prior probabilities of choosing a coin

There are 20 coins in all (19 unbiased + 1 double-headed).

$$P(U)=\frac{19}{20},\qquad P(D)=\frac{1}{20}$$

Step 2: Probabilities of getting head from each type

For an unbiased coin, head appears with probability $$\frac12$$: $$P(H\,|\,U)=\frac12$$.
For a double-headed coin, head always appears: $$P(H\,|\,D)=1$$.

Step 3: Use Bayes’ theorem

Substituting the values into $$(1)$$:

$$P(U\,|\,H)=\dfrac{\frac{19}{20}\times\frac12}{\frac{19}{20}\times\frac12+\frac{1}{20}\times1}$$

Simplify numerator and denominator separately:

Numerator  $$=\frac{19}{20}\times\frac12=\frac{19}{40}$$

Denominator $$=\frac{19}{40}+\frac{1}{20}=\frac{19}{40}+\frac{2}{40}=\frac{21}{40}$$

Hence

$$P(U\,|\,H)=\frac{\frac{19}{40}}{\frac{21}{40}}=\frac{19}{21}$$

Step 4: Identify $$m$$ and $$n$$

The required probability is $$\frac{m}{n}=\frac{19}{21}$$ with $$\gcd(19,21)=1$$, so $$m=19$$ and $$n=21$$.

Step 5: Compute $$n^2-m^2$$

$$n^2-m^2=21^2-19^2=(21-19)(21+19)=2\times40=80$$

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

Option A is correct.