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 m/n where gcd(m, n) = 1, then n² − m² is equal to:

 Question

A bag contains 19 unbiased coins and one coin with heads on both sides.
One coin is drawn at random and tossed; a head turns up.
If the probability that the drawn coin was unbiased is $\dfrac{m}{n}$ where $\gcd(m,n)=1$, then find:

$$n^2-m^2$$

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Question Image

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Short Solution

1. Define events:

   • $U$: chosen coin is unbiased

   • $D$: chosen coin is double-headed

   • $H$: head turns up

2. Use Bayes’ theorem:

   $$P(U\mathrm{\mid }H)=\frac{P(U)P(H\mathrm{\mid }U)}{P(U)P(H\mathrm{\mid }U)+P(D)P(H\mathrm{\mid }D)}$$

3. Compute each term:

   • $P(U)=\dfrac{19}{20}$

   • $P(D)=\dfrac{1}{20}$

   • $P(H|U)=\dfrac{1}{2}$fair coin)

   • $P(H|D)=1$(double-headed always gives head)

4. Plug values, simplify to find $m$ and $n$.

5. Compute $n^{2}-m^{2}$
   

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Image Solution

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Conclusion

Step 1: Apply Bayes

$$P(U\mathrm{\mid }H)=\frac{\frac{19}{20}\cdot \frac{1}{2}}{\frac{19}{20}\cdot \frac{1}{2}+\frac{1}{20}\cdot \mathrm{1<span\; class="vlist"><span\; class="mord"><span\; class="mord"><span\; class="mfrac"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span><span\; class="vlist-r"><span\; class="vlist"></span></span></span></span><span\; class="mclose\; nulldelimiter"></span></span></span></span><span\; class="vlist-s"></span>}}$$

Step 2: Simplify

• Numerator:

  $$\frac{19}{20}\cdot \frac{1}{2}=\frac{19}{\mathrm{40}}$$

• Denominator:

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

So:

$$P(U\mathrm{\mid }H)=\frac{19\mathrm{/}40}{21\mathrm{/}40}=\frac{19}{\mathrm{21}}$$

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Step 3: Identify $m$ and $n$

$$\frac{m}{n}=\frac{19}{21},m=19,n=21$$

Step 4: Compute $n^{2}-m^{2}$

$$n^2-m^2={21}^2-{19}^2=(21-19)(21+19)=2\times 40=80$$

✅ Final Answer:

$$\boxed{n^{2}-m^{2} = 80}$$

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Video Solution