Bayes Theorem Coin Probability Problem Solution

 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