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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability2 Marks
Question
There are three coins. One is a two headed coin $($having head on both faces$),$ another is a biased coin that comes up heads $75\%$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
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Answer

Let $E_{1 }$ : Two headed coin,
$E_2$ : Biased coin
and $E_3$ : Unbiased coin.
Let $A$ : Coin shows head.
Then $P (E_1) = P (E_2) = P (E_3) = \frac{1}{3}$
As the two $-$ headed coin has head on both sides so it will show head only.
Also $P(\frac{A}{E_1}) = P ($correct answer given that he knows$) = 1$
And $P(\frac{A}{E_2}) = P ($coin shows head given that the coin is biased$) = 75\% = \frac{75}{100}=\frac{3}{4}$
And $P(\frac{A}{E_3}) = P ($coin shows head given that the coin is unbiased$) = \frac{1}{2}$
Now the probability that the coin is two headed, being given that it shows head, is $P(\frac{E_1}{A}$).
By using Bayes’ theorem, we have:
$\mathrm{P}\left(\mathrm{E}_{1} | \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} | \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} | \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} | \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\mathrm{A} | \mathrm{E}_{3}\right)}$
$= \frac{\frac{1}{3} \cdot 1}{\frac{1}{3} \cdot 1+\frac{1}{3} \cdot \frac{3}{4}+\frac{1}{3} \cdot \frac{1}{2}}$
$= \frac{\frac{1}{3}}{\frac{1}{3}\left(1+\frac{3}{4}+\frac{1}{2}\right)}$
$= \frac{\frac{1}{9}}{\frac{9}{4}}=\frac{4}{9}$
$\therefore  P(\frac{E_1}{A}) = \frac{4}{9}$

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