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Two coins are available, one fair and the other two-headed. Choose a coin and toss it once; assume that the unbiased coin is chosen with probability $\frac{3}{4}$, Given that the outcome is head, the probability that the two-headed coin was chosen, is
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Verified Answer
The correct answer is:
$\frac{2}{5}$
Let $F$ denotes fair coin $T$ denotes two headed $H$ denotes head occurs
$$
\begin{array}{lc}
\therefore \quad P(F)=\frac{3}{4}, P(T)=1-\frac{3}{4}=\frac{1}{4} \\
\therefore \quad P\left(\frac{T}{H}\right)=\frac{P\left(\frac{H}{T}\right) \cdot P(T)}{P\left(\frac{H}{T}\right) \cdot P(T)+P\left(\frac{H}{F}\right) \cdot P(F)}
\end{array}
$$
By Baye's theorem
$$
=\frac{1 \cdot \frac{1}{4}}{1 \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4}}=\frac{2}{5}
$$
$$
\begin{array}{lc}
\therefore \quad P(F)=\frac{3}{4}, P(T)=1-\frac{3}{4}=\frac{1}{4} \\
\therefore \quad P\left(\frac{T}{H}\right)=\frac{P\left(\frac{H}{T}\right) \cdot P(T)}{P\left(\frac{H}{T}\right) \cdot P(T)+P\left(\frac{H}{F}\right) \cdot P(F)}
\end{array}
$$
By Baye's theorem
$$
=\frac{1 \cdot \frac{1}{4}}{1 \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4}}=\frac{2}{5}
$$
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