Question

A biased coin which comes up heads three times as often as tails is tossed. If it shows heads, a chip is drawn from urn-I which contains $2$ white chips and $5$ red chips. If the coin comes up tail, a chip is drawn from urn-II which contains $7$ white and $4$ red chips. Given that a red chip was drawn, what is the probability that the coin came up heads?

Answer 1

Let $E_1$ be the event of getting a head, and $E_2$ be the event of getting a tail in the biased coin.

If $P\left(X\right)$ is a probability function on event $X$,

$P\left(E_1\right)=3P\left(E_2\right)=3(1-P(E_1\left)\right)$

$\Rightarrow P\left(E_1\right)=\frac{3}{4}$ and $P\left(E_2\right)=\frac{1}{4}$

Let $A$ be the event of getting a red chip.

So, $P\left(A/E_1\right)=\frac{5}{7}$ ($5$ red chips in a total of $7$ chips)

$P\left(A/E_2\right)=\frac{4}{11}$ ($4$ red chips in a total of $11$ chips)

So, according to Baye's theorem,

$P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)}$

Where $P\left(E_1/A\right)$ is the probability of getting a heads given a red chip was drawn.

ie, $P\left(E_1/A\right)=\frac{\frac{3}{4}\frac{5}{7}}{\frac{3}{4}\frac{5}{7}+\frac{1}{4}\frac{4}{11}}=\frac{15\times 11}{15\times 11+28}=\frac{165}{193}$