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 $E_2=\frac{1}{4}$

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

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

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

So, according to Baye's theorem,

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

Where, $P\left(\frac{E_1}{A}\right)$ is the probability of getting a heads a red chip was drawn

$=\frac{\frac{3}{4}\times \frac{5}{7}}{\frac{3}{4}\times \frac{5}{7}+\frac{1}{4}\times \frac{4}{11}}$

$=\frac{\frac{15}{28}}{\frac{15}{28}+\frac{1}{11}}$

$=\frac{15\times 11}{15\times 11+28}$

$=\frac{165}{193}$