Question

Urn I contains two white chips and one red chip; urn II has one white chip and two red chips. One chip i drawn at random from urn I and transferred to urn II. then onechip is drawn from urn II. Suppose that a red chip is selected from urn II. what is the probability that the chip transferred was white?

Answer 1

Let,

The event that a red chip is selected from urn II be denoted by \(B\)

The event that the chip transferred from urn I is white be denoted by \(A_{1}\).

The event that the chip transferred from urn I is red be denoted by \(A_{2}\).

The probability that the chip transferred from urn I is white is denoted as \(P\left(A_{1}\right)\)

Therefore,

\(P\left(A_{1}\right)=\frac{2}{3} \ldots \ldots(1 .)\)

This implies, probability that the chip transferred from urn I is red is denoted as \(P\left(A_{2}\right)\) and is obtained as

\(P\left(A_{2}\right)=1-\frac{2}{3}\)

\(P\left(A_{2}\right)=\frac{1}{3} \ldots \ldots\) (2.)

The probability that a red chip is selected from urn II when a white chip is transferred from urn I is denoted by \(P\left(B \mid A_{1}\right)\). Therefore,

\(P\left(B \mid A_{1}\right)=\frac{2}{4} \ldots \ldots\) (3.)

The probability that a red chip is selected from urn II when a red chip is transferred from urn I is denoted by \(P\left(B \mid A_{2}\right)\). Therefore,

\(P\left(B \mid A_{2}\right)=\frac{3}{4} \ldots \ldots\) (4.)

We need to determine the probability that a white chip is transferred from urn I given that a red

chip is selected from urn II. In other words we need to determine \(P\left(A_{1} \mid B\right)\).

By Baye's theorem,

$$ P\left(A_{j} \mid B\right)=\frac{P\left(B \mid A_{j}\right) P\left(A_{j}\right)}{\sum_{i=1}^{n} P\left(B \mid A_{i}\right) P\left(A_{i}\right)} $$

Therefore,

$$ P\left(A_{1} \mid B\right)=\frac{P\left(B \mid A_{1}\right) P\left(A_{1}\right)}{P\left(B \mid A_{1}\right) P\left(A_{1}\right)+P\left(B \mid A_{2}\right) P\left(A_{2}\right)} \cdots \cdots (5.) $$

Substituting the values from (1.), (2.), (3) and (4.) in (5.) we get

$$ \begin{array}{l} P\left(A_{1} \mid B\right)=\frac{\left(\frac{2}{4}\right)\left(\frac{2}{3}\right)}{\left(\frac{2}{4}\right)\left(\frac{2}{3}\right)+\left(\frac{3}{4}\right)\left(\frac{1}{3}\right)} \\ =\frac{\left(\frac{4}{12}\right)}{\left(\frac{4}{12}\right)+\left(\frac{3}{12}\right)} \\ =\frac{\left(\frac{4}{12}\right)}{\left(\frac{7}{12}\right)} \end{array} $$

\(P\left(A_{1} \mid B\right)=\frac{4}{7}\)