Question

From a bag containing n balls, all either black or white, all numbers for each being equally likely, a ball is drawn and turns out to be white. This is replaced and another ball is drawn, which is also white. If the ball is replaced prove that chance of drawing black ball in next try is $\dfrac{1}{2}\left( n-1 \right){{\left( 2n+1 \right)}^{-1}}$.

Answer 1

First of all, we do not know how many of the n balls are white or black. So, to begin with, we will find the probability of the bag having k balls. Then, we will apply the Bayes’ Theorem to find the probability of drawing a black ball in the third draw, given that the first and the second ball is white with replacement.

Complete step by step answer:
Let W(i) be the event that a white ball is drawn in the ${{i}^{th}}$ draw and let B(i) be the event that a black ball is drawn in the ${{i}^{th}}$ draw.
Now, it is given that the bag has only two types of balls, either black or white, so, B(i) + W(i) = 1 and B’(i) = W(i), where B’ is the complement of B.
We do not know the number of white balls in the bag.
Suppose, there are k white balls in the bag, where k can be any integer from 0 to n.
Therefore, k = 0 or k = 1 or so on k = n. So, there are (n + 1) ways in which white balls can be there in the bag.
So, if A(k) is the probability of having k white balls in the bag, so $P\left( A\left( k \right) \right)=\dfrac{1}{n+1}$, as it can be any one way of the (n + 1) ways.
We know that $P\left( B|A \right)=\dfrac{P\left( A,B \right)}{P\left( A \right)}$, where $P\left( B|A \right)$ is the probability of B happening given that A has happened already and $P\left( A,B \right)$ is the probability of A and B happening.
We need to find the probability that the third ball is black, when first and second balls are white. Therefore, we need to find $P\left( B\left( 3 \right)|\left( W\left( 2 \right),W\left( 1 \right) \right) \right)$.
$\Rightarrow P\left( B\left( 3 \right)|W\left( 2 \right),W\left( 1 \right) \right)=\dfrac{P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right) \right)}{P\left( W\left( 2 \right),W\left( 1 \right) \right)}$
We will evaluate the numerator and denominator separately.
The numerator is $P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right) \right)$.
We shall iterate $P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right) \right)$ with the varying number of white balls in the bag.
$P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right) \right)=P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right),A\left( 0 \right) \right)+P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right),A\left( 1 \right) \right)+...+P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right),A\left( n \right) \right)$
We can say that $P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right),A\left( 0 \right) \right)=P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right),A\left( n \right) \right)=0$
Because if there are no white balls or no black balls, in both events, our desired event will be impossible.
Therefore, $P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right) \right)=\sum{P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right),A\left( k \right) \right)}$

& P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right) \right)=\sum{P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right),A\left( k \right) \right)} \\\ & =\sum{P\left( A\left( k \right) \right)\times P\left( W\left( 1 \right)|A\left( k \right) \right)\times P\left( W\left( 2 \right)|A\left( k \right)W\left( 1 \right) \right)\times P\left( B\left( 3 \right)|A\left( k \right)W\left( 1 \right)W\left( 2 \right) \right)} \end{aligned}$$ Now, probability of drawing a white ball in the first if there are k white balls in the bag containing total n ball will be given as $P\left( W\left( 1 \right)|A\left( k \right) \right)=\dfrac{k}{n}$. Probability of drawing another white ball in the same conditions is $P\left( W\left( 2 \right)|A\left( k \right)W\left( 1 \right) \right)=\dfrac{k}{n}$, as it is given that there is a replacement. Probability of drawing a black ball from a bag with (n – k) black balls and total n balls is given as $P\left( B\left( 3 \right)|A\left( k \right)W\left( 1 \right)W\left( 2 \right) \right)=\dfrac{n-k}{k}$. $$\begin{aligned} & P\left( B\left( 3 \right),W\left( 2 \right),W\left( 1 \right) \right)=\sum{P\left( A\left( k \right) \right)\times P\left( W\left( 1 \right)|A\left( k \right) \right)\times P\left( W\left( 2 \right)|A\left( k \right)W\left( 1 \right) \right)\times P\left( B\left( 3 \right)|A\left( k \right)W\left( 1 \right)W\left( 2 \right) \right)} \\\ & =\sum{\dfrac{1}{n+1}\times \dfrac{k}{n}\times \dfrac{k}{n}\times \dfrac{\left( n-k \right)}{n}} \end{aligned}$$ Now, we will simplify the denominator $P\left( W\left( 2 \right),W\left( 1 \right) \right)$ $P\left( W\left( 2 \right),W\left( 1 \right) \right)=P\left( W\left( 2 \right),W\left( 1 \right),A\left( 0 \right) \right)+P\left( W\left( 2 \right),W\left( 1 \right),A\left( 1 \right) \right)+...+P\left( W\left( 2 \right),W\left( 1 \right),A\left( n \right) \right)$ Now, $P\left( W\left( 2 \right),W\left( 1 \right),A\left( 0 \right) \right)=0$, as for drawing white balls, there should be at least one white ball in the bag. $\begin{aligned} & P\left( W\left( 2 \right),W\left( 1 \right) \right)=\sum{P\left( W\left( 2 \right),W\left( 1 \right),A\left( k \right) \right)} \\\ & =\sum{P\left( A\left( k \right) \right)\times P\left( W\left( 1 \right)|A\left( k \right) \right)\times P\left( W\left( 2 \right)|A\left( k \right),W\left( 1 \right) \right)} \\\ & =\sum{\dfrac{1}{n+1}\times \dfrac{k}{n}\times \dfrac{k}{n}} \end{aligned}$ $\Rightarrow P\left( B\left( 3 \right)|W\left( 2 \right),W\left( 1 \right) \right)=\dfrac{\sum{\dfrac{1}{n+1}\times \dfrac{k}{n}\times \dfrac{k}{n}\times \dfrac{\left( n-k \right)}{n}}}{\sum{\dfrac{1}{n+1}\times \dfrac{k}{n}\times \dfrac{k}{n}}}$ Terms which do not have k can be taken outside the summation and can be cancelled. $\Rightarrow P\left( B\left( 3 \right)|W\left( 2 \right),W\left( 1 \right) \right)=1-\dfrac{\sum{{{k}^{3}}}}{n\sum{{{k}^{2}}}}$ We know that $\sum{{{k}^{3}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}}$ and $\sum{{{k}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ $$\begin{aligned} & \Rightarrow P\left( B\left( 3 \right)|W\left( 2 \right),W\left( 1 \right) \right)=1-\dfrac{\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}}{\dfrac{{{n}^{2}}\left( n+1 \right)\left( 2n+1 \right)}{6}} \\\ & \Rightarrow P\left( B\left( 3 \right)|W\left( 2 \right),W\left( 1 \right) \right)=1-\dfrac{3\left( n+1 \right)}{2\left( 2n+1 \right)} \\\ & \Rightarrow P\left( B\left( 3 \right)|W\left( 2 \right),W\left( 1 \right) \right)=\dfrac{\left( n-1 \right)}{2\left( 2n+1 \right)} \\\ \end{aligned}$$ **Note:** $P\left( A,B \right)$ can also be written as $P\left( A\cap B \right)$. Therefore, $P\left( B \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$. When A and B are independent events, $P\left( A\cap B \right)=P\left( A \right)P\left( B \right)$.