Two urns contain respectively 'a white and b black' and 'b white and a black' balls. A series of drawings is made according to the following rules:
(i) Each time only ball is drawn and immediately returned to the same urn it came from.
(ii) If the ball drawn is white, the next drawing is made from the first urn.
(iii) If it is black, the next drawing is made from the second urn.
(iv)The first ball drawn comes from the first urn.
What is the probability that nth ball drawn will be white ?

My Attempt
Probability of White ball from urn 1: $\frac{a}{a+b}$

Probability of Black ball from urn 1: $\frac{b}{a+b}$

Probability of White ball from urn2: $\frac{b}{a+b}$

Probability of Black ball from urn 2: $\frac{a}{a+b}$

I don't know how to proceed mathematically from here, but from trial and error: $$ \sum_{i=0}^n \binom {n}{i} (\frac{a}{a+b})^{n-i} (\frac{b}{a+b})^i $$

I couldn't find this question anywhere online. Sorry if repost.. Thanks in advance!


This can be modelled as a Markov chain with two states, each representing an urn we are to draw from next. Let State $0$ be Urn $1$, and State $1$ be Urn $2$.

The transition matrix is then:

$$ P = \begin{bmatrix} \frac{a}{a+b} & \frac{b}{a+b} \\ \frac{b}{a+b} & \frac{a}{a+b} \\ \end{bmatrix} $$

The event "draw a white ball on the $n^{th}$ draw" is the same as "we are in state $0$ after $n$ steps". The $n$-step transition matrix can be derived by standard methods such as diagonalisation (or Wolfram Alpha :-):

$$ P^n =\dfrac{1}{2} \begin{bmatrix} 1 + \left(\frac{a-b}{a+b}\right)^n & 1 - \left(\frac{a-b}{a+b}\right)^n \\ 1 - \left(\frac{a-b}{a+b}\right)^n & 1 + \left(\frac{a-b}{a+b}\right)^n \\ \end{bmatrix} $$

Since we begin with Urn $1$ (State $0$), the probability we require is:

$$p_{00}^n = \dfrac{1}{2} + \dfrac{1}{2}\left(\frac{a-b}{a+b}\right)^n.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.