1. ## involving Baye's rule

Each of k jars contains m white and n black balls. A ball is randomly chosen from jar 1 and transferred to jar 2 then a ball is randomly chosen from jar 2 and transferred to jar 3 etc. Finally, a ball is chosen from jar k. Show that the probability that the last ball is white is the same as the probability that the first ball is white, i.e. it is m/(m+n)

2. Originally Posted by ashimb9
Each of k jars contains m white and n black balls. A ball is randomly chosen from jar 1 and transferred to jar 2 then a ball is randomly chosen from jar 2 and transferred to jar 3 etc. Finally, a ball is chosen from jar k. Show that the probability that the last ball is white is the same as the probability that the first ball is white, i.e. it is m/(m+n)
Probabilities for Jar 2:

Probability Jar 2 gets a black is n/(m + n)
Probability Jar 2 gets a white is m/(m + n).

If Jar 2 gets a black then it has m white and n+1 black. If Jar 2 gets a white then it has m+1 white and n black.

Probabilities for Jar 3:

If a black got put into Jar 2,then:

Probability Jar 3 gets a black is (n+1)/(m+n+1)
Probability Jar 3 gets a white is m/(m+n+1).

If a white got put into Jar 2, then:

Probability Jar 3 gets a black is n/(m+n+1)
Probability Jar 3 gets a white is (m+1)/(m+n+1).

So the probability of getting a white from Jar 3 is:

$\displaystyle \left( \frac{n}{m+n} \right) \, \left( \frac{m}{m+n+1} \right) + \left( \frac{m}{m+n} \right) \, \left( \frac{m+1}{m+n+1} \right)$

$\displaystyle = \frac{n m}{(m+n)(m+n+1)} + \frac{m (m+1)}{(m+n)(m+n+1)}$

$\displaystyle = \frac{nm + m (m+1)}{(m+n)(m+n+1)} = \frac{m (n + m+1)}{(m+n)(m+n+1)} = \frac{m}{m+n}$.

Similarly the probability of getting a black from Jar 3 is $\displaystyle \frac{n}{m+n}$.

A tree diagram makes it all very clear.

Obviously exactly the same argment can now be used to get the probaility of getting a white or black black from Jar 4, 5, 6, ...... k.

3. thankyou very much for your help.