Meyer probability 3.37 problem question

meyer probability

Any one has solved the 3.37 problem of the meyer probability and application, it says like this :

3.37 you have n baskets, each one has w white balls and b black balls, if you take out a ball in the firts basket and put in the second basket, and take out a ball of the second basket and put in the third basket, and so on until the n basket. What is the probability of take out a white ball in the n basket, if in the first basket you take out a white ball.

Hi snaider!
Following is the solution.

Let P[k] = Probability of taking a white ball from k-th basket
So, 1 - P[k] = Probability of taking no white ball from k-th basket

So,
P[k] = P{a white ball taken from (k-1)-th basket AND then a white ball taken from k-th basket} OR P{no white ball taken from (k-1)-th basket AND then a white ball taken from k-th basket}

So,
P[k] = (P{a white ball taken from (k-1)-th basket}*P'{white}) + (P{no white ball taken from (k-1)-th basket}*P"{white})

So,
P[k] = P[k-1]* {(w+1)/(w+1+b)} + (1 - P[k-1])*{w/(w+b+1)}

P[k] = (w + P[k-1])/(w+b+1), after simplification.

And, P[1] = (w)/(w+b).


So, P[2] = {w + (w)/(w+b)}/(w+b+1)

= (w)/(w+b)

and so on ...

So, P[n] = (w)/(w+b). : ANSWER.

The result according to the book is :


P=[w/(w+b)]+[b/[(w+b)*(w+b+1)**n-1]]

** exponent

In the result you give, you did not realize that you put a extra ball in each basket, so the total balls are w+b+1, and also that this is conditional probability.