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Originally Posted by SEBOL
Hi guys. I need help to do those problems:
1) A bag contains m white and 3 black balls. Balls are drawn one by one without replacement till all the black balls are drawn. What is the probability that this procedure for drawing balls will come to an end at the rth draw?
2) A family consists of a grandfather, m sons and daughters and 2n grandchildren. They are to be seated in a row for dinner. The grandchildren wish to occupy the n seats at each end and the grandfather refuses to have a granchildren on either side of him. In how many ways can the family be made to sit?
3)There are 15 seats in the first row of a cinema hall. Find the number of ways in which 4 seats of the first row can be alloted so that no two of them are consecutive and the seat number 6 must be occupied.
4)Let F denote the set of all onto functions from A={a1,a2,a3,a4} to B={x,y,z}. A function f is chosen at random from F. Find the probability that f^(-1) (x) consists of exactly one element.
Thanks for any help |
please verify if u have answers with u
Sol1: at r th draw we shud have drawn the 3rd black ball..right?
so at r-1 th draw ,we shud have 2 black ball out and r-3 white ball out
hence the probability goes
lets suppose we take out white balls first and in the last 3 draw we take out those 3 black balls
[m*(m-1)*(m-2)....m-(r-3)*3*2*1] / [ (m+3)*(m+2)...(m-r)*(m-r+1)*(m-r+2)*(m-r+3)]
== [6* m!/(m-r+4)!] / [ (m+3)!/ (m-r+4)! ]
== [6*m!/(m+3)!]
now see the white n black balls can be permuted into each other
and the permutation will be (r-2)*(r-1)/2;
hence ans is [(r-2)*(r-1)/2]*[6*m!/(m+3)!]
regards
nitin