probability question (1 Viewer)

[duy.le]

can some one explain it to me?

A box contains n jellybeans, some white and some black. Anne and Bruce take turns picking a jellybean from the box without looking until the box is empty. Anne picks first.

(i) If there are 1 black and n-1 white jellybeans and n is odd, find the probability that Anne picks the black jellybean.

(ii) If there are 2 black and n-2 white jellybeans and n is even, find the probability that Anne is the first to pick a black jellybean.

(iii) If there are 2 black and n-2 white jellybeans and n is odd, find the probability that Anne is the first to pick a black jellybean.

ok its a bit easy but im really confused. If your gonna provide an answer can you please explain each step.

 

[Templar]

(i) The probability that Anne picks the black jelly bean is the sum of her picking it up on the first go, her second go, etc.
First go: 1/n
Second go: (n-1)/n*(n-2)/(n-1)*1/(n-2)=1/n (first two fractions are Anne and Bruce pick up white)
Third go: (n-1)/n*(n-2)/(n-1)*(n-3)/(n-2)*(n-4)/(n-3)*1/(n-4)=1/n
etc

Since by the (n+1)/2 go all jelly beans has been picked up, her chance of picking up black is (n+1)/2n

(ii) Anne can pick up black jelly bean on her first, second etc up to n/2 th go.
First go: 2/n
Second go: (n-2)/n*(n-3)/(n-1)*2/(n-2)=2/n*(n-3)/(n-1)
Third go: (n-2)/n*(n-3)/(n-1)*(n-4)/(n-2)*(n-5)/(n-3)*2/(n-4)=2/n*(n-5)/(n-1)
etc
n/2 th go: (n-2)/n*...*1/3*2/2=2/n*1/(n-1)

Summing up we get 2/n*(n-1+n-3+...+1)/n-1=2/n*n^2/4=1/2n

I'll leave (iii) for now (too much bother typing it all up, need LaTeX)