13if i don't consider the random division in two equal parts...still i m getting the answer...n-1/2n-1....the ans given in the book
A pack contains n distinct red cards and n distinct white cards.This pack of 2n cards is divided into 2 equal parts at random.A card is drawn from each part at random ...find the probability that both the cards are of same colour?
-
UP 0 DOWN 0 2 11
11 Answers
62suppose that one pack has r red cards
Probability of this is
nCr.nC(n-r)/2nCn
probability of chosing similar colored card from both the packs is
{r(n-r)+(n-r).r}/n2
Now i think this should be a "simple" (not very ) series to sum.
62btw i am sure there must be a simpler "trick" that i am unable to think... :(
1362celestine i tried doing it by symmetry first .... but could not! :(
Could you give me ur reasoning.. it will be interesting to see.. cos i could nto figure how the two cases are the same....
33I think drawing 1 card from each part is same as choosing two cards from the entire set..so the answer should be 2nC1/2nC2
I think it is the same case which you told for
http://targetiit.com/iit_jee_forum/posts/a_pack_of_cards_236.html
62This is what i first thought too... but i have reservations... I am personally not fully convinced that they will be the same..
if you are it is gr8.. may be u can tell how!
33Don't know how but only a little instinct... [11]
If we divide the entire set after reshuffling it and select first card from both sets.... it is same as the probability of 1st card and (n+1)th card of being same color..
[7]
13I am also personally not fully convinced...thaz y i ve asked for it.....but it is giving the right ans.....but wat should be right way to do this....???
9sry nish for replyin so late jus resaw this post in my bookmarks now
yup ur exp correctly leads to final ans as 2nCn-2/ 2nCn
another way is by using symmetry
consider that 1 card is chosen , prob of this =1
now rest of the cards hve eql prob of bein chosen as 2nd card regarding of their arrangement wen u consider all cases together .
so n-1 cards give fav output against 2n-1 tot cards hence
P = n-1 / 2n - 1