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....???
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?
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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... :(
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
62celestine 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..
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13
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