Probability Q's (2 Viewers)

[Master E]

Hi, a few questions:

1) A bunch of 6 mixed roses is to be made out of 12 red and 8 yellow roses. What is the probability there will be 4 red and 2 yellow roses?
What is the probability that a bunch consists of one couloured flower only?

2) 8 cards made up of 4 pairs are chosen. the 8 cards are shuffled and placed on the table in four groups of 2. what is the probability the cards are properly paired?

can anyone help? thanks

 

[aakash]

1) Total no. of ways to choose 6 roses out of 20 (possible outcomes)= 20C6=38760
no. of ways to choose 4 out of 12 red ones = 12C4
no. of ways to choose 2 out of 8 yellow ones=8C2

favourable outcomes = 12C4 X 8C2 = 13860

Req. probability = 13860/38760 = 231/646

P(all 6 are red) = 12C6 / 20C6
P(all 6 are yellow) = 8C6 / 20C6
hence P(all same color) = P(all 6 are red) + P(all 6 are yellow) = 7/285

2)Consider A1,A2,B1,B2,C1,C2,D1,D2
no. of ways to arrange them = 8!
now consider A1,A2 as single entity (coz we want them to be together) also consider B1,B2 ; C1,C2 and D1,D2 as single entity
no. of ways to arrange these = 4! * 2* 2* 2*2 (the 2's - coz A1A2 or A2A1 are both acceptable)
req. prob = (4!*2*2*2*2)/8! = 1/105


Another way : consider AABBCCDD
no. of ways to arrange = 8!/(2!2!2!2!)
(2! - since there are 2 A's and similarly for B,C,D)

possible outcomes = 4! ( considering AA, BB, CC, DD as 4 entities)

Req prob = 1/105