Question

an interesting probability question...

Answer 1

This looks like one of those "Art of Problem Solving" Math Olympiad style questions. :) How far have you gotten with it before getting stuck?

Answer 2

one thing is for sure ..
n>m

Answer 3

read it and find its value and put after getting the value of it ,put this equal to m/n. And find m and n after comparing both side ...and find m+n...

Answer 4

hey Experiment....have you solved it?

Answer 5

@experiment r u der?

Answer 6

haven't tried yet.

Answer 7

well, if you want to collaborate ... i'll join.

Answer 8

ok where is your problem?

Answer 9

well, how did you do that?? I mean how did you multiply that?? on the second step??

Answer 10

because of these are independent event... hence by using multiplication theorem ,i have do as..

Answer 11

In my opinion, the probability might be higher.
let's try for p(g1)....(g4)r(1) != r(1)p(g1)....(g4)

Answer 12

The required probability will be \[64\div 495\]
hence m=64 and n=495 i.e m+n=559

Answer 13

@artofspeed,hey r u der?

Answer 14

yes

Answer 15

im trying to understand it lol

Answer 16

@experimentX yes GGGGG are missing there Experiment by mistake ok correct that..

Answer 17

also ...we can say that we should consider every possiblies ... accounting the factor that arrangement of two types of candy.

Answer 18

so.. what's the final solution? :S

Answer 19

I'll answer tonight.

Answer 20

thanks

Answer 21

If we take total number of ways to pick 5 candies as \(\Large_{12}P_5\)

Out of these acceptable arrangements are:
GRRRR ---> 8*4*3*2*1
GGRRR ---> 8*7*4*3*2
GGGRR ---> 8*7*6*4*3
GGGGR ---> 8*7*6*5*4
GGGGG ---> 8*7*6*5*4
\[\frac mn = \Large{8\cdot 4\cdot6\left( 1+7+21+35+35 \right)\over 12\cdot11\cdot10\cdot9\cdot8}=\huge\frac15\]

m+n=6