A bag contains red and blue marbles totaling between 50 and 100 in number. If two marbles are drawn at random, the chances of them both being red are 25%. One third of the marbles are picked from the bag at random, and thrown away. If one marble is now drawn at random from the bag, what is the probablitity of it begin blue (in %)?
please help
thinking
at first glance it looks like half are red and half are blue, so if you throw out one third it is still half red and half blue. but this is wrong
i think we have to figure out how many are in the bag. part 2 says it is divisible by 3, since you can throw out 1/3 of them
suppose there are 99 marbles in the bag. and let x by red. then part one says \[\frac{x}{99}\times \frac{x-1}{98}=\frac{1}{2}\]
no integer solution to this
hi, thanks, could you let me know how you go 1/2
oh damn. i meant \[\frac{x}{99}\times \frac{x-1}{98}=\frac{1}{4}\]
could you please help with, why have you taken x/99* x-1/98
ok lets say to make it easy that there are 51 marbles in the bag, with x red. the probability that you select a red one is \[\frac{x}{51}\]
then select another red one. the probability that it is also red is \[\frac{x-1}{50}\] because there are x - 1 red ones left and 50 left in the bag
nice! since its just presentage at the end we can assume some x=3n=total number of balls in the bag...
so the probability that they are both red is \[\frac{x}{51}\times \frac{x-1}{50}\]\
yeah but i think there is something wrong here, because i don't think that there are integer solutions to this
maybe it means the probability they are both red are about 1/2
i mean about 1/4
since i don't think you can solve \[\frac{x}{3n}\times \frac{x-1}{3n-1}=\frac{1}{4}\] for integer x
of course i could be wrong, but i really really doubt it
ok, thanks for your help
if it means "about 1/4" then half are red and half are blue, and the answer would be 1/2
no you cannot. not possible.
only can say "about half" are red
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