Question

Balls are randomly removed from a bag without replacement. If the probability that the first 5 balls drawn from the bag are all green is 1/2, what is the least possible number of balls in the bag at the start?

the answer would be 10, right?

Answer 1

I believe 10 is correct

Answer 2

!!! thx...
can you help me with something else?

Answer 3

...?

Answer 4

what?

Answer 5

What ordered triple of integers (a,b,c), with c>b>a>0, satisfies
(root a + root b + root c)^2=10+ root 24 +root 40 + root 60 ?

Answer 6

which, i know the last three roots can be simplified, and i finally managed to remember (i think) how to simplify that whole left side, so now i have
a+b+c+2rootab+2rootbc+2rootac+20+2root6+2root10+2root15

Answer 7

btw im having trouble typing, so i type stuff out the long way like roots... sorry about that

Answer 8

and i was thinking... ab, bc, and ac could be in some order equal to 6, 10, and 15, couldn't they? so wouldnt i make a system of equations for that?

Answer 9

sorry ... i got pulled away from the computer...this question is actually pretty easy

Answer 10

\[(\sqrt{a} + \sqrt{b} + \sqrt{c})^2=10+\sqrt{24} +\sqrt{40} + \sqrt{60}\]

\[a+b+c+2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})=10+2\sqrt{6} +2\sqrt{10} + 2\sqrt{15}\]

\[a+b+c+2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})=10+2(\sqrt{6} +\sqrt{10} + \sqrt{15})\]

so we can have \(a+b+c=10\) and \(ab=6, ac=10\) and \(bc=15\)

a solution is \(a=2, ~b=3\) and \(c=5\)

Answer 11

sorry, i left too... so basically there's no system involved, really, you just... oh. ok. i see now. thanks!!!

Answer 12

umm, how did you guys end up with 10?

Answer 13

um... was i anywhere near the right track previously?

Answer 14

oh, sorry, kevin person, that was a different problem... are you talking about the balls one? scrath the "different problem"... ok...

Answer 15

so i think it's because the probability of there being all green balls in there five drawn is one half, you would double the amount of balls at hand to get the total amount of balls.

Answer 16

i think.

Answer 17

from the 10 balls in the bag originally, how many of them are green?

Answer 18

five... the problem said that, i thought...

Answer 19

waaaiiiiit....

Answer 20

then, probability of first five balls being all green = 5/10*4/10*3/10*2/10*1/10 which clearly is not 1/2

Answer 21

ok. so then what'd you do?
i am perfectly helpless when it comes to probability, so it looks like i need instruction

Answer 22

well I just know the answer is not 10, I'm still thinking about it, gimme a sec

Answer 23

I'm sorry the above should be 5/10*4/9*3/8*2/6*1/5

Answer 24

wow it's 10 balls with 9 of them being green. Nice coincidence!! Then the required probability would be\[P=\frac{ 9 }{ 10 }\times \frac{ 8 }{ 9 }\times \frac{ 7 }{ 8 }\times \frac{ 6 }{ 7 }\times \frac{ 5 }{ 6 }\]

which simplifies nicely to 5/10 which is 1/2

Answer 25

wait, how did you do that? first off, how'd you figure 9 out of 10? or did you just do the above work backwards? that cant be what you did... what did you do?

Answer 26

I really don't know! The pattern of cancelling the numerator of one term with the denominator of the other came to my mind and I chose the first term such that the denominator of the first term would be twice the numerator of the last.

Answer 27

there are five balls so there should be five terms

Answer 28

well... ok, that makes sense... but how did you get those terms?! HOW DID YOU COME UP WITH 9 OUT OF 10?!

Answer 29

so 10 and 5 should be the denominator of the first term and numerator of the last term, right? Since I need numerator of every term to cancel the denominator of the next term I had to choose 9, since removing one ball would mean the denominator of the second term is 9

Answer 30

ohhh... ok, i get that. ok, thanks! makes better sense, now, i think...
so the answer IS 10? or isnt it?

Answer 31

yeah, it seems so

Answer 32

ok, cool.
so i was right before, just did it wrong. lucky.