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Re: The last one probe for today: There are 8 balls in a box; [#permalink]
I agree with your criticism. Forgive me. :pray
I can only say that that day I was pretty tired inventing questions, and that this one was the 20th or 21st one.

Eliminate this question?
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Re: The last one probe for today: There are 8 balls in a box; [#permalink]
stolyar wrote:
I agree with your criticism. Forgive me. :pray
I can only say that that day I was pretty tired inventing questions, and that this one was the 20th or 21st one.

Eliminate this question?


Just rewrite the questions so that the assumptions are clear. BTW, i KNOW firsthand how hard it is to write good questions so kudos to you for even trying.
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Re: The last one probe for today: There are 8 balls in a box; [#permalink]
What is that?

Also four balls are taken without replacement, if we said four at one chunk, it would be 1/2.

:lol:

Please respond to message I left you on PS:Shoes,
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Re: The last one probe for today: There are 8 balls in a box; [#permalink]
Four balls are taken at random without repetition: You wrote Stolyar.

How about four balls taken without replacement.

C ( 1 ball, 1 colorless) * ..... C( 1 ball, 1 colorless)
One colorless ball cannot be replaced, but there are three others bringing the total to four !-

VT
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Re: The last one probe for today: There are 8 balls in a box; [#permalink]
4/8 * 3/7 * 2/6 * 1/5

24/(48*35)

1/70

Which is what anandk has, but expressed differently.

anand-- I just took the probability of each turn producing a colored ball, assuming the previous one was successful.

What's your logic?
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Re: The last one probe for today: There are 8 balls in a box; [#permalink]
Hell :oops:

I couldn't come out with 4/8 * 3/7 * 2/6 * 1/5
such a shame. This is the classical way of solving such problems.
Thanks to stoolfi for reminding me.
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Re: The last one probe for today: There are 8 balls in a box; [#permalink]
I wasn't trying to shame you, I want you to tell me how you came up with your answer, which, as I already wrote, is the same as mine.

I see people using combinatorics a lot when I don't use them, and I usually can't figure out why they are working...
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Re: The last one probe for today: There are 8 balls in a box; [#permalink]
Well there is one ball of each color. So picking each with out replacement is 1/8 * 1/7 * 1/6 * 1/5
But you can arrange these 4 colored balls in 4! ways right
so you have 4! * 1/8 * 1/7 * 1/6 * 1/5

This is the same technique you use when you toss 5 coins and want to have exactly 3 Heads. You just find the probablity that 3 Hs will come and then multiply the P with no of combinations that give you 3Hs. This gives you total probability.

In the above example you treated 4 colored balls the same. This is perfect because the other balls are non colored. I liked your application of probability.



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