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Simple Probability problem - I am idiot ??

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Director
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Simple Probability problem - I am idiot ?? [#permalink] New post 19 Oct 2004, 07:20
Hi everyone.
I am actually trying to go beyond basic prob tests questions to be able to manage a possible confusing one during the real test.
In this way I am trying to solve each problem, if relevant, by the prob or the comb paths.
One of them, really simple, confuses me :

5 marbles in a bag, 2 red, 3 blue, 2 picks without replacement.
Prob(at least one red) ?

By the prob way Prob = prob(RR)+prob(RB)+prob(BR) = 2/20+6/20+6/20 = 7/10

Now if we consider the various way to pick 2 marbles from the bag : 5C2

2C1 to select 1 red marble and then 4 ways to select the remaining one...

result : 4*2C1/5C2 = 8/10

I have the feeling i am back 3 months ago : that is stupid but I do not succeed in pointing out what is wrong... Well maybe not at all I know that in the comb way the RR case is overweighted (2/10 in place of 1/10) but I do not have any idea to express it in a different way
CIO
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 [#permalink] New post 19 Oct 2004, 08:41
I think you've got two flaws here.

The first is that you're considering this to be a combinations question, where the order doesn't matter. So you're saying there are 5C2 ways to pick the balls, which comes out to 10, but in fact, there are 20 ways to do it. Then you're applying the same thing to the at least one red part, and minimizing the number of pics. The way you're doing it, there are actually just four pics, not 8.

You can't confuse "choosing" with "order matters". When you worked out your numerator, you figured this case: Red first, then either red, or blue, or blue, or blue, of which there are 4 possibilities:

R1 R2
R1 B1
R1 B2
R1 B3

Then you multiplied this by 2, because it could also have started with R2. That's right, and so now you've also got:

R2 R1
R2 B1
R2 B2
R2 B3

And so you've got 8/10. But what you're neglecting are all the ways that this could happen if blue comes first, of which there are 6. So there are 14 all together. But now you've got 14 out of 10, because you used combinations in the denominator, when you should have used permutations.

Otherwise, if you did it with combinations, you should see that the double R's are counted twice in that method, so you should eliminate one of them, getting 7/10.

Personally, I would use the first method you wrote over the second method any day.
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 [#permalink] New post 19 Oct 2004, 09:31
twixt, remember one thing:
Whenever you see "at least" in this kind of question,
Think of "1-" (one minus).

This is how you could solve the problem:
1-P(Only blue) = 1 - (3/5)*(2/4) = 7/10
Or if you insist on counting combinations:
1- (3C2 / 5C2 ) = 7/10.

Hope this helps..
CIO
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 [#permalink] New post 19 Oct 2004, 09:43
Dookie wrote:
twixt, remember one thing:
Whenever you see "at least" in this kind of question,
Think of "1-" (one minus).

This is how you could solve the problem:
1-P(Only blue) = 1 - (3/5)*(2/4) = 7/10
Or if you insist on counting combinations:
1- (3C2 / 5C2 ) = 7/10.

Hope this helps..


Great point, Dookie...
Director
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 [#permalink] New post 19 Oct 2004, 13:03
Thank you guys. That's crystal clear !
  [#permalink] 19 Oct 2004, 13:03
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