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# There are 2 blue and 3 red balls in a bag. If two balls are

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Director
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There are 2 blue and 3 red balls in a bag. If two balls are [#permalink]

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06 Feb 2007, 22:33
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There are 2 blue and 3 red balls in a bag. If two balls are picked w/o replacement, what is the prob of getting at least 1 blue?

I tried it this way and messed up:

All outcomes
10C5 = 10

Fav outcomes
2C1 = 2

2C2 = 2(1)/(2)(0) (there's obviously something wrong here)

Can anyone explained where I messed up and whether this method is applicable? I've since realized it's not the quickest way but I need to know why this didn't work.
Intern
Joined: 20 Jan 2007
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06 Feb 2007, 23:26
2C2= 2!/2!0!

What I assume you did not new is that 0!=1

Anyways... what I would do is the following:

the probability of having at least one blue is the complementary probability of getting 2 red balls withouth replacement. That is:

P(at least 1 blue ball)=1 - P(two red balls)

P(two red balls)=P(1st ball=red)Â·P(2nd ball=red)

Since there is no replacement:

P(1st ball=red)= 3/5
P(2nd ball=red)=2/4 (2 red balls left out of 4 balls)

P(at least 1 blue ball)= 1-3/5Â·2/4 = 1-6/20 = 14/20=7/10=70%

Hope it helps!
Director
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06 Feb 2007, 23:44
Yes. Your solution is the quicker solution I soon realized. Thanks very much. But can anyone tell me what is wrong w/the combo method?

All outcomes
10C5 = 10

Fav outcomes
2C1 = 2

2C2 = 2(1)/(2)(0!) = 1

2+1/10 = 3/10 - what am I missing?
Senior Manager
Joined: 04 Jan 2006
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08 Feb 2007, 08:35
ggarr wrote:
There are 2 blue and 3 red balls in a bag. If two balls are picked w/o replacement, what is the prob of getting at least 1 blue?

I tried it this way and messed up:

All outcomes
10C5 = 10

Fav outcomes
2C1 = 2

2C2 = 2(1)/(2)(0) (there's obviously something wrong here)

Can anyone explained where I messed up and whether this method is applicable? I've since realized it's not the quickest way but I need to know why this didn't work.

Umm, 2C2 = (2!)/(2! x 0!)
and 0! = 1

Therefore, 2C2 = 2/(2 x 1) = 1
Senior Manager
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08 Feb 2007, 08:43
Now I will solve this question by using your approach. Your method is correct but need a little bit of adjustment in logic.

Question: Find Prob[ Blue >= 1]

This is kind of advance probability question. You can split Prob[ B >= 1] into 2 cases.

Prob[Blue >=1 ] = Prob[Blue = 1] + Prob[Blue = 2]

Prob[Blue=1] = (pick 1 blue) x (pick 1 red) / (pick 2 ball randomly)
= (2C1) x (3C1)/(5C2)
= (2 x 3) / (10) = 6/10

Prob[Blue=2] = (pick 2 blue) / (pick 2 ball randomly)
= (2C2) / (5C2)
= 1/10

Therefore; Prob[Blue >= 1] = 6/10 + 1/10 = 7/10

Or, you can use another method,which has already been shown by one of our friend above me, to solve this question.
Prob[Blue>=1] = 1 - Prob[Red=2]

Prob[Red = 2] = (3C2)/(5C2)
= 3/10

Prob[Blue>=1] = 1 - Prob[Red=2] = 1 - 3/10 = 7/10
Manager
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08 Feb 2007, 23:05
ggarr wrote:
There are 2 blue and 3 red balls in a bag. If two balls are picked w/o replacement, what is the prob of getting at least 1 blue?

I tried it this way and messed up:

All outcomes
10C5 = 10

Fav outcomes
2C1 = 2

2C2 = 2(1)/(2)(0) (there's obviously something wrong here)

Can anyone explained where I messed up and whether this method is applicable? I've since realized it's not the quickest way but I need to know why this didn't work.

I will solve it like this:

1 blue ball: 2C1 * 3C1 / 5C2 = 6/10 = 0.6
both balls are blue: 2C2/10 = 1/10 = 0.1

total p = 0.6 + 0.1 = 0.7
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Re: Prob prob   [#permalink] 08 Feb 2007, 23:05
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