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# Probability - Concept clarification - URGENT pls...

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Director
Joined: 15 Aug 2005
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Probability - Concept clarification - URGENT pls... [#permalink]

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29 Aug 2005, 22:18
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hey guys,

I am wondering if any one you guys can explian to me a very basic concept in probablity?

A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

A. 21/50
B. 3/13
C. 47/50
D. 14/15
E. 1/5

Method used to solve this is by determining the probability that no marbles are blue, and then by subtracting that probability from 1 and determining that probability that at least one marble is blue.

WHAT IS WRONG IF -

you do this in two steps like this -

1. atleast 1 marble is blue means either 1 marble is blue and the other is red or both of them are blue.
2. so, determine the probablity of in both cases and then add them as per the additive law.

I will really appreciate if someone were to throw some light on this.

Thanks.
_________________

Cheers, Rahul.

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Intern
Joined: 13 Aug 2005
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30 Aug 2005, 00:26
In General, you are right. Both ways would work.
As long as you remember that you need to calculate all options: R-B, B-R, B-B
The questions you should ask your self are: what is the quickest way to solve, and where you would be subjected to less mistakes.
It looks like the (1-prob) way is better way for solving this, from both time management and probable mistakes.

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30 Aug 2005, 00:32
rahul, you can do this in 2 ways

Comb/perm way:

There are 90 different possibilities if 2 marbles are drawn from 10 total.
10!/(10-2)! = 90.

Now number of combinations of at least one blue marble. there are 3 possibilities here:

1 first marble red, second marble blue
2 first marble blue, second marble red
3 both blue.

calculate possibilites for 1,2,3 and add them

1. 3*7 +
2. 7*3 +
3. 7*6

You get 84. so answer is 84/90 = 14/15

Probability way:

same as 1,2,3 above, calculate probabilites for 1,2,3 and use additive law to add them.

so you get P(B,R) + P(R,B) + P(B,B):

1. 7/10 * 3/9 = 21/90
2. 3/10 * 7/9 = 21/90
3. 7/10 * 6/9 = 42/90

adding them gives 84/90 = 14/15

good god, please tell me im correct?

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Director
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30 Aug 2005, 01:02
Hey Yaron and Decibel,

You guys are awesome! Thanks a ton! 14/15 is the right answer.
_________________

Cheers, Rahul.

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Manager
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31 Aug 2005, 03:34
The best way to tackle this question is to use the 1 - Prob method.
What is the probability of at least 1 blue? It is 1 - the prob. of having
both marbles red. In this sense:

3/10 *2/9 = 6/90

1 - 6/90 = 84/90 = 14/15
It is a much quicker way to solve this question .

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Manager
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31 Aug 2005, 03:36
tobiastt wrote:
The best way to tackle this question is to use the 1 - Prob method.
What is the probability of at least 1 blue? It is 1 - the prob. of having
both marbles red. In this sense:

3/10 *2/9 = 6/90

1 - 6/90 = 84/90 = 14/15
It is a much quicker way to solve this question .

yup I would go for this method as well.

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31 Aug 2005, 03:36
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