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A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls

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A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls [#permalink] New post 30 Sep 2012, 06:36
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

79% (01:42) correct 21% (01:12) wrong based on 108 sessions
A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls and 7 Turquoise balls. Two balls are chosen from the jar. What is the probability that both balls chosen are Tan?

A. 1/70
B. 2/49
C. 1/21
D. 6/441
E. 1/49

[Reveal] Spoiler:
Don't we have to assume that the balls are identical? If they are, then the number of ways of choosing 2 Tan identical balls out of 3 tan balls = 1. Isn't it?
If I assume that the balls are not withdrawn "at a time", OA is correct. However, how do I know whether the balls are withdrawn at a time or one by one?
[Reveal] Spoiler: OA

Last edited by Bunuel on 01 Oct 2012, 06:22, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A jar contains 6 Magenta balls [#permalink] New post 30 Sep 2012, 09:12
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voodoochild wrote:
A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls and 7 Turquoise balls. Two balls are chosen from the jar. What is the probability that both balls chosen are Tan?

1) 1/70
2) 2/49
3) 1/21
4) 6/441
5) 1/49

Don't we have to assume that the balls are identical? If they are, then the number of ways of choosing 2 Tan identical balls out of 3 tan balls = 1. Isn't it?
If I assume that the balls are not withdrawn "at a time", OA is correct. However, how do I know whether the balls are withdrawn at a time or one by one?


The assumption is that balls of the same color are identical.
OA answer is correct anyway. It doesn't matter that you stick both your hands in a jar and remove two balls together or get them one-by-one.
What matters is the final result: the types/colors of the two chosen balls.

If you use probabilities: first ball Tan 3/21, second ball Tan 2/20, altogether (3/21)*(2/20) = 1/70.

Combinatorics: total number of ways to choose 2 balls out of 21 is 21C2 = 21*20/2 = 21*10. Total number of ways to choose 2 Tan balls out of 3 is 3C2 = 3*2/2 = 3.
Therefore, the required probability is 3/(21*10) = 1/70.

Answer A.
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Re: A jar contains 6 Magenta balls [#permalink] New post 30 Sep 2012, 10:08
Ok. But what about this one:
a-bag-has-4-blue-3-yellow-and-2-green-balls-the-balls-of-139818.html

In this example, the number of ways of choosing any one out of 3 Yellow balls = 1+1+1+1 (no ball is chosen) instead of (3C0=1)+(3C1=3)+(3C2=3)+(3C3=1)=8 ways

Can you please explain the difference between these two questions?
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Re: A jar contains 6 Magenta balls [#permalink] New post 21 Oct 2012, 11:45
EvaJager wrote:
voodoochild wrote:
A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls and 7 Turquoise balls. Two balls are chosen from the jar. What is the probability that both balls chosen are Tan?

1) 1/70
2) 2/49
3) 1/21
4) 6/441
5) 1/49

Don't we have to assume that the balls are identical? If they are, then the number of ways of choosing 2 Tan identical balls out of 3 tan balls = 1. Isn't it?
If I assume that the balls are not withdrawn "at a time", OA is correct. However, how do I know whether the balls are withdrawn at a time or one by one?


The assumption is that balls of the same color are identical.
OA answer is correct anyway. It doesn't matter that you stick both your hands in a jar and remove two balls together or get them one-by-one.
What matters is the final result: the types/colors of the two chosen balls.

If you use probabilities: first ball Tan 3/21, second ball Tan 2/20, altogether (3/21)*(2/20) = 1/70.

Combinatorics: total number of ways to choose 2 balls out of 21 is 21C2 = 21*20/2 = 21*10. Total number of ways to choose 2 Tan balls out of 3 is 3C2 = 3*2/2 = 3.
Therefore, the required probability is 3/(21*10) = 1/70.

Answer A.



Eva,
Sorry to open this old thread. But, in this post : math-combinatorics-87345.html , the poster has mentioned that "Number of ways to pick 0 or more objects from n identical objects = n + 1" -- Hence, the number of ways to pick 2 balls out of say 5 identical balls must NOT be 5C2 -- Correct? Please help me.

thanks
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Re: A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls [#permalink] New post 21 Oct 2012, 12:28
voodoochild wrote:
A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls and 7 Turquoise balls. Two balls are chosen from the jar. What is the probability that both balls chosen are Tan?

A. 1/70
B. 2/49
C. 1/21
D. 6/441
E. 1/49

[Reveal] Spoiler:
Don't we have to assume that the balls are identical? If they are, then the number of ways of choosing 2 Tan identical balls out of 3 tan balls = 1. Isn't it?
If I assume that the balls are not withdrawn "at a time", OA is correct. However, how do I know whether the balls are withdrawn at a time or one by one?


If it is not stated, we should assume without replacement,
Therefore: 3/21 x 2/20 = 1/70

Answer A
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Re: A jar contains 6 Magenta balls [#permalink] New post 21 Oct 2012, 14:49
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voodoochild wrote:
EvaJager wrote:
voodoochild wrote:
A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls and 7 Turquoise balls. Two balls are chosen from the jar. What is the probability that both balls chosen are Tan?

1) 1/70
2) 2/49
3) 1/21
4) 6/441
5) 1/49

Don't we have to assume that the balls are identical? If they are, then the number of ways of choosing 2 Tan identical balls out of 3 tan balls = 1. Isn't it?
If I assume that the balls are not withdrawn "at a time", OA is correct. However, how do I know whether the balls are withdrawn at a time or one by one?


The assumption is that balls of the same color are identical.
OA answer is correct anyway. It doesn't matter that you stick both your hands in a jar and remove two balls together or get them one-by-one.
What matters is the final result: the types/colors of the two chosen balls.

If you use probabilities: first ball Tan 3/21, second ball Tan 2/20, altogether (3/21)*(2/20) = 1/70.

Combinatorics: total number of ways to choose 2 balls out of 21 is 21C2 = 21*20/2 = 21*10. Total number of ways to choose 2 Tan balls out of 3 is 3C2 = 3*2/2 = 3.
Therefore, the required probability is 3/(21*10) = 1/70.

Answer A.



Eva,
Sorry to open this old thread. But, in this post : math-combinatorics-87345.html , the poster has mentioned that "Number of ways to pick 0 or more objects from n identical objects = n + 1" -- Hence, the number of ways to pick 2 balls out of say 5 identical balls must NOT be 5C2 -- Correct? Please help me.

thanks


The wording "Number of ways to pick 0 or more objects from n identical objects = n + 1" is not clear.
It is meant that from n identical objects, we can choose none (which is 0), 1, 2,..., or all n objects - therefore we have n+1 choices.
For example, if we have 5 identical balls, we can choose 0, 1, 2, 3, 4, or 5 balls. We have a total of 5 + 1 choices.

Once we made up our mind how many balls we want to choose, say 2, then the number of possibilities to choose them is 5C2 = 5*4/2.
Because, for the first ball 5 possibilities, for the second 4 possibilities, and then we have to divide by 2!, because the balls are identical, order doesn't matter (I would say it is meaningless, red is red, so why should I care which red I picked first?).
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Re: A jar contains 6 Magenta balls [#permalink] New post 09 Aug 2013, 21:26
[/quote]


Eva,
Sorry to open this old thread. But, in this post : math-combinatorics-87345.html , the poster has mentioned that "Number of ways to pick 0 or more objects from n identical objects = n + 1" -- Hence, the number of ways to pick 2 balls out of say 5 identical balls must NOT be 5C2 -- Correct? Please help me.

thanks[/quote]

The wording "Number of ways to pick 0 or more objects from n identical objects = n + 1" is not clear.
It is meant that from n identical objects, we can choose none (which is 0), 1, 2,..., or all n objects - therefore we have n+1 choices.
For example, if we have 5 identical balls, we can choose 0, 1, 2, 3, 4, or 5 balls. We have a total of 5 + 1 choices.

Once we made up our mind how many balls we want to choose, say 2, then the number of possibilities to choose them is 5C2 = 5*4/2.
Because, for the first ball 5 possibilities, for the second 4 possibilities, and then we have to divide by 2!, because the balls are identical, order doesn't matter (I would say it is meaningless, red is red, so why should I care which red I picked first?).[/quote]

Hi,

i have a small doubt in this.

This example of 5 balls and pick up 2 ball can be written as like this also right
Probablity to take first ball is : 2/5
Probablity to take second ball: 1/4
so independent event = 2/5* 1/4= 1/10
which is reciprocal of wat u got. . is this correct answer.? i am trying to understand this in probablity ways/ pleas healp me


Thanks,
Rrsnathan
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Re: A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls [#permalink] New post 09 Aug 2013, 22:49
Expert's post
Probability = \frac{Desired Outcomes}{Total Outcomes}

Desired Outcomes = Number of ways of choosing 2 tan balls from 3 Tan balls = 3C2

Total Outcomes = Number of ways of choosing any 2 balls from 21 balls = 21C2

Thus Probability = \frac{3C2}{21C2} = \frac{1}{70}
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Re: A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls [#permalink] New post 10 Aug 2013, 00:33
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voodoochild wrote:
A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls and 7 Turquoise balls. Two balls are chosen from the jar. What is the probability that both balls chosen are Tan?

A. 1/70
B. 2/49
C. 1/21
D. 6/441
E. 1/49

[Reveal] Spoiler:
Don't we have to assume that the balls are identical? If they are, then the number of ways of choosing 2 Tan identical balls out of 3 tan balls = 1. Isn't it?
If I assume that the balls are not withdrawn "at a time", OA is correct. However, how do I know whether the balls are withdrawn at a time or one by one?


Both balls are tan = 3/21 × 2/20 = 1/70
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Asif vai.....

Re: A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls   [#permalink] 10 Aug 2013, 00:33
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