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Probability

by 4meonly » Fri Nov 14, 2008 6:04 am
There are 20 red marbles, 20 blue marbles, 5 white marbles and 5 green marbles in a bag. If two marbles are to be selected at random, without replacement, what is the probability of getting two marbles with the same color.

[spoiler]OA 16/49[/spoiler]
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by jimmiejaz » Fri Nov 14, 2008 6:23 am
so, we want balls of the same color, which can be any one of the events:
2 red or 2 blue or 2 white or 2 green

In this case
the probability without replacement...
p = 20/50 * 19/49 + 20/50 * 19/49 + 5/50 * 4/49 + 5/50 * 4/49
Solve, we get probability = 16/49.
Hope it helps.
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by cramya » Fri Nov 14, 2008 8:06 am
If u do it like u did yes we get 16/49. Agreed...



Lets say we did:

20C1*19C1+20C1*19C1+5C1*4C1+5C1*4C1 / 50C2 why do we end up getting 32/49 and not 16/49??

What am I missing here? Any ideas?
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by logitech » Fri Nov 14, 2008 9:24 am
cramya wrote:If u do it like u did yes we get 16/49. Agreed...



Lets say we did:

20C1*19C1+20C1*19C1+5C1*4C1+5C1*4C1 / 50C2 why do we end up getting 32/49 and not 16/49??

What am I missing here? Any ideas?
50C2 = 50x49/2

it is a listing rather than comb. so 50C2 doubles your chances. and you have 32/49 instead of 16/49
LGTCH
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by cramya » Fri Nov 14, 2008 9:27 am
it is a listing rather than comb
Pl explain more on wat u exactly mean by this
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by logitech » Fri Nov 14, 2008 9:52 am
cramya wrote:
it is a listing rather than comb
Pl explain more on wat u exactly mean by this
Cramya:

Lets say we have 3 marbles one red, one blue, and one yellow

what is the probability of picking 2 marbles, Red and Blue, without replacing them.

Well

1/3x1/2=1/6

In your solution

1c1 x 1c1/3c2 = 1/3

Because you are actually finding the probability of A marble/ 2 marbles

Going back to our original problem the probability you found is not:

Probability of X / SET X

BUT;

Probability of X / (CHOOSEN SET X )

Is it clear now ?
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