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Difficult Math Problem #117 - Probability

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Difficult Math Problem #117 - Probability

by 800guy » Wed Apr 11, 2007 12:42 pm
A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16

oa coming after some people respond, from diff math doc
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Re: Difficult Math Problem #117 - Probability

by guynoor » Wed Apr 11, 2007 1:26 pm
800guy wrote:A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16

oa coming after some people respond, from diff math doc
I am a bit confused on this question too. I understand that the P of not selecting a Blue on the first draw is 3/4. But how is it that on the second draw the P of not selecting blue is 5/7 ???
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by jayhawk2001 » Wed Apr 11, 2007 9:21 pm
Probability of 1 Blue = 2/8 * 6/7 * 2 = 24 / 56
Probability of 2 Blue = 1 / 8C2 = 2/56

Hence probability of atleast 1 blue = Sum of above 2 prob = 26 / 56

Hence probability of no blue = 1 - 26/56 = 30/56 = 15/28

Is it A ?
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by Neo2000 » Wed Apr 11, 2007 9:47 pm
I like jayhawk's approach.

I actually worked it the other way around.

If NOT blue then you can get
1. Two Red
2. TWo Green
3. Two Yellow OR
4. 1R and 1G OR
5. 1G and 1Y OR
6. 1Y and 1R

Now picking 2 cards from 8 = 8C2 = 28

2R = 2C2 = 1
2G = 2C2 = 1
2Y = 2C2 = 1
1R and 1G = 2C1 x 2C1 = 4
1Y and 1G = 2C1 x 2C1 = 4
1R and 1Y = 2C1 x 2C1 = 4

So total = 15

Therefore probability = 15/28
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by Cybermusings » Thu Apr 12, 2007 3:23 am
First one is blue and second one is blue = 2/8 * 1/7 = 1/28
First one is blue and second one is other color = 2/8 * 6/7 = 6/28
First one is other color and second one is blue = 6/8 * 2/7 = 6/28

Thus neither is blue = 1 - (1/28 + 6/28 + 6/28)
= 1 - 13/28
= 15/28
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by RAGS » Thu Apr 12, 2007 5:14 am
the Q asks probability of both r not blue
I think the solutions provided here r for nither card drawn blue
Also the probability of both cards not blue is 27/28 which is not one of the options
Please reply if i am wrong
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oa

by 800guy » Fri Apr 13, 2007 9:28 am
oa:

Chance of drawing a blue on the first draw = 2/8, so chance of not drawing a blue on the first draw is 6/8

similarly chance of not drawing blue on second draw = 5/7

Therefore probability of not drawing blue in 2 draws = 6/8*5/7 = 15/28
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my approach

by alugoya » Fri Apr 13, 2007 12:57 pm
I would do it the following way

total out comes = 8c2 = 28

if none of the 2 cards drawn are blue(2 cards are drawn from the other 6 cards)...

so favourable events = 6c2 = 15

probability of not getting the blue card = 15/28
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by Scott@TargetTestPrep » Mon May 20, 2019 6:09 pm
800guy wrote:A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16
The probability that the first card will not be blue is 6/8 and that for the second is 5/7. Therefore, the probability that both cards will be non-blue is 6/8 x 5/7 = 30/56 = 15/28.

Answer: A

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