Difficult Math Problem #117 - Probability

[800guy]

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

[guynoor]

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 ???

[jayhawk2001]

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 ?

[Neo2000]

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

[Cybermusings]

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

[RAGS]

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

[800guy]

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

[alugoya]

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

[Scott@TargetTestPrep]

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