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probability again :(

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by mehaksal » Mon Aug 27, 2012 7:40 am
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

I know it's a simple question, with a simple method of solving, but with probability I tend to go very confused about which way to go.
Eg. Here I thought of going 8C2 - 2c2 way, which is very wrong!
Help me out!
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by cypherskull » Mon Aug 27, 2012 8:08 am
P(Not Blue) = 1 - P(Blue)

1st card

P(Blue) = 2/8 = 1/4
So, P(Not Blue) = 3/4 -------(1)

2nd card having selected one non-blue card
P(Blue) = 2/7 [Since there are still 2 blue cards]
So, P(Not Blue) = 5/7 -------(2)

Therefore, probability of selecting both blue cards = (3/4)*(5/7) = [spoiler]15/28 Ans: A[/spoiler]

Let me know the OA.
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by cypherskull » Mon Aug 27, 2012 8:21 am
Another easier approach could be as follows:

P(not blue) = (Number of ways a non-blue card can be selected)/(Total # of ways 2 cards can be selected
= 6C2/8C2
[spoiler]= 15/28[/spoiler]
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Sunit

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