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Probability

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Probability

by ritula » Wed Aug 13, 2008 11:06 pm
I know that this one is pretty easy but I am getting answer as 27/28.Pls help


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
Philosophers have interpreted world in various ways, the point is to change it!
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by amitansu » Wed Aug 13, 2008 11:52 pm
Without blue selection, there are total 6 of them can be chosen.

So for first pick, 6/8
next pick, 5/7

both=> 6/8*5/7=15/28
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by sudhir3127 » Thu Aug 14, 2008 12:54 am
hey i think u probably did

1- (1/28) = 27/28..

even for a second i also did it that way and then saw the answers and realised its not the way..

ur right in ur approach though that...

probably of not blue is 1- ( probability of blue )

but u need to careful when u do this...

heres how u shud do it..

Probability that 1st is blue = 2/8

But 1st will not be blue 1 - 1/4 = 3/4 of the time;
then getting a blue has probability 2/7 (2 blues still left)
since that happens 3/4 of the time,
then: 3/4 * 2/7 = 6/28 =

So add them up: 2/8 + 6/28= 7/28 + 6/28 = 13/28

thus probability of not blue is

1- prob. of blue..

which is 1- 13/28 = 15/28..

though u get an answer this way .. i will still prefer the follow the other way of doing it...

P(1st card not blue) = 6/8 [6 non-blue available out of 8 cards]
P(2nd card not blue given 1st card was not blue) = 5/7 [5 remaining non-blue out of 7]

P(both not blue) = (6/8 )*(5/7) = 15/28

Hope it helps..
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by bluementor » Thu Aug 14, 2008 1:16 am
Hi I think this is a matter of understanding exactly what the question wants, which I'm not sure myself. It could be either one of the following:

1. P(of no blue cards at all) = 1-P(at least 1 blue)

2. P(of no 2 blue cards) = 1 - P(2 blue cards)

From your answers, it seems like its probably (1) that the question is asking. But I'm not sure how do you know that from the question. Could the wording of the question been improved?

BlueMentor
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by pepeprepa » Thu Aug 14, 2008 1:20 am
I read back the question and I do not see 2 ways of understanding it.

"what is the probability that they will both are not blue". For me it is clear, 0 blue in the total.
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Both are not blue vs None is blue

by Ramp » Wed Feb 25, 2009 5:47 pm
Bringing up an old topic .

Both are not blue = P (None is blue )+ P(one is blue and other not)
= 30/56 + 12/56 = 27/28

It can also be solved by
1- P(both are blue) = 1- (2/56) = 27/28

However from the OA (15/28) the question seems to be interpreted as 'None is blue ' case which is just a subset of 'Both are not blue case'.
I think answer should be 27/28

Feedback pls.
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