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Probabilty question: where m i wrong??

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by him1985 » Sat Jan 28, 2012 10:23 pm
Q. 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


We can have two approches here:
(1). direct find probabilty that both balls are not blue.
6/8 * 5/7 = 15/28

(2). We can find the prob. that both balls are blue and can substract from 1

1- (2/8*1/7) = 27/28

I am getting two different answer but according to probability rule i should get answer same in both way. Please help....
Himanshu Chauhan
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Source: — Problem Solving |

by rijul007 » Sat Jan 28, 2012 11:00 pm
him1985 wrote:Q. 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


We can have two approches here:
(1). direct find probabilty that both balls are not blue.
6/8 * 5/7 = 15/28

(2). We can find the prob. that both balls are blue and can substract from 1

1- (2/8*1/7) = 27/28

I am getting two different answer but according to probability rule i should get answer same in both way. Please help....
Analyze the appraoches yourself, and answer this ques for each appraoch
does the computed probability include the selection of one blue and one green?

Did you see the difference?
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by LalaB » Sun Jan 29, 2012 1:28 am
him1985,

the probability that both are not blue=1-(probability that one or both are blue)

the probability that they will both are not blue=1-(2C2+2C1*6C1)/8C2=1-13/28=15/28

One more approach-

6C2/8C2=15/28
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by ronnie1985 » Sun Jan 29, 2012 10:59 am
No blue = 6C2
2 balls = 8C2
P = 6c2/8C2 = 15/28.
Follow your passion, Success as perceived by others shall follow you
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by him1985 » Sun Jan 29, 2012 6:13 pm
LalaB wrote:him1985,

the probability that they will both are not blue=1-(2C2+2C1*6C1)/8C2=1-13/28=15/28
LalaB: Can you please elaborate this calculation. Probability that both are blue =(2C2+2C1*6C1)/8C2
Himanshu Chauhan
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by [email protected] » Sun Jan 29, 2012 9:06 pm
him1985 wrote:
LalaB wrote:him1985,

the probability that they will both are not blue=1-(2C2+2C1*6C1)/8C2=1-13/28=15/28
LalaB: Can you please elaborate this calculation. Probability that both are blue =(2C2+2C1*6C1)/8C2
I can elaborate.
selecting 2 blue out of 2 blue =2C2=1
selecting 1 blue out of 1 blue and selecting any other out of 6 different=2C1*6C1=12

So Prob is =1-13/8C2=1-13/28=15/28
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