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I am simply amazed.. great going guys...

Here's another one:

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

Hi myself Ravikant from Aurangabad. I want to take GMAT in 2009. This is my first post... please reply... thanks for the good work... The question is:

My name is PREET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?

a) 5%
b) 10%
c) 20%
d) 25%
e) 12.5%

I am not sure of the answer But here goes my way of understanding the problem.

As 2 cards are taken together, there are in total 2^4 possibilities.

Total outcomes hence = 16
Taking out the blue cards from the solution set, the total number of ways the rest of the three set of cards can be arranged among themselves in 3C1*3C1 ways = 9 Ways.

So the prob that both cards will not be blue is = 9/16 ---- Option C

My name is PREET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?

a) 5%
b) 10%
c) 20%
d) 25%
e) 12.5%
Following is the probable solution:
PREET Five alphbets. Now selecting a pair means selecting 2 out of 5. This is possible by 5C2 ways i.e. 5*4/1*2=10 or 5!/(3!*2!) =10.
Now if EE is selected then the the name will not change.
Hence probabillity=favorable outcomes/Total possible outcomes
=1/10 = 10%.

Rerobability 3 Hours, 39 Minutes ago Karma: 0
I am simply amazed.. great going guys...

Here's another one:

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
Please note, although no boardcode and smiley buttons are shown, they are still useable

Following is the probable answer:
Probability of drawing two things at a time is same as drawing one by one and not replacing it.
P(Both not Blue)= P(First one not Blue)*(Second one not Blue)
= (6/8)*(5/7)= (6*5)/(8*7)=15/28 Hence A.

rchuck21 wrote:
I am not sure of the answer But here goes my way of understanding the problem.

As 2 cards are taken together, there are in total 2^4 possibilities.

Total outcomes hence = 16
Taking out the blue cards from the solution set, the total number of ways the rest of the three set of cards can be arranged among themselves in 3C1*3C1 ways = 9 Ways.

So the prob that both cards will not be blue is = 9/16 ---- Option C

hi rahul...well tried ...but that is not the solution...
2 cards are drawn without replacement...so after u draw the first card...you just have 7 cards to choose from..

look at the solution which has been posted above...that will help to clear your doubt...