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 of getting at least one blue?
CB
Quote:
Originally Posted by vino.born2win
well !!!i should definitely thank all of you for replying 2 my post and giving logical answers...as i don knw the answers i should say all of the answers seem logically correct...except for the gumballs problem..
why do we need to choose 20 gumballs...arent 10 +1 balls enough???
...
for n gum balls of different colors it will be 10(n-1)+1.
we're taking the worst case scenario i.e for taking out 1 color out of 10 we're doing the whole cycle once b4 the next same color gumball comes out.
Quote:
Originally Posted by jess_84
2. If 2n+1 is the median of seven consecutive integers, then what is the mean of the integers?
the set is of arithmetic series(since the integers are cosecutive) therefore the middle term will be it`s mean and already the question has mentioned the middle term(i.e the median ...
1)
S is the set of all integers from 1 to 48, incluside.
V is the sum of all the even integers in S
D is the sum of all the odd integers in S.
Column A: V
Column B: D + 23
2)
Column A: 100,210 X 90,021
Column B: 100,021 X 90,210
I know there's a quick way to solve #2, but I'm just not seeing it.
1) Sum of evens = 2 + 4 + 6 + ... 44 + 46 + 48
note that 2+48 = 50 and 4+46 = 50 and so on. So V = 12*50 = 600
Same with D: 1 + 3 + 5 + ... + 43 + 45 + 47, note that 1+47 = 48 and 3+45 = 48 and so on. So D = 12*48 = 576.
Col A: 600
Col B: 576+23 = 599
another way ...
Sum of even numbers = m/2 * (m/2 + 1) (I googled this formula )
48/2 * (48/2+1) = 24*25 = 600
Sum of odd numbers =...
For the second one:
in both multiplications the sum of first+second is the same.
If a+b = const, then a*b is bigger for the case where a and b are closer to (a+b)/2.
So, B is the answer.
Amount of A left/ Amount of A originally present = (1-x/M)^n
We know the above formula. But what if the M is a ratio of two elements say 3:2 or 1:9. Then how should we proceed.
CB
Code:
Amount of A left/ Amount of A originally present = (1-x/M)^n
We know the above formula. But this is only useful when we have only a single element in the beginning as M and later another element is added. But what if M is already a mixture in ratio of two elements 3:2 or 1:9 and some of the mixture is removed and one of the element from the mixture is added. Then how can we ...
thx ace gre for conerning but my gre is on tuesday the upcomin one... so wish me "good luck "..lol iam so excited ... any tips for the last days ...???
You're right mx+b, I was too hasty in my thinking...n must be multiple of 6. (though in my defense, I was talking about n>10) though for 14, Rita is certain of winning if and only if she plays initially, and then plays (6-S) on her 2nd turn.
Then I'd suggest drilling them until you can solve them faster. I don't think that DI's are any more time-intensive than any other type of problem if you know what you're doing. You will almost certainly see a DI within the first six problems and probably see two within the first twelve. I'd take up to 5 minutes each to solve those. After the 12th or 15th problem, if you want to conserve time by skipping any remaining...
Hi I've just taken a GRE practice test from the Princeton Review, and received 750 for the Quantitative section, even though I made five mistakes in the last set of questions.
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Can you por favor, El Señor suggest ,Whatever Probability is there in the Quantitative Aptitude will be sufficient and i need to do all exercises .I have exam next month 1st week i have only 15 days time and now i got QP book.Can you por favor, El Señor suggest which chapters will be more helpful in QP. Thanks
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