Princeton CATS

Hey Kryzak...

Just took GMAT PREP(2).. got a 730... Q(47) V(44).... Had some really difficult SC missed 4/16 on this one well 3 really the last 1 I had to choose as I was running out of time... I owe you one for that RC tip.... i have been able to finish my tests since u gave that to me...

wow, great job tlntd! Congrats! I'll let you know how I do tomorrow.

I also just took a Power Prep test today, got a 730 (45/45). This was significantly higher than the GMAT Prep I took 2 weeks ago (700 - 38/49) But I have studied alot of math in the past 2 weeks. Does anyone have an idea how the Power Prep compares to GMAT Prep? I have 3 weeks till test day, and want to save my last GMATPrep until one week before.

I would like to post attachments for some problems I was working on can anyone tell me why I am not allowed to ?

There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines. If 3 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazines will be selected ?

A)1/2

B) 2/3

C) 32/35

D) 11/12

E)13/14


I know its combinations but i cant seem to get the answer

You might want to post this question in the math section. But here's how I'd solve it -
In this case I think it's easier to solve for the probability that none of the magazines selected will be fashion, then subtract that number from 1. If there are 8 magazines, 4 of each, the probability that the first magazine selected is sports is 4/8, or 1/2. That leaves 7 magazines to choose from, 3 of which are sports. So the probability that the second one chosen is sports is 3/7. That leaves 6 with 2 sports, so the third probability is 2/6, or 1/3. Multiply those together: 1/2x3/7x1/3, for a total probability of getting all sports magazines = 1/14. So the opposite probability, i.e. that at least one of the magazines is fashion, = 1 - 1/14, or 13/14.

Clear as mud?

I was thinking along these lines...

Prob= (#comb satisfactory/total # of combinations)


so out of 8 magazines we choose 3 so we have 56 possible combinations of magazines. then I was trying to figure out the numerator ... I understand how u are solving it but it feels like a combination question...

Probability and combinations are really two different animals, with different formulas, etc. The biggest difference is that with a probability problem, the answer is going to be less than 1, whereas with a combinations problem you are multiplying. The formula you are using for probability is confusing because you are using the word "combinations" where it should be "events".

If you wanted to know how many different arrangements of the three magazines you chose there could be, that would be a combinations problem (8C3). But the question didn't ask that, it asked the probability of one of the magazines being a fashion magazine. This seems like an area where the GMAT might try to trip us up!

Robin,

Here is how its done using combinations

There are 8C3=56 ways to choose 3 magazines. Within these 56 ways, there are 4C3=4 ways to chose the sports only. So, the probability that we select no fashion magazines is 4/56. Thus, the probability that at least one fashion magazine is chosen is 1-(4/56)=52/56=13/14.

That's what I love about this forum - I learn something new every day! Thanks.