A complete guide to Permutation, Combination and Probability. All you need for GMAT

[lionheart7]

[quote=almas]

Originally Posted by TwinnSplitter

Thanks for these great resources.

My answers to that document are as follows, anybody else who did them let me know what you got:
B
C
A
C

A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

D (the way I did this took kind of long, if somebody else could let me know how they did it I'd appreciate it)


(7c1*3c1+7c2)10c2= 42/45=14/15
answer D

= 1 - Probability that both marbles are red
= 1 - (3/10)*(2/9)
= 1 - 6/90
= 1 - 1/15
= 14/15

[Da_Gr8_Mperor]

Well both the ways are correct.
Pick whichever way suits you fine.

[Cloudnineast]

I think its easy to miscalculate the no. of students in the sailing club in the business school problem. I just did it in my head & got 150/400 instead of 250/400. Both the answers are up there. Good Catch to write it out.

[shingivyas]

Hello guys!

http://www.mansw.nsw.edu.au/members...vol23no4yen.htm
the link was awesome but am stuck-
Q 1.In how many different ways can 5 boys and 4 girls sit together in a row?
a. What if Katie and Christine want to sit together

answer: Let both girls be considered as one person/seat.

Now there are only '8 people' to arrange on 8 seats. 8! = 40 320.

But for the '2-girls seat' GG, the two girls can sit in 2 different ways: G1G2 or G2G1 (Katie and Christine can swap seats). So for each of the 8! arrangements, there are 2 possibilities for GG.

a similar question says-

6 persons-3 asians, 2 european, 1 american.. they have to stand for a pic so that 3 asians are together and 2 europeans are together.. the possible no of combinations are???

[mevz]

Let the 3 Asians b considered as one person/seat
Similarly let 2 Europeans b considered as one person/seat
So now we have 3 people (1 Asian,1 European and 1 American) to arrange on 3 seats.
=3!=6
But the 3 Asians and 2 Europeans can sit in 3! and 2! different ways

So for each 3! arrangements,there r 6 possiblities for Asains and 2 possiblities for Europeans.

i.e. 3!* 3!* 2!=6*6*2=72

Let me know the correct answer...

[shingivyas]

hey Mevz...

That correct 72, is right.
and the logic, got it clearly.


Thanks buddy

[hesmala]

that's a very good site for beginers

[chaim]

Anyone got the article emailed? I would also like to receive it.

This is a great thread, but same here, I can't download the article, and am in a time contstrain. please upload it to FileHo.com it is a file hosting and you can just post the link.

thank you so much, we all appreciate it

[MonkeyGirl]

Can someone send me the Probability.doc?
My email is .....


I'm taking the GMAT in 2 weeks, so there is no chance I'll post 100 new posts at this time ...


Thanks

[siri2gmat]

A good link for the beginners of permutation and combinations