Probability problems

Forum for the GRE subject test in mathematics.
Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Probability problems

Postby Nameless » Sun Sep 07, 2008 12:04 am

Hi Dear,
In my opinion, probability problems are the most time consuming problems. I hate these problems so much because we have to pay much attention for every word when reading . My goal for this topic is we should review some basic formulas and post some problems to get used to the techniques

Okie, I post the first problems so that we can discuss :

Problem 35 - GR0568

At a banquet, 9 women and 6 men are to be seated in a row of 15 chairs. If the entire seating arrangement is to be chosen random, what is the probability that all the men will be seated next to each other in 6 consecutive positions?

a. 1/ (15 C 6) - where n C k = n!/[k!(n-k)!]
b. 6!/(15 C 6)
c. 10!/15!
d. 6!9!/14!
e. 6!10!/15!

lovemath
Posts: 12
Joined: Fri Sep 05, 2008 9:02 pm

Postby lovemath » Sun Sep 07, 2008 7:31 am

yeah, i hate probability problems too.

my technique on this one is :
1. if the men must sit together, then we can assume them as one.. so the possibility is 9 women + 1 'men' = 10!
2. within this 6 men there are 6! possibility, so it brings total to 6! * 10!
3. so the probability is 6!10!/15!

I have a question taken from schaum college maths:
How many signals can be made with 3 white, 3 green, and 2 blue flags by arranging them on a mast five at a time?
The answer is 170.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Postby Nameless » Sun Sep 07, 2008 9:52 am

Hi lovemath,

Are there any additional conditions for your problem? :) like what does a signal mean in this case?
Last edited by Nameless on Tue Sep 09, 2008 7:06 pm, edited 1 time in total.

lovemath
Posts: 12
Joined: Fri Sep 05, 2008 9:02 pm

Postby lovemath » Sun Sep 07, 2008 12:06 pm

this is the full question :
How many signals can be made with 3 white, 3 green, and 2 blue flags by arranging them on a mast:
a). all at a time.
b). three at a time
c). 5 at a time.


a). 3+3+2=8 flag at a time. The answer is : 8!/3!3!2!, because there are flags that has the same color.
b). this is tricky. If there are 3 blue flags, to arrange the signal is simple: 3.3.3=27 ways, but if 2 flags, confusing. The answer is 26, so i assumed it's from 27 - 1 where :
* 27 is when there are 3 blue flags
* 1 is when all the blue is used on the mast..
how to do it without subtracting ?
c). ???




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