# Permutations and Combinations in Probability

• Feb 18th 2009, 05:44 PM
overduex
Permutations and Combinations in Probability
Hi, can someone please explain what to do, stats really isn't my thing, so if you could please explain your answer, I'd really appreciate it!

3 red cubes, 4 blue, 6 yellow are arranged in a row. find the probability that (a) the cubes at each end are the same color
and (b) the cubes at each end are of a different color.

Thanks a lot!
• Feb 18th 2009, 06:15 PM
Plato
Quote:

Originally Posted by overduex
Hi, can someone please explain what to do, stats really isn't my thing, so if you could please explain your answer, I'd really appreciate it!
3 red cubes, 4 blue, 6 yellow are arranged in a row. find the probability that (a) the cubes at each end are the same color

The total number of ways the arrange thes cubes is $N=\frac{13!}{(3!)(4!)(6!)}$ that becomes your denominator.
The number of ways to have a reds at each end is $\frac{11!}{(1!)(4!)(6!)}$

Now you do it for the other two colors.
• Feb 18th 2009, 06:27 PM
overduex
Quote:

Originally Posted by Plato
The total number of ways the arrange thes cubes is $N=\frac{13!}{(3!)(4!)(6!)}$ that becomes your denominator.
The number of ways to have a reds at each end is $\frac{11!}{(1!)(4!)(6!)}$

Now you do it for the other two colors.

after you do it for the other 2 colors, do you multipy them together? or add?
• Feb 18th 2009, 06:40 PM
Plato
Quote:

Originally Posted by overduex
after you do it for the other 2 colors, do you multipy them together? or add?

• Feb 18th 2009, 07:17 PM
overduex
Quote:

Originally Posted by Plato

sorry to bother again, but could you explain how to get the answer?
It is supposed to be 4/13 but i keep getting ridiculous numbers as answers. I basically just added what you had written
11!/(1!)(4!)(6!) + 11!/(3!)(2!)(6!) + 11!/(3!)(4!)(4!)

do you see anything I may be doing wrong?
• Feb 19th 2009, 02:15 PM
Iunno whats wrong. I tried it out, I i seemed to get 4/13. Maybe you're simplifying wrong?

11!/(1!x4!x6!) = 2310

11!/(3!x2!x6!) = 4620

11!/(3!x4!x4!) = 11550

13!/(3!x4!x6!) = 60060

So, (2310/60060) + (4620/60060) + (11550/60060) = (18480/60060) = (4/13).

If your entering the numbers into your calculator like this 11!/3!x4!x4!, then you might be getting the wrong answer, be in reality it'd be this:
(11!/3!) x 4! x 4! which might explain the random numbers.
• Feb 19th 2009, 02:21 PM
overduex
Quote:

Iunno whats wrong. I tried it out, I i seemed to get 4/13. Maybe you're simplifying wrong?

11!/(1!x4!x6!) = 2310

11!/(3!x2!x6!) = 4620

11!/(3!x4!x4!) = 11550

13!/(3!x4!x6!) = 60060

So, (2310/60060) + (4620/60060) + (11550/60060) = (18480/60060) = (4/13).

If your entering the numbers into your calculator like this 11!/3!x4!x4!, then you might be getting the wrong answer, be in reality it'd be this:
(11!/3!) x 4! x 4! which might explain the random numbers.

yeah, I figured out my mistake. I was wayyy off. haha, thanks for trying though!