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Probabilities of Picking Colored Balls out of an Urn

Date: 03/13/2006 at 22:34:10
From: Nathaniel
Subject: combinations and permutations, with different color balls

An urn contains 8 white, 6 blue, and 9 red balls.  How many ways can 
6 balls be selected to meet each conditon?

  a) all balls are red
  b) 3 are blue, 2 are white, and 1 is red
  c) 2 are blue and 4 are red
  d) exactly 4 balls are white

There are 8 + 6 + 9 = 23 balls total.  For the first question I tried
(9/23)*(8/22)*(7/21)*(6/20)*(5/19)*(4/18) but I think I did it wrong
because it didn't really work out.  Please help me--I don't understand 
this!



Date: 03/15/2006 at 01:16:43
From: Doctor Wilko
Subject: Re: combinations and permutations, with different color balls

Hi Nathaniel,

Thanks for writing to Dr. Math!

I'm going to go into some depth and explain two different ways to 
think about each of your problems.

If you're not familiar with "choose notation, i.e., nCr (n objects, 
choose r of them)" please visit our archives for the details.  

  Permutations and Combinations
    http://mathforum.org/dr.math/faq/faq.comb.perm.html 

Below I'll show you how to calculate these answers as probabilities, 
but if you just want the number of ways to pick the balls in each of 
the problems then the answer is simply in each numerator.

=======
Part a
=======

An urn contains 8 white, 6 blue, and 9 red balls.  How many ways can 6 
balls be selected so that all 6 balls are red?

You have 23 balls total.  These calculations assume each ball as being 
numbered or distinct.

======================================
Method 1, Choose Notation (Unordered)
======================================
An easy way to do these problems is with choose notation because this 
way automatically takes into account the order in which the balls 
could be selected.

The denominator is usually the easiest.  How many balls total are 
there and how many do you want?  23C6 = 100947 ways to choose any 6 
balls from the 23 balls total.  

The numerator can be a little trickier.  Think about how many total 
red balls there are and how many you want?  9C6 = 84 ways to choose 
6 red balls from 9 red balls total.  

Finally the probability of choosing 6 balls and getting all red is,

  9C6         84
 ------ =   ------- =~ 0.000832
  23C6      100947

======================================
Method 2, Probabilities (Ordered)
======================================
An alternate way to do this question without choose notation is as 
follows.  This looks like what you tried, and you did it correctly!

Find the probabilities of pulling 6 consecutive red balls.  Because 
you don't put the balls back in the urn after each draw, decrease the 
denominators accordingly.

What's the probability of pulling a red? 9/23
What's the probability of pulling another red?  8/22

Continuing like this, you get,

  9 *  8 *  7 *  6 *  5 * 4         60480
 ---------------------------- =    -------  =~ 0.000832 (same)
 23 * 22 * 21 * 20 * 19 * 18       72681840


=======
Part b
=======

An urn contains 8 white, 6 blue, and 9 red balls.  How many ways can 6 
balls be selected so that 3 are blue, 2 are white, and 1 is red?

======================================
Method 1, Choose Notation (Unordered)
======================================
Using similar reasoning like we did for part a we get,

  6C3 * 8C2 * 9C1
 ----------------- =~ 0.049927
       23C6

======================================
Method 2, Probabilities (Ordered)
======================================
Without choose notation, it becomes more important to pay attention 
to order.

We want BBBWWR

The probability of this is (same argument as for part a)

   (6 *  5 * 4)  *  (8 * 7)  * 9         60480
  -------------------------------  =    ---------  =~ 0.000832 
  (23 * 22 * 21) * (20 * 19) * 18       72681840

This is not the final answer!  This is the probability of getting 
this specific order (BBBWWR).

But we don't care about this specific order of the balls, we want 
ANY order of these six balls!

The total number of orderings of these six balls is

    6!         720
 --------  =  ----- = 60 different orders
 3!*2!*1!       12

So the actual calculation we want should look like (multiply by 60),

       6 * 5 * 4   *  8 * 7  * 9       60480
 60 * --------------------------  =  ---------  =~ (0.000832 * 60) 
      23 * 22 * 21 * 20 * 19 * 18     72681840


 =~ 0.049927 (same answer we got above with choose notation)


=======
Part c
=======

An urn contains 8 white, 6 blue, and 9 red balls.  How many ways can 6 
balls be selected so that 2 are blue and 4 are red?

This problem is solved exactly like part b.

If you choose either method, you should get =~ 0.018723

=======
Part d
=======

An urn contains 8 white, 6 blue, and 9 red balls.  How many ways can 6 
balls be selected so that exactly 4 balls are white?

We are choosing 6 balls from the urn.  So, this question is really 
asking, how many ways can you choose 4 white balls and 2 non-white 
balls?

There are 8 white balls and 15 non-white balls.

Now the problem follows exactly as the others above.

If you try to calculate this, the number of ways of choosing exactly 
4 white balls (and 2 non-whites) is 7350.

The probability of this happening is =~ 0.072811

----------------------

With these kinds of problems be very clear to define what you want, be 
aware of whether order will matter, and if you can, try to calculate 
the answer more than one way as a double-check.

Probability, permutation, and combination problems can definitely be 
tricky, but will get easier with practice!

Does this help?  Please write back if you have questions on any of this.

- Doctor Wilko, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Permutations and Combinations

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