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Listing Sample Spaces

Date: 02/13/2003 at 11:45:29
From: Diann
Subject: Equally Likely Outcomes

An urn contains 2 red, 2 black and 1 green ball, from which a sample 
of two is chosen without replacement. Determine a sample space of 
equally likely outcomes and assign probabilities.

My problem with this is that when you assign the probabilities that 
the sample space for red is (1,2 1,3 1,4 1,5 2,3 2,4 2,5), the sample 
space for blue is (3,4 3,5 4,5), but because 2 balls are being drawn 
there can be no sample space for green; you can't draw 2 green balls 
because there is only one. What does this mean, that there is 0 
probability for green?

Date: 02/18/2003 at 22:20:20
From: Doctor Wallace
Subject: Re: Equally Likely Outcomes

Hello Diann,

Remember that a "sample space" is just another name for "all possible 
outcomes for an event."

In your example, the urn contains 5 balls - 2 red, 2 black, 1 green.

Now, before we can have a sample space, we have to have an event.

For example, if I said the event was "Selecting 1 ball from the urn," 
then the sample space would contain the following:

  red, red, black, black, green

So the probability of choosing 1 red ball would be 2/5, since 2 
outcomes are favorable, and the sample space contains 5 possibilities.

Your example said you are to draw 2 balls, without replacement. That 
is enough to determine the sample space of outcomes for this event.  
There is no separate "sample space for red" or "sample space for 
green."  There is only ONE sample space for "drawing 2 balls without 
replacement" - and it contains a list of ALL possible outcomes for 
this event. To make it easier to list them all, think of the red balls 
and the black balls with numbers on them so we can distinguish them.  
Red1, Red2, Black1, Black2.  Here it is:

       First pick             Second pick
   -----------------         ---------------
(1)       red1                    red2
(2)       red2                    red1
(3)       red1                    black1
(4)       red1                    black2
(5)       red2                    black1
(6)       red2                    black2
(7)       red1                    green
(8)       red2                    green
(9)       black1                  red1
(10)      black1                  red2
(11)      black2                  red1
(12)      black2                  red2
(13)      black1                  black2
(14)      black2                  black1
(15)      black1                  green
(16)      black2                  green
(17)      green                   red1
(18)      green                   red2
(19)      green                   black1
(20)      green                   black2

The advantage of a list like this is that each of these 20 outcomes is 
equally likely to occur. This is why it is VITAL that we distinguish 
between the 2 red balls and the 2 green ones. If we don't, we miss 
some of the possibilities.

Now, armed with the chart, you can go through and make a list of the 
probabilities of various things. I think this is what the "assign 
probabilities" part of your question means.

For example, what is the probability that you draw green, then black 
in that order?

Well, we know there are 20 possible outcomes, so the denominator of 
the probability fraction is 20. The numerator will be the number of 
"favorable outcomes," or how many ways we can draw green, then black.  
Looking at the chart, green then black is represented by numbers (19) 
and (20). This is 2 chances out of 20, or 2/20, which we will reduce 
to 1/10. So the probability of drawing green then black is 1/10.

You can use the sample space chart to answer ANY question you want 
regarding this problem now. Let's do one more. What is the probability 
that you draw red then black?  In the chart, this is represented by 
numbers (3), (4), (5), and (6). This is 4 chances out of 20, or 4/20 
which is 1/5.

Finally, to answer the question you mentioned about green, the 
probability that you draw green then green is indeed zero. This is 
because there is only 1 green in the urn, so there are NO entries in 
the sample space that contain green then green. So there are 0 chances 
out of 20 or 0/20 which, of course, is 0.

The strategy of listing the sample space will work for probability 
problems where the space is small and manageable, like this one.  
Other problems have sample spaces with thousands or even millions of 
possibilities, and to solve those we must turn to the formulas.

I hope this helps you, Diann.  Thanks for writing to Dr. Math!

- Doctor Wallace, The Math Forum 
Associated Topics:
High School Probability

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