Expert: Steve Holleran - 7/24/2008

Question
I am about to take the praxis exam and I am stuck on probability. Can you explain to me the difference between "with replacement" and "without replacement" when choosing for two things of the same color. Thankyou.

Answer
Hi Jessica,

Okay, I'll try to give an example of what I believe you are asking about:

Let's say a container has 5 red and 8 blue marbles.  You will make two successive draws from the container, and note the color drawn each time.

Let's find the probability that you draw 2 red marbles.


With replacement:

When you draw the first time, there are 5 red marbles out of a total of 13 marbles (5 red + 8 blue), so the probablility or a red is

P(red) = 5/13.

Now, you replace the marble you drew on the first draw, then take your second draw.  Well, the situation is now exactly the same.  There are 13 marbles in the container and 5 are red, so this probability is also
5/13.  Then the probability of both draws being red is the product:

P(red, red) = 5/13 * 5/13 = 25/169=0.148 or about 14.8%


Without replacement:

Here, the first draw is exactly the same as above, so the probability of red on the first draw is once again 5/13.

BUT, now you do not replace the marble in the container, and for the second draw you assume that your first one was red (what you wanted).
So now the container has only 12 marbles, and only 4 of them are red
(because you assume you took a red out on the first draw).  

The probability of red on the second draw is now 4/12.  So the probability of

P(red, red) = 5/13 * 4/12 = 20/156 = 0.128 or about 12.8%


Just to make sure its clear, here's what it would look like if you did a third draw:

With replacement
P(red,red,red) = 5/13 * 5/13 * 5/13 = 125/2197 =  0.0059

Without replacement
P(red,red,red) = 5/13 * 4/12 * 3/11 = 60/1716 = 0.0349


I hope this helps, and good luck on the exam!

Steve