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About Steve Holleran
Expertise
I can help with all math questions from basic math to Calculus. Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.

Experience
33 years teaching experience in NJ public schools

Education/Credentials
B.S. Mathematics : Wake Forest University 1972 M.S. Mathematics : Monmouth University 1981

 
   

You are here:  Experts > Science > Mathematics > Advanced Math > Probability

Topic: Advanced Math



Expert: Steve Holleran
Date: 7/24/2008
Subject: Probability

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

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