Probability With/ Without Replacement - Lesson
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Think about this...
Suppose you have 2 bags with the exact same sets of marbles inside.  Let's say there are 4 red, 5 blue and 9 yellow marbles in each bag.
From the first bag, you reach in and  make a selection.  You record the color and then drop the marble back into the bag.  You then repeat the experiment a second time.
From the second bag you do exactly the same thing EXCEPT, after you select the first marble and record it's color, you do NOT put the marble back into the bag,  You then select a second marble, just like the other experiment.
The first experiment involves a process called
"with replacement".  You put the object back into the bag so that the number of marbles to choose from is the same for both draws. 
The second experiment involves a process called
"without replacement".  You do not put the object back in the bag so that the number of marbles is one less than for the first draw.
As you might imagine, the probabilities for the 2 experiments will not be the same.  In this lesson we will illustrate a variety of these types of problems and explain how to arrive at the correct solutions.
 

 


An Important Note
Sometimes a problem will not specifically state whether it is a problem "with or without replacement".  In these cases it is very important to ask yourself this question:
"Is this problem with replacement?"
or
"Is this problem without replacement?"
Let common sense and a little intuition guide you through these
types of problems.
 


Let's try a few...


A player is dealt 2 cards from a standard deck of 52 cards.
What is the probability of getting a pair of aces?
 

Think about it....obviously this is an experiment "without replacement" because the player was given 2 cards.
To calculate the probability of a pair of aces you use the rules for compound events:
P(ace on first card) = 4/52
(remember,there are 4 aces in the deck)
P(ace on the second card) = 3/51!
(the first card drawn was an ace!)
So the probability of getting 2 aces is:
P(ace,ace) = (4/52)(3/51) = 12/2652 = 1/221!
 

Try this one...


A jar contains 2 red and 5 green marbles.  A marble is drawn, it's color noted, and put back in the jar.  This process is repeated a total of 4 times.
What is the probability that you selected 4 green marbles?
 

Since you put the marble back in the jar after each selection this is an experiment "with replacement".
So,the probability for each draw will be exactly the same:
P(green) = 5/7
Therefore:
P(green,green,green,green)= (5/7)(5/7)(5/7)(5/7)
P(g,g,g,g) = 625/2401
 

Do you have the idea??
Remember...Let common sense be your guide!!!

 For some more practice on your own....
Click here!

 
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