What does it mean for the probability of something if an object is chosen without replacement?

For a finite number of discrete objects and a given event occuring, without replacement reduces the probability of that event occuring but increases the probability of the other events occuring.

Example:

If there are 5 red discs and 2 blue discs in a bag, then the probabilities of selecting a disc at random are:

P(red disc) = 5/7

P(blue disc) = 2/7

a) If one takes a red disc from the bag and is NOT REPLACED back into the bag, then the probabilities of selecting a disc at random are:

P(red disc) = 4/6 < 5/7

P(blue disc) = 2/6 > 2/7

b) If one takes a blue disc from the bag and is NOT REPLACED back into the bag, then the probabilities of selecting a disc at random are:

P(red disc) = 5/6 > 5/7

P(blue disc) = 1/6 < 2/7

Hi Ivy,

Let's assume we have N objects and the chance of selecting any one of these objects is equal. Then the probability of randomly selecting any one of these objects would 1/N. That is, 1 divided by the total number of objects.

If we were to add an object, we would have N+1 objects, and the probability of randomly selecting one would decrease (as it would now be 1/(N+1) ).

Similarly, if we select an object and don't replace it, we would have N-1 objects to choose from. So the probability of randomly selecting any given object will again be 1 divided by the total number of objects.

I hope that helps,

Zach

Suppose we have a class of 8 girls and 5 boys, and from the class 3 students are chosen, at random, one at a time.

If the students are chosen “without replacement” then

the 1^st student is chosen from the class of 13,

the 2^nd student is chosen from the remaining 12,

and the 3^rd student is chosen from the remaining 11.

Consequently the probability of, for example, the 2^nd student chosen being a girl depends on whether the 1^st student was a girl or boy. 

If an object is chosen without replacement this will affect the probability when the process is repeated.

If there are 4 balls, 2 red, 2 white on the first selection P(red) would be 2/4 (1/2), repeating the process would give a probability of 1/3 ( only 1 red ball remains in the bag of 3 balls). The probability of 2 red balls being selected would therefore be 1/2 x1/3 = 1/6.

It is usually easiest to do calculations of probabilities without replacement using tree diagrams.

It means that once an object of the total number of objects is chosen in the first event, the same event is repeated with the object chosen in the first event being left out.

You will notice the total number of objects at the end of each event(branch) in a tree diagram with a denominator one less than the previous.

It means that once you have taken out one of the objects from a group you don't put it back. This reduces the number of objects by one for your second event.

This means, if you represent what happens on a tree diagram, the outcome of the second event is conditional on the outcome of the first event, so if your first event is known as A and your second event is known as B then e.g. P(AnB) = P(A)xP(B|A), the first branch is P(A) the second branch is P(B|A). Similarly P(A'nB) = P(A')xP(B|A') . Hence you are required to multiply along the relevant pair of branches to find the probability of both events happening. Your denominator will be one less on the second layer of the tree diagram.