REPLACEMENT or WITHOUT REPLACEMENT

~Finding probabilities of selecting so many objects in a group without replacement, can be solved by dividing two combinations.

~The top combination gives the number of ways you can select the objects & the bottom combination gives the total number of ways of selecting the same number of ANY of the objects (sample space).

~This is nothing more than the basic definition of probability.

~This technique (dividing combinations) is equivalent to selecting that number of preferred objects (one at a time, without replacement).

~Let's look at an example:

~A bin contains 123 light bulbs of which 12 are defective. If 4 light bulbs are randomly selected from the bin without replacement, find the probability that all four bulbs selected are good ones.

~Note that there are 111 good bulbs & 12 defective bulbs.

~The number of ways of selecting 4 good bulbs must come from the 111 good ones & is given by 111C4.  

~The number of ways of selecting any 4 bulbs (sample space) is given by 123C4.

~By division, we get an answer of .6597

~The TI-83 does this easily by entering 111 Math, over to PRB, down to menu 3, enter 4 divide, enter 123 Math, over to PRB, down to menu 3, enter 4, enter.

~The same answer can be found by selecting one at a time without replacement, then using the multiplication rule.

~That would look like this: (111/123)(110/122)(109/121)(108/120) gives .6597 also.

~However, I recommend using the TI-83 & dividing the combinations (it's faster). Remember, this is used when selecting without replacement.

~However, if we do the same problem with replacement, the answer comes out differently.

~In that case, the probability of selecting each good bulb remains the same for each of the four selections. This gives, using the multiplication rule, (111/123)(111/123)(111/123)(111/123) or better yet, (111/123)⁴ or .6632 (a different answer)

~So, be aware.  Make sure you read these types of problems very carefully to see which case applies. This could be the difference between a correct answer or a wrong answer on a test.

~Note: Sampling without replacement can have an affect on the standard deviation of the sample means. This occurs, if the sample size, n, is greater than 5% of the population size, N. In that case, the standard deviation of the sample means must be adjusted by multiplying by the finite population correction factor: square root of [(N-n)/(N-1)].

~However, when we sample with replacement, the population size is considered infinite and this correction does not apply.