Probability Counting techniques

The need for counting techniques When the outcomes of an experiment are equally likely, computing probability reduces to counting. If N is the number of outcomes in the sample space and N(A) the number of outcomes in the event A, then P(A)=N(A)/N. If a list of outcomes is easily obtained and N is small, then N and N(A) may be determined without the benefit of counting techniques. However, this isn’t always the case.

Multiplication principle If an experiment consists of k stages, where stage j can be carried out in ways, the number of ways to carry out the experiment is

Sampling with replacement After selecting an object it is replaced before the next object is taken If a sample of size k is taken from a set of n objects, the number of possible ordered samples is . Example: If a die is rolled five times, the number of ordered samples is .

Sampling without replacement: permutations and combinations Each of the seven departments has one representative on the student council. From these seven, one is selected as chair, another as vice-chair, and a third to be the secretary. How many ways are there to select the three officers? Now suppose three of the seven representatives are to be selected to attend a convention. In how many ways can the three be chosen? In the first situation, order matters, in the second it doesn’t.

Order matters: permutations If denotes the number of ways of choosing k objects of n, where order matters, then For the selection of chair, vice-chair, and secretary, there are (7)(6)(5)=7!/(7-3)! =210 permutations.

Order doesn’t matter: combinations If order doesn’t matter, we have to remember that each of the orderings have been counted separately. We need to divide by this number, so that For the selection of the committee to go to the convention, there are combinations.

Application An iPod playlist contains 100 songs, 10 of which are by the Beatles. What is the probability that the first Beatles song heard is the fifth song played? The total number of ways to play the first five songs is 100(99)(98)(97)(96). The number of these where the first Beatles song is the fifth song is 90(89)(88)(87)(10). The answer is the ratio of the two.

Application (alternate solution) If we don’t consider order, there are ways to choose the location of the Beatles songs (the denominator). Of these, we have ways to choose the location of the last 9 Beatles songs, and one specific way to choose the first five selections. Taking the ratio, we get the same answer as before.