[1] "4.3.1, 4.3.2, 4.3.5-12, 4.3.13,4.3.19, 4.21, 4.24, 4.3.28, 4.3.29"Lecture 6/7
Useful Counting Rules
Section 4.3 Useful Counting Rules
Suppose that an experiment contains a large number N of simple events and you know that all the simple events are equally likely - each simple event has probability \(\frac{1}{N}\).
The probability of an event A is just the number of simple events in the event A (\(n_A\)) divided by N.
The mn rule
Suppose that an experiment is performed in two stages. If there are m possible outcomes for the first stage and, for each of these outcomes, there are n possible outcomes for the second stage then there are \(mn\) possible outcomes.
For example, car can be ordered in three styles and one of four paint colors. There are 12 possible options.
Example 4.9
Two dice are are tossed - how many simple events are in sample space S?
Example 4.10
A candy dish contains one yellow and two red candies - if you draw two candies without replacement, how many simple events are there.
Extended mn rule
If there are k stages with \(n_i\) possible outcomes at each stage then the number of possible outcomes for the experiment is
\[ n_1n_2n_3\dots n_k = \prod_{i=1}^k n_i \]
Example 4.12
A truck driver can take:
3 routes from city A to B
4 routes from B to C
3 routes from C to D
How many total routes can they take from A to B to C to D?
Counting rule for permutations
Orders or permutations, Let’s say you have 3 books how can you arrange two of them on a shelf? How about 3 books?
Factorial Notation:
The number of ways we can arrange *n_ distinct objects, taking them r at a time is
\[ P^{n}_r = \frac{n!}{(n-r)!} \]
\[ n! = n(n-1)(n-2)\dots(3)(2)(1) \]
Note that \(0! \equiv 1\)
Example 4.13
Three lottery tickets are drawn from a total of 50 by three employees - How many simple events are associated with the experiment?
A counting rule for combinations
What happens when you don’t care about the ordering of some items?
For example, we talked about how to arrange 2 books of 3 on a bookshelf. What if we didn’t care about the order and just wanted to pick 2 of 3 books?
\[ C^{n}_r = \frac{n!}{r!(n-r)!} \]
Example 4.15
A printed circuit board may be purchased from five suppliers - how many ways can three suppliers be chosen?
Example 4.16
Five manufacturers produce a certain electronic device, whose quality varies. from manufacturer to manufacturer. If you were to select three manufacturers at random, what is the chance that the selection would contain exactly two of the best three?
What is the event of interest?
\(P(A) = \frac{n_A}{N}\)
Step 1 - How many ways are there to select three manufacturers from a group of five ?
\(N = C^5_3\)How to find \(n_a\)
How many ways are there to select two of the best three?
How many ways to select of the two “not best”
Excel
- Combin
- permut
Homework
Answers: Chapter 4 - Section 3