Day 18 Basic Counting Rule

Probabilities Related concepts: Experiment, Event, Sample Space If we assume all sample points are equally likely, the probability of interest is therefore: { # of supporting sample points } / { Total # of sample points in the sample space }

Our Routine We have been calculating probabilities by: 1. List all sample points in the sample space 2. List all supporting sample points 3. Count 1 and 2 and do the division

But the truth is…. There are problems that the outcomes are just too many for us to count…. Example: Toss a coin 6 times Roll a fair die 3 times 3 People in the elevator in Math building Choose what to include in your sandwich at Subway

Some other examples There are 2,598,960 different ways to deal a poker hand. There are 13,983, 816 different ways to select 6 out of 49 lottery numbers. The area code 765 can include as many as 10^7 different telephone numbers.

Sometimes listing all the sample points in the sample space may be impossible, so is listing all the supporting sample points. We have to go a little abstract….

Basic Count Rule Suppose that are r actions (choices, experiments) are to be performed in a definite order, further suppose that there are m 1 possibilities for the first action, m 2 possibilities for the second action, etc, then there are m 1 *m 2 *…*m r total possibilities altogether.

Example Assuming Mary has 6 pairs of shoes, 10 different tops, 8 different bottoms and 4 different outwears, then how many combinations can she have for outfit? 6*8*10*4=1920

Example I bought a lock for my new bike since my last one was stolen. There are 4 slots numbered 0 to 9, how many possible combinations are there? 10^4

Subway example (simple version) At a Subway, you have to decide what you want to put in your sandwich, the choices you have are: Four types of bread Five types of cheese Six types of veggies Seven types of meat Assuming you can only choose one from each of the above categories, how many total possible combinations could we get?

Try to solve some problems 1. Mary’s outfit problem Mary is having a job interview and she wants to decide what to wear. If there are 2 pairs of shoes, 3 tops, 2 bottoms and 2 outwears that are appropriate for an interview and she randomly picks what to wear for the interview among all she has, what is the probability that she wears an interview-appropriate outfit?

My bike lock problem 1. If the combination only includes even numbers, 2. If the first number can not be 0 and all four numbers can not be the same, 3. If the combination must have at least one 4 and at least one 5. What are the probabilities for those specified combination?

Subway problem If I don’t like white bread, only like Swiss cheese, don’t like onion and am allergic to seafood and chicken, what is the chance that I get a sandwich that I actually like?

Birthday Problem You are at a bar and it is a quiet night, only 22 people, including you, are there. You bet with the bartender that if no two persons at the bar have the same birthday, you will buy everybody a drink. The bartender seems to know some probability and he wants to count himself in to do the math. Would you let him do that?

22 people: P{no one shares the same birthday}= (365*364*363*…*344)/(365^22)= people P{no one shares the same birthday}= (365*364*363*…*344*343)/(365^23)= In this case, you want to bet on two people have the same birthday.