Download presentation

Presentation is loading. Please wait.

Published byImani Punch Modified about 1 year ago

1
MAT 1000 Mathematics in Today's World

2
Last Time We talked about computing probabilities when all of the outcomes of a random phenomenon are “equally likely” We also looked at a way to experimentally determine probabilities using “simulations”

3
Today When all outcomes are equally likely, finding probabilities only requires counting. We will learn methods to count even when there are more outcomes then we could write down in a list.

4
Counting

5
Equally likely outcomes All that these formulas require is counting. For example, if I want to know the probability of any of the outcomes from rolling two dice, it’s enough to know that there are 36. I don’t need to list them all. This is especially useful when there are too many outcomes to list.

6
Formulas for counting

7

8

9

10

11

12
In this case the number turns out to be small enough to count without a formula. Our “event” is getting a PIN all of whose digits are the same. For example 1111, or How many such four- digit PINs are there? There are ten. The digits could all be any one of the ten digits 0, 1, …, 9.

13
Formulas for counting

14
Formulas for counting Here’s a different counting problem: how many four-digit PINs are there if we are not allowed to repeat digits? In other words a PIN like 1234 is allowed, but not a PIN like 2452 or 7674, where a digit is repeated. We already know that the number of such PINs must be less than 10,000, but how many are there? We can use the fundamental principle of counting.

15
Formulas for counting We are “choosing” four digits. How many choices are there for the first? There are 10 choices. What about the second? Now, we don’t have 10 choices—we can’t repeat the first digit. Suppose the first digit was 7. If we pick 7 for the second digit, we’ve repeated ourselves. So the second digit can be any number except 7. Then how many choice are there? There are 9 choices—anything except our first digit.

16
Formulas for counting So we have 10 choices for the first digit and 9 for the second. How many choices are there for the third? We have already picked two digits, and we can’t reuse either of those. So we only have 8 choices. We want a four-digit PIN, so we need to pick one more number. Since it can’t be the same as the first, second, or third choice we’ve made, we’ve only got 7 options.

17
Formulas for counting

18

19
Counting

20
How many outcomes are in the event? In other words, how many PINs are there where no digits are repeated and the first digit is 1? We have to count. Unlike earlier, we don’t have four spaces to fill. ___ ___ because we know what goes in the first space…

21
Counting How many outcomes are in the event? In other words, how many PINs are there where no digits are repeated and the first digit is 1? We have to count. Unlike earlier, we don’t have four spaces to fill. _1_ ___ ___ ___ because we know what goes in the first space… … a 1.

22
Counting So we only have three spaces to fill. How many choices do we have for the first space? Even though there are 10 digits, remember that the first digit must be a 1, and we aren’t allowed to repeat. So we only have 9 choices. How about for the second digit? There are 8 choices.

23
Counting

24

25

26

27
Suppose 10 runners are competing in a race. The winner gets a gold medal, second place gets a silver medal, and third place gets a bronze medal. How many different arrangements of gold, silver, bronze are possible? Here we are “filling” three “spaces” ______ ______ _______ gold silver bronze How many possible gold medal winners are there? 10. Once we know who wins the gold, we’ve only got 9 choices for silver, and then 8 for the bronze.

28
Counting

29

30

31
If I was one of the runners, I might like to try and find the probability of getting a medal. This is really three “events” 1.I get the gold medal 2.I get the silver medal 3.I get the bronze medal These events are disjoint (I can’t win more than one medal), so to find the probability of any one of these three events occurring, I can add the probabilities of each.

32
Counting Let’s count the number of possible arrangements in which I am the gold medal winner. __Me___ ______ _______ gold silver bronze There are 9 runners other than me, and we can put any one of them in the silver medal spot. That leaves 8 choices for the bronze medal spot. So there are 72 arrangements in which I am the gold medal winner

33
Counting

34

35
Summary

36

37

38

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google