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Overview
The treatment is organized in a way that permits readers to advance sequentially or skip around between chapters. An essential companion volume to the author's Attacking Trigonometry Problems and Attacking Problems in Logarithms and Exponential Functions, this book will equip students with the skills they will need to successfully approach the problems in probability and statistics that they will encounter on exams.
Product Details
| ISBN-13: | 9780486816517 |
|---|---|
| Publisher: | Dover Publications |
| Publication date: | 10/21/2016 |
| Series: | Dover Books on Mathematics |
| Sold by: | Barnes & Noble |
| Format: | eBook |
| Pages: | 160 |
| File size: | 9 MB |
About the Author
Read an Excerpt
Attacking Probability and Statistics Problems
By David S. Kahn
Dover Publications, Inc.
Copyright © 2016 David S. KahnAll rights reserved.
ISBN: 978-0-486-81651-7
CHAPTER 1
Basic Probability
In this unit, we are going to learn some of the basics of probability. Statistics, as we will learn, is a way of describing and analyzing events that have already occurred. Probability, on the other hand, is a way of predicting events that might occur. In order to understand the probability that a specific event might occur, we need to know about how often that event might occur relative to how often related events might occur. For example, if we roll a die, how often will the number 1 occur compared to how often any of the numbers from 1 to 6 might occur. Given that there are six numbers on a die and, assuming that it is an honest die, any of those numbers could occur when we roll the die, we expect the number 1 to 1/6 occur of the time. Suppose, instead, that we had a 10-sided die, with the numbers 1 to 10 on it. If we roll that die, we would expect the number 1 to occur 1/10 of the time. Going back to our regular die, how often would we expect an even number to occur? Well, there are 3 even numbers, {2, 4, 6}, and 3 odd numbers, {1, 3, 5}. When we roll the die, we would expect an even number to occur 3/6 or 1/2 of the time.
This leads us to our first rule:
Rule #1: The probability that an event will occur is
number of ways the specific event can occur/ total number of possible outcomes
Note that when we use the term event, what we mean is a roll of the die, a toss of the coin, etc. The denominator, or total number of possible outcomes is referred to as the sample space. For example, if we roll a 6-sided die, the sample space is the set of numbers that we can roll, namely {1, 2, 3, 4, 5, 6}. If we toss a coin twice, the sample space is {HH, HT, TH, TT}, where H represents the head side of a coin (heads) and T represents the tail side of a coin (tails). A sample space is very useful for enumerating the possible outcomes when that number is relatively small. We would not want to write out the sample space for tossing 10 coins. There would be 210 = 1024 elements!
Let's look at some more examples.
Example 1: We roll a 6-sided die. What is the probability that a multiple of 3 will occur?
There are two multiples of 3, namely 3 and 6. There are 6 possible outcomes. Thus, the probability that a multiple of 3 will occur is 2/6, or 1/3.
Example 2: We toss a fair coin. What is the probability that we get heads?
There are 2 possible outcomes – heads and tails. There is one way to get heads. Thus, the probability that we will get heads is 1/2.
By the way, a fair coin means that the chances of getting heads are the same as the chances of getting tails.
These examples are very simple so let's look at some that are a little more complicated. Suppose that we toss a fair coin twice. What is the probability that we get 2 heads?
Let's think about what could happen when we toss a coin twice. We could get heads followed by heads. We could get heads followed by tails. We could get tails followed by heads, or we could get tails followed by tails. To represent these in a convenient shorthand, the possible outcomes are: {HH, HT, TH, TT}, as noted earlier. Note that there are 4 outcomes and that only one of them is the one that we want, so the probability is 1/4.
Notice that heads followed by tails is not the same as tails followed by heads. We will see later that sometimes the order in which events happen is important and sometimes it is not.
Example 3: If we toss a fair coin twice, what is the probability that we get at least 1 tail?
If we refer to the previous set of outcomes and sample space, we can see that tails occurs in 3 of the 4 possible outcomes, so the probability is 3/4.
Example 4: We roll a red die and a green die simultaneously. What is the probability that we roll a 7?
There are many possible outcomes, so let's make a table of them. The numbers are in the order (red, green).
Note that there are 6 ways that we could roll a 7: {(1, 6), (2, 5), (3, 4), (4, 3), (5,2), (6, 1)}. There are 36 possible outcomes, so the probability of getting a 7 is 6/36, or 1/6. You should also notice that, for example, the roll (1, 6) is different from the roll (6, 1). The former consists of the red die showing a 1 and the green die showing a 6, whereas on the latter, the red die shows a 6 and the green die shows a 1. This distinction is important, as we will see later.
Let's look at the table again, this time totaling the result of each roll. We get:
Let's look at the probability of each possible sum. If we roll the two dice, we could get any integer sum ranging from 2 to 12.
What is the probability of rolling a 2? There are 36 possible outcomes and only one of them is a 2, namely a 1 on each die, so P(2) = 1/36. (By the way, we will use the notation "P(x) =" to signify the probability of getting outcome x.)
What is the probability of rolling a 3? There are two ways to get a 3, namely rolling a (1, 2) or a (2, 1), so P(3) = 2/36.
Let's list them all:
P(2) = 1/36
P(3) = 2/36
P(4) = 3/36
P(5) = 4/36
P(6) = 5/36
P(7) = 6/36
P(8) = 5/36
P(9) = 4/36
P(10) = 3/36
P(11) = 2/36
P(12) = 1/36
Let's look at some of the results. The highest probability is that one will roll a 7; the lowest probability is that one will roll either a 2 or a 12. The probabilities start at 1/36, go up to 6/36, and then go back down to 1/36. The probability of rolling a 2 is the same as the probability of rolling a 12; the probability of rolling a 3 is the same as the probability of rolling an 11; the probability of rolling a 4 is the same as the probability of rolling a 10; and so on. There are all sorts of patterns with probabilities and we will see more of these as we explore the subject. Most important, let's add up the probabilities. We get 1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36 + 5/36 + 4/36 + 3/36 + 2/36 + 1/36 = 36/36 = 1. This leads us to a very important rule:
The sum of the probabilities of a set of all possible outcomes is 1.
In other words, if we choose an event, say tossing 3 coins, and we list all of the possible outcomes, the sum of the probabilities of these outcomes is always 1. Also:
The probability of any outcome is 0 ≤ P(x) ≤ 1.
That is, we cannot get a probability that is negative or greater than 1.
These rules will prove very useful later on. Let's do an example.
Example 5: If we toss 3 coins, what are all of the possible probabilities for getting heads?
We could get the following outcomes: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. The probabilities are:
P(one head) = 3/8
P(two heads) = 3/8
P(three heads) = 1/8.
Note that if we add these up, we get 3/8 + 3/8 + 1/8 = 7/8, but we are supposed to get a total of 1. What are we forgetting? We are forgetting the probability for the outcomes where we get tails only. That is, P (0 heads) = 1/8. Now if we add up the probabilities, we get 3/8 + 3/8 + 1/8 + 1/8 = 1.
How did we calculate these probabilities? We listed all of the outcomes and then we looked at the number of ways we could get the outcome we want. This is called the Counting Principle. It is an excellent way to calculate probabilities when the number of possible outcomes is small.
Example 6: What is the probability of drawing one queen from an ordinary deck of cards?
There are 52 cards in a deck. Four of the cards are queens, so P (queen) = 4/52 = 1/13.
The Counting Principle is very useful for simple probabilities. However, suppose we are rolling 3 dice. The number of possible outcomes is 216, and it would become unwieldy to write them out all of the time. Thus, there must be other ways to calculate probabilities that are not as tedious. One such way is the Multiplication Rule.
The Multiplication Rule says that if P(A) = x and P(B) = y, then P (A and B) = xy.
That is, if we want to find the probability of 2 events occurring in a row, we multiply the 2 probabilities. Furthermore, if we want to find the probability of 3 events occurring in a row, we multiply the 3 probabilities, and so on. Let's do an example.
Example 7: What is the probability of tossing a fair coin twice and getting heads both times?
The probability of getting heads on the first toss is P (heads on the first toss) = 1/2. The probability of getting heads on the second toss is also P (heads on the second toss) = 1/2. Thus, the probability of getting heads twice is P(HH) = 1/2.1/2 = 1/4. By the way, go back to where we listed the 4 outcomes and notice that we could have also used the Counting Principle to get the answer.
Example 8: What is the probability of tossing a coin 6 times in a row and getting heads each time?
Although we could write out all of the possible outcomes, there are 64 of them (we will learn where this comes from later) and it would get tedious. Using the Multiplication Principle, we get P(HHHHHH) = 1/2. 1/2. 1/2. 1/2. 1/2. 1/2. = )1/2)6 = 1/64.
Example 9: What is the probability of drawing 2 queens in a row from a deck of playing cards?
This question actually has two possible answers. Why? It depends on what happens after one draws the first card in the deck. Is it replaced in the deck or not? Let's look at the difference.
We draw the first card. The probability of getting a queen is 4/52. Now, if the card is replaced in the deck, then the probability of getting a queen is again 4/52, so the probability of getting 2 queens in a row is 4.52. 4.52 = 16/2704 = 1/169. Suppose instead that after drawing the first card it is not replaced in the deck. Now there are only 51 cards left and only 3 queens (because the first card was a queen), so the probability of getting 2 queens in a row is 4/52.3/51 = 12/2652 = 1/221. Note that this is less probable. This should make intuitive sense because the overall number of cards has been reduced by 1, which is around 2% of the total number of cards. The number of queens has also been reduced by 1, but 1 queen is 25% of the total number of queens.
The first variation of this question is referred to as "with replacement" and the second as "without replacement." As we can see, whether the card is replaced or not makes a difference. Let's do another example.
Example 10: What is the probability of drawing 2 spades in a row (a) with replacement and (b) without replacement?
(a) There are 52 cards in a deck and 13 of them are spades, so the probability of getting a spade on the first draw is 13/52. After we replace the card, the probability of getting a spade on the second draw is again 13/52. Therefore, the probability of getting 2 spades in a row is 13/52.13/52 = 169/2704 = 1/16.
(b) This time, the probability of getting a spade on the first draw is again 13/52. But, after drawing the first card, we do not replace that card in the deck. Now, there are 51 cards left in the deck and 12 spades, so the probability of getting a spade on the second draw is 12/52. Thus, the probability of getting 2 spades in a row is 13/52.12/51 = 156/2652 = 1/17.
Remember that the sum of the probabilities of a particular set of events occurring is always 1. Thus, if the probability of an event occurring is given as p, then the probability that the event does not occur is 1 - p. This can be a very useful shortcut to finding a probability. Take a look at the next example.
Example 11: If we toss a fair coin 4 times, what is the probability of getting at least 1 head?
We could find the probabilities of getting 1, 2, 3, or 4 heads, and then add up the probabilities. Or, we could find the probability that we get 0 heads, and subtract that from 1. Why can we do that? Because there are only 5 possibilities, 0, 1, 2, 3, or 4 heads. The sum of the probabilities of these 5 outcomes must be 1. So, if we subtract the probability of 0 heads from 1, we will be left with the sum of the other 4 probabilities. What is the probability of getting 0 heads? The probability that we do not get heads (that is, tails) on any toss is 1/2, so the probability of getting 0 heads on 4 tosses is 1/2. 1/2. 1/2. 1/2. = 1/16. Thus, the probability of getting at least 1 heads is 1 - 1/16 = 15/16. Work this out for yourself to see that it is true.
Let's do another example.
Example 12: We toss a fair coin 10 times. What is the probability that we get at least 1 head?
We could find the probabilities of getting 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 heads and then add them up. However, it would be much simpler to find the probability of getting zero heads and subtracting that from 1. The probability of getting 0 heads [MATHEMATICAL EXPRESSION OMITTED]. Thus, the probability of getting at least one head is 1 - 1/1024 = 1023/1024.
Let's do some practice problems.
Practice Problems
Practice Problem 1: If we roll a 6-sided die, what is the probability that we will get a prime number?
Practice Problem 2: If we toss a fair coin 3 times, what is the probability of getting 3 tails?
Practice Problem 3: If we toss a red and a green die simultaneously, what is the probability of rolling either a 7 or an 11?
Practice Problem 4: If we toss a red and a green die simultaneously, what is the probability of not rolling either a 7 or an 11?
Practice Problem 5: A jar contains 6 red, 8 blue, and 4 green marbles. What is the probability of drawing two red marbles in a row (a) with replacement; (b) without replacement?
Practice Problem 6: We draw 2 cards in a row, without replacement, from a standard deck of cards. What is the probability that they are both aces?
Practice Problem 7: We draw 3 cards in a row, without replacement, from a standard deck of cards. What is the probability that we get 3 hearts?
Practice Problem 8: We toss a coin 5 times. What is the probability that we get at least 1 tail?
Practice Problem 9: If the probability of having a male baby is 0.51, what is the probability of a family having three babies, all of them male?
Practice Problem 10: If the probability of having a male baby is 0.51, what is the probability of a family having three babies, at least one of which is male?
Solutions to Practice Problems
Solution to Practice Problem 1:If we roll a 6-sided die, what is the probability that we will get a prime number?
The possible rolls are 1, 2, 3, 4, 5, and 6. Of these rolls, 2, 3, and 5 are prime numbers. So the probability of rolling a prime number is 3/6 = 1/2.
Solution to Practice Problem 2:If we toss a fair coin 3 times, what is the probability of getting3 tails?
The probability of getting tails when we toss a fair coin is 1/2, so the probability of getting 3 tails is 1/2. 1/2. 1/2. = 1/8.
Solution to Practice Problem 3:If we toss a red and a green die simultaneously, what is the probability of rolling either a 7 or an 11 ?
When we roll the two dice, there are many possible outcomes, so let's make a table of them. The numbers are in the order (red, green). We could get:
[TABLE OMITTED]
If we total the results, we get:
[TABLE OMITTED]
There are 6 ways to roll a 7 and 2 ways to roll an 11, giving a total of 8 possibilities. There are 36 possible rolls, so the probability of rolling either a 7 or an 11 is 8/36 = 2/9.
Solution to Practice Problem 4:If we toss a red and a green die simultaneously, what is the probability of not rolling either a 7 or an 11?
As we saw in the previous problem, the probability of rolling either a 7 or an 11 is 2/9. We know that the sum of the probabilities of all possible rolls is 1, so the probability of not rolling a 7 or an 11 is 1 - 2/9 = 7/9.
Solution to Practice Problem 5:A jar contains 6 red, 8 blue, and 4 green marbles. What is the probability of drawing 2 red marbles in a row (a) with replacement; (b) without replacement?
(a) There are 18 marbles in the jar, so the probability of drawing a red marble on the first draw is 6/18. Now, we replace the marble after we have drawn it, so the probability of drawing a red marble on the second draw is again 6/18. Thus, the probability of drawing 2 red marbles in a row is 6/18.6/18 = 1/9.
(b) The probability of drawing a red marble on the first draw is 6/18. Because we do not replace the marble after drawing it, there are now only 5 red marbles in the jar and a total of only 17 marbles. The probability of drawing a red marble on the second draw is 5/17. Thus, the probability of drawing 2 red marbles in a row is 6/18.5/17 = 5/51.
Solution to Practice Problem 6:We draw 2 cards in a row, without replacement, from a standard deck of cards. What is the probability that they are both aces?
There are 52 cards in a standard deck and there are 4 aces, so the probability that the first card is an ace is 4/52. After we draw an ace, there are only 51 cards left in the deck and only 3 aces, so the probability that the second card is an ace is 3/51. Thus, the probability of drawing both aces is 4/52.3/51 = 1/221.
Solution to Practice Problem 7:We draw 3 cards in a row, without replacement, from a standard deck of cards. What is the probability that we get 3 hearts?
There are 52 cards in a standard deck and 13 of them are hearts, so the probability that the first card is a heart is 13/52. After we draw the first heart, there are only 51 cards left in the deck and only 12 hearts, so the probability that the second card is a heart is 12/51. Now, after drawing 2 hearts, there are only 50 cards left in the deck and only 11 hearts, so the probability that the third card is a heart is 11/50. Thus, the probability that all 3 cards are hearts is 13/52.12/51.11/50 = 1716/132600 = 11/850.
(Continues...)
Excerpted from Attacking Probability and Statistics Problems by David S. Kahn. Copyright © 2016 David S. Kahn. Excerpted by permission of Dover Publications, Inc..
All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.
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Table of Contents
Table of Contents
Unit One Basic Probability
Unit Two Permutations and Combinations
Unit Three More Probability
Unit Four Conditional Probability
Unit Five Statistics Terms and Experimental Design
Unit Six Some Basics of Statistics
Unit Seven Center and Spread
Unit Eight Exploring Data and Introducing the z Score
Unit Nine Probability Distributions and More About the z Test
Unit Ten Confidence Intervals
Unit Eleven Hypothesis Testing
Unit Twelve Working with Two Samples, Correlation, and Regression
Appendix