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2.1: Calculating Basic Probabilities

Created by: Bruce DeWItt

Learning Objectives

  • Understand how to calculate and write a probability
  • Understand what constitutes chance behavior
  • Understand the concept of the Law of Large Numbers

Probabilities give us an idea of how likely it is for a certain event to happen. For example, when a coin is flipped, the chance that it comes up heads is 50%. Probabilities can be expressed as decimals, fractions, percents, or ratios. We could have said the probability of flipping heads is , 0.5, \frac{1}{2}, 50% or 1:2. Each of these conveys the idea that we should expect to get a heads half of the time. Probabilities only give us an idea of what to expect in the long run. However, they do not tell us what will happen in the short term.

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Suppose we flip a coin 10 times in a row and get heads each time. The next coin flip is still a random event because while we cannot tell for certain what the next flip will be, we can be certain that about 50% of all tosses over a long set of tosses will be heads. Some people think that we are on a roll so we are more likely to get another heads. Others will say that getting tails is more likely because we are due to get tails. The truth is that we cannot tell what will happen on the next flip. The only thing we know for certain is that there is a 50% chance that the coin will be heads on its next flip. If we continue to flip this same coin hundreds of times, we would expect the percent of heads to get closer and closer to 50%.

Chance Behavior is not predictable in the short term, however, it has long term predictability. The Law of Large Numbers tells us that despite the results on a small number of flips, we will eventually get closer to the theoretical probability. The outcomes in any random event will always get close to the theoretical probability if the event is repeated a large number of times. We might roll a die 4 times in a row and get a 6 each time, however, if we rolled this die hundreds of times, the percent of time that we get a 6 will get closer and closer to the theoretical probability of \frac{1}{6}.

When calculating a probability, we divide the number of favorable outcomes (outcomes we are interested in) by the total number of outcomes. In other words, the probability that outcome 'A' occurs is found by the formulaP(A)=\frac{\#\;of\;favorable\;outcomes}{total\; \#\;of\;outcomes}.

Consider a standard deck of 52 playing cards.

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If we asked the question "What is the probability of being dealt a face card (jack, queen, or king)?", we would need to count how many cards are face cards and then divide by the total number of cards, 52.

In this situation there are 12 face cards and 52 cards overall so our probability of getting a face card is \frac{12}{52}=\frac{3}{13}\approx0.23.


In probability, there are outcomes that are sure to happen and there are outcomes that are impossible. If we are once again dealing with a standard 52 card deck, the chance of being dealt either a red card or a black card if one card is dealt is 100%. The chance of being dealt a blue card is 0% since there are no blue cards in a standard deck. All random events have probabilities between 0 and 1. In addition, the sum total of the probabilities for all possible outcomes in the sample space is equal to 1. In other words, if an event occurs, there is a 100% chance that one of the possible outcomes will happen. The list below summarizes these rules.

a) The probability of a sure thing is 1.

b) The probability of an impossible outcome is 0.

c) The sum of the probabilities of all possible outcomes is 1.

d) The probability for any random event must be somewhere from 0 to 1.

As shown earlier, we notate the probability of event 'A' happening as P(A). For example, the probability of rolling a three on a six-sided die can be written P(3)=\frac{1}{6}. Sometimes we are interested in the probability of an event not occurring. This is called the complement of the event. We can write the probability of the complement of event 'A' happening as P(~A), P(not A), or P(A^c). The formula for the complement of an event is P(not\;A)=1-P(A). On our die rolling question, P(~3)= 1-P(3)=1-\frac{1}{6}=\frac{5}{6}. In other words, there is a \frac{5}{6} chance of the dice not landing on a 3. It is important to notice that the probability of an event happening and the probability of its complement always add up to 1.

Example 1

Which of the following situations are random events?

i) A student looks through their closet to decide what shirt to wear to school.

ii) A student labels each of their 6 pairs of shoes 1 through 6 and then rolls a single die to decide which pair to wear.

iii) The state legislature decides to increase funding to schools by 3%.

iv) A professional golfer makes a hole-in-one on a 200 yard hole.


Solution

Situations i) and ii) are not random events. In both cases, there are additional factors that are influencing the decision. The day of the week or the temperature outside might influence your shirt choice and how much money the state legislature happens to have might influence funding.

Both situations iii) and iv) are random events because while we can't predict what will happen in this particular instance, we can make long term predictions. We can predict the percent of the time the student might end up with the shoes labeled #2 and we can predict the percent of the time that the golfer will make a hole-in-one based upon previous performance.

Example 2

In the game of pool, there are a total of 15 balls. Balls numbered 1-8 are solid and balls 9-15 are striped. There are two pool balls of each color, for example, there are two yellow pool balls. One of those are solid and one of those are striped. The only exception to this is that there is only 1 black pool ball, the eight ball, and it is solid.

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Suppose the pool balls were put in a bag and a single pool ball is pulled out of the bag. What is the probability that the ball:

a) is yellow?

b) is striped?

c) has a number on it that is greater than 10?

d) is not striped?


Solution

a) P(Y)=\frac{2}{15}\approx0.13

b) P(Striped)=\frac{7}{15}\approx0.47

c) P(>10)=\frac{5}{15}=\frac{1}{3}\approx0.33

d) P(~Striped)=1-P({Striped})=1-\frac{7}{15}=\frac{8}{15}\approx0.53

In addition to these types of questions, we can also calculate probabilities by incorporating our counting methods from Chapter 1. Recall that the probability of an event occurring is the number of favorable outcomes divided by the number of total possible outcomes.

Example 3

A jury of 12 people is to be selected from a group of 12 men and 8 women. What is the probability that the jury has at least 6 women on it?

Solution

The total number of outcomes possible is based upon selecting 12 members from a pool of 20. Since order will not matter, there are _{20} C _{12}=125,970 ways to pick a jury of 12. We now want to have at least 6 women on the jury. This means we could have 6 women and 6 men or 7 women and 5 men or 8 women and 4 men on the jury. Mathematically, this would be _8 C _6\times_{12} C_6+_8C_7\times_{12}C_5+_8C_8\times_{12}C_4=28\times924+8\times792+1\times495=25,872+6,336+495=32,703. There are 32,703 ways to have at least 6 women on the jury out of a possible total of 125,970 different juries or \frac{32,703}{125,970}\approx0.26. There is about a 26% chance that the jury will have at least 6 women on it.

Sometimes, data is organized in a Venn diagram, as shown in Example 4 on the following page. We will examine these in greater depth in section 2.3 but for now, it is important to understand that a Venn diagram is an organizational tool that makes it easier to interpret a situation and answer basic probability questions.


Example 4

A class of 30 students is surveyed to see whether or not they had a science class and/or a math class this trimester. There are 18 students that have a math class, 14 students who have a science class, and 4 students who have neither. It also turns out that this includes 6 students who currently have both classes. The results of the survey are shown in the Venn diagram below.

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a) How many total students are taking a math class this trimester?

b) What is the probability that a randomly selected student is taking a math class this trimester?

c) What is the probability that a randomly selected student is taking both a math and science class this trimester?

d) What is the probability that a randomly selected student is not taking either a math or science class this trimester?

Solution

a) There are 12 kids who only have a math class and 6 kids who have both a math a science class this trimester for a total of 18 kids.

b) P(Math)=\frac{18}{30}=\frac{3}{5}=0.6

c) P(Math \And Science)=\frac{6}{30}=\frac{1}{5}=0.2

d) P(No Math or Science)=\frac{4}{30}=\frac{2}{15}\approx 0.13


Problem Set 2.1

Exercises

For problems 1-5, express your answer both as a fraction (reduce if possible) and as a decimal to the nearest hundredth.

1) Suppose a single card is dealt from a standard deck of 52 cards. Find the probability that the card is:

a) a red card.

b) a face card.

c) an ace.

d) a three.

e) a club.

f) the three of clubs.

g) a black king.

h) not a spade.

2) A bag contains some jelly beans. There are a total of 6 red jelly beans, 4 green jelly beans, 2 black jelly beans, 5 yellow jelly beans, and 3 orange jelly beans in the bag. Suppose one jelly bean is drawn from the bag.

a) Find P(purple).

b) Find P(yellow).

c) Find P(~red).

3) A single 6-sided die is rolled one time. Find the probability that the result is:

a) a three

b) a seven

c) an even number

d) a prime number

e) a number equal to or greater than 5.


4) The game Scattegories® uses a 20-sided die. It has all the letters of the alphabet on it except Q, U, V, X, Y, and Z. Find each probability below if the die is rolled one time.

a) P(Vowel)

b) P(~Vowel)

c) P(Q)

d) P(Qc)

e) P(a letter alphabetically after Q)

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[Figure9]

5) The month of October in a 2011 calendar has 31 days with October 1st being a Saturday as shown in the calendar on the following page. Suppose a day is randomly selected. Find each probability.

a) P(weekend)

b) P(not a weekend)

c) P(October 31st)

d) P(October 32nd)

e) P(~October 31st)

f) P(an odd-numbered day)


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6) A roulette wheel contains 38 slots. When the wheel is spun, a ball is dropped onto the wheel and the ball will stop on one of the slots. There are 18 black slots, 18 red slots, and 2 green slots. Suppose the ball on a roulette wheel has landed on red four times in a row. What is the chance that the ball will drop on red on the next spin?

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7) A coin has been flipped 10 times. Suppose that it has come up heads on only 2 out of those ten times.

a) What percent of the time has the coin come up heads?

b) Suppose we flip the coin 90 more times and 45 of those 90 flips come up heads. Of the 100 flips completed so far, what percent of the time has the coin come up heads?

c) Suppose we continue to flip the coin an additional 900 times and that 450 of those 900 flips come up heads. Of the 1000 flips completed, what percent of the time has the coin come up heads?

d) As we flipped the coin more and more, the percentage of heads got closer and closer to 50% despite the fact that only 2 of the first 10 flips were heads. What rule does this illustrate?


8) Two 6-sided dice are rolled and we keep track of the total on the two dice.

a) Make a 6 by 6 grid showing the different totals that you can get when rolling the two dice.

b) What is the probability that you get doubles?

c) What is the probability that you get a total of 7?

d) What is the probability that you get a total of at least 8?

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9) The high school concert choir has 7 boys and 15 girls. The teacher needs to pick three soloists for the next concert but all of the members are so good she decides to randomly select the three students for the solos.

a) In how many ways can the teacher select the 3 students?

b) What is the probability that all three students selected are girls?

c) What is the probability that at least one boy is selected?

10) A test begins with 5 multiple choice questions with four options on each question. It then has 5 true/false questions.

a) How many answer keys are possible?

b) What is the probability of getting every question correct if a student guesses on each question. Leave your answer as a fraction.


11) A lawn and garden store is moving locations and needs to move its riding lawn mowers to the new store. They have 8 mowers with 36-inch decks, 15 mowers with 42-inch decks, and 6 mowers with 48-inch decks that need to be moved. The trailer they are using can move a total of 8 mowers on each load so several trips will have to be made.

a) In how many ways can 8 mowers be randomly selected for the first load?

b) What is the probability that all the mowers with 48-inch decks get selected for the first load? Leave your answer as a reduced fraction.

c) What is the probability that the first load has exactly two 36-inch deck mowers, four 42-inch deck mowers, and two 48-inch deck mowers?

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Review Exercises

12) In how many ways can three students be selected for a committee if there are 11 students from which to select?

13) A hockey player needs new skates, a new helmet, and a new stick. Hockey Central has 5 brands of skates, 6 brands of helmets, and 8 brands of sticks. In how many different ways can the player select one of each item?

14) Two standard 6-sided dice are rolled and the results from the two dice are added together. Build a grid to determine which outcome is most likely to occur.

15) On a TV game show, three contestants must each pick a box which they believe contains the grand prize based upon clues given about each box. In how many different ways can this be done if there are 10 boxes from which to choose?


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Date Created:

Jun 14, 2011

Last Modified:

Sep 30, 2014
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