5.1: Introduction on Probability
- Page ID
- 91506
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Students will be able to:
- Understand and use probability terminology correctly.
- Calculate the probability of a given event for equally likely outcomes.
- Determine the compliment of a given event and find its probability.
Toss a thumb tack one time. Do you think the tack will land with the point up or the point down?
We cannot predict which way the tack will land before we toss it. Sometimes it will land with the point up and other times it will land with the point down. Tossing a tack is a random experiment since we cannot predict what the outcome will be. We do know that there are only two possible outcomes for each trial of the experiment: lands points up or lands point down. If we repeat the experiment of tossing the tack many times we might be able to guess how likely it is that the tack will land point up.
- A random experiment is an activity or an observation whose outcome cannot be predicted ahead of time.
- The result of an experiment is called an outcome.
- A trial is one repetition of a random experiment.
- The sample space is the set of all possible outcomes for a random experiment.
- An event is a subset of the sample space. An event can be thought of as a collection of outcomes that are grouped together.
- A simple event is an event that consists of only one outcome. For example, rolling a die and getting a "2" is a simple event.
- The probability of an event is the measure of how likely it will happen. It can be expressed as a number between 0 and 1, or as a percentage between 0% and 100% The probability of an impossible event is 0 (or 0%), and the probability of a certain event is 1 (or 100%).
Do you think chances of the tack landing point up and the tack landing point down are the same? This is an example where your intuition may be wrong. Having only two possible outcomes does not mean each outcome has a 50/50 chance of happening. In fact we are going to see that the probability of the tack landing point up is about 66%.
To begin to answer this question, toss a tack 10 times. For each toss record whether the tack lands point up or point down.
|
Trial |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|---|---|---|---|---|---|---|---|---|---|---|
|
Up/Down |
Up |
Down |
Up |
Up |
Down |
Up |
Up |
Down |
Up |
Down |
The random experiment here is tossing the tack one time. The possible outcomes for the experiment are that the tack lands point up or the tack lands point down so the sample space is S = {point up, point down}. We are interested in the event E that the tack lands point up, E = {point up}.
Based on our data we would say that the tack landed point up six out of 10 times. The fraction, , is called the relative frequency. Since
we would guess that probability of the tack landing point up is about 60%.
Let’s repeat the experiment by tossing the tack ten more times.
|
Trial |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|---|---|---|---|---|---|---|---|---|---|---|
|
Up/Down |
Up |
Up |
Up |
Down |
Down |
Up |
Up |
Down |
Up |
Up |
This time the tack landed point up seven out of 10 times or 70% of the time. If we tossed the tack another 10 times we might get a different result again. The probability of the tack landing point up refers to what happens when we toss the tack many, many times. Let’s toss the tack 150 times and count the number of times it lands point up. Along the way we will look at the proportion of landing point up.
|
Trials |
Number Up |
Total Number of Up |
Total Number of Trials |
Proportion |
|---|---|---|---|---|
|
0-10 |
6 |
6 |
10 |
6/10=0.60 |
|
11-20 |
7 |
13 |
20 |
13/20=0.65 |
|
21-30 |
8 |
21 |
30 |
21/30=0.70 |
|
31-40 |
6 |
27 |
40 |
27/40=0.68 |
|
41-50 |
8 |
35 |
50 |
35/50=0.70 |
|
51-60 |
8 |
43 |
60 |
43/60=0.72 |
|
61-70 |
4 |
47 |
70 |
47/70=0.67 |
|
71-80 |
7 |
54 |
80 |
54/80=0.68 |
|
81-90 |
5 |
59 |
90 |
59/90=0.66 |
|
91-100 |
6 |
65 |
100 |
65/100=0.65 |
|
101-110 |
10 |
75 |
110 |
75/110=0.68 |
|
111-120 |
5 |
80 |
120 |
80/120=0.67 |
|
121-130 |
8 |
88 |
130 |
88/130=0.68 |
|
131-140 |
4 |
92 |
140 |
92/140=0.66 |
|
141-150 |
7 |
99 |
150 |
99/150=0.66 |
When we have a small number of trials the proportion varies quite a bit. As we start to have more trials the proportion still varies but not by as much. It appears that the proportion is around 0.66 or 66%. We would have to do about 100,000 trials to get a better approximation of the actual probability of the tack landing point up.
The tack landed point up 99 out of 150 trials. The probability P of event E is written as:
\[P(E)=\frac{\# \text { of trials with point up }}{\text { total number of trials }}=\frac{99}{150} \approx 0.66\]
We would say the probability that the tack lands point up is about 66%.
Equally Likely Outcomes
In some experiments all the outcomes have the same chance of happening. If we roll a fair die the chances are the same for rolling a two or rolling a five. If we draw a single card from a well shuffled deck of cards, each card has the same chance of being selected. We call outcomes like these equally likely. Drawing names from a hat or drawing straws are other examples of equally likely outcomes. The tack tossing example did not have equally likely outcomes since the probability of the tack landing point up is different than the probability of the tack landing point down.
An experiment has equally likely outcomes if every outcome has the same probability of occurring.
For equally likely outcomes, the probability of event A, P(A), is:
\[P(A)=\dfrac{ \text {number of ways for A to occur}}{\text { total number of outcomes }}.\]
Round Off Rule: Give probabilities as a fraction or as a decimal number rounded to three decimal places.
A standard deck of 52 playing cards consists of four suits (hearts, spades, diamonds and clubs). Spades and clubs are black while hearts and diamonds are red. Each suit contains 13 cards, each of a different rank: An Ace (which in many games functions as both a low card and a high card), cards numbered 2 through 10, a Jack, a Queen, and a King.
The image below gives an example of a complete deck of 52 cards.
Draw a single card from a well shuffled deck of 52 cards. Each card has the same chance of being drawn so we have equally likely outcomes. Find following probabilities of the given events.
- P(card is red)
P(card is red) = \(\dfrac{ \text {number of red cards}}{\text { total number of cards }} = \dfrac{26}{52} = \dfrac{1}{2}\)
The probability that the card is red is \(\dfrac{1}{2}\).
- P(card is a heart)
P(card is a heart) = \(\dfrac{ \text {number of hearts}}{\text { total number of cards }} = \dfrac{13}{52} = \dfrac{1}{4}\)
The probability that the card is a heart is \(\dfrac{1}{4}\).
- P(card is a red 5)
P(card is a red 5) = \(\dfrac{ \text {number of red fives}}{\text { total number of cards }} = \dfrac{2}{52} = \dfrac{1}{26}\)
The probability that the card is a red five is \(\dfrac{1}{26}\).
Roll a fair die one time. The sample space is S = {1, 2, 3, 4, 5, 6}. Find the following probabilities of the given events.
- P(roll a four)
P(roll a four) = \(\dfrac{ \text {number of ways to roll a four}}{\text { total number of ways to roll a die }} = \dfrac{1}{6}\)
The probability of rolling a four is \(\dfrac{1}{6}\).
- P(roll an odd number)
The event roll an odd number is E = {1, 3, 5}.
P(roll an odd number) = \(\dfrac{ \text {number of ways to roll an odd number}}{\text { total number of ways to roll a die }} = \dfrac{3}{6} = \dfrac{1}{2}\)
The probability of rolling an odd number is \(\dfrac{1}{2}\).
- P(roll a number less than five)
The event roll a number less than five is F = {1, 2, 3, 4}.
P(roll a number less than five) = \(\dfrac{ \text {number of ways to roll number less than five}}{\text { total number of ways to roll a die }} = \dfrac{4}{6} = \dfrac{2}{3}\)
The probability of rolling a number less than five is \(\dfrac{2}{3}\).
A small bookcase contains five math, three English and seven science books. A book is chosen at random. What is the probability that a math book is chosen?
Solution
Since the book is chosen at random each book has the same chance of being chosen and we have equally likely outcomes.
\[P(\text { math book })=\frac{\text { number of ways to choose a math book }}{\text { total number of books }}=\frac{5}{15}=\frac{1}{3}\]
The probability a math book was chosen is \(\dfrac{1}{3}\).
Three Ways of Finding Probabilities
There are three ways to find probabilities. In the tack tossing example we calculated the probability of the tack landing point up by doing an experiment and recording the outcomes. This was an example of an empirical probability. The probability of getting a red jack in a card game or rolling a five with a fair die can be calculated from mathematical formulas. These are examples of theoretical probabilities. The third type of probability is a subjective probability. Saying that there is an 80% chance that you will go to the beach this weekend is a subjective probability. It is based on experience or guessing.
- A theoretical probability is based on a mathematical model where all outcomes are equally likely to occur.
- An empirical probability is based on an experiment or observation and is the relative frequency of the event occurring.
- A subjective probability is an estimate (a guess) based on experience or intuition.
Law of Large Numbers
When studying probabilities, many times the law of large numbers will apply. If you want to observe what the probability is of getting tails up when flipping a coin, you could do an experiment. Suppose you flip a coin 20 times and the coin comes up tails nine times. Then, using an empirical probability, the probability of getting tails is 9/20 = 45%. However, we know that the theoretical probability for getting tails should be 1/2 = 50%. Why is this different? It is because there is error inherent to sampling methods. However, if you flip the coin 100 times or 1000 times, and use the information to calculate an empirical probability for getting tails up, then the probabilities you will observe will become closer to the theoretical probability of 50%. This is the law of large numbers.
The law of large numbers means that with larger numbers of trials of an experiment the observed empirical probability of an event will approach the calculated theoretical probability of the same event.
Complements
If there is a 75% chance of rain today, what are the chances it will not rain? We know that there are only two possibilities. It will either rain or it will not rain. Because the sum of the probabilities for all the outcomes in the sample space must be 100% or 1.00, we know that
P(will rain) + P(will not rain) = 100%.
Rearranging this we see that
P(will not rain) = 100% - P(will rain) = 100% - 75% = 25%.
The events E = {will rain} and F = {will not rain} are called complements.
The complement of event E, denoted by \(\overline{E}\), is the set of outcomes in the sample space that are not in the event E. The probability of \(\overline{E}\) is given by \(P(\overline{E}) = 1- P(E)\).
Draw a single card from a well shuffled deck of 52 cards.
- Look at the suit of the card. Here the sample space S = {spades, clubs, hearts, diamonds}. If event E = {spades} the complement \(\overline{E}\) = {clubs, hearts, diamonds}.
- Look at the value of the cards. Here the sample space is S = {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K}. If the event E = {the number is less than 7} = {A, 2, 3, 4, 5, 6} the complement \(\overline{E}\) = {7, 8, 9, 10, J, Q, K}.
A train arrives on time at a particular station 85% of the time. Does this mean that the train is late 15% of the time? The answer is no. The complement of E = {on time} is not \(\overline{E}\) = {late}. There is a third possibility. The train could be early. The sample space is S = {on time, early, late} so the complement of E = {on time} is \(\overline{E}\) = {early or late}. Based on the given information we cannot find P(late) but we can find P(early or late) = 15%.
Impossible Events and Certain Events
Recall that \(P(A)=\dfrac{ \text {number of ways for A to occur}}{\text { total number of outcomes }}\). What does it mean if we say the probability of the event is zero? \(P(A)=\dfrac{ \text {number of ways for A to occur}}{\text { total number of outcomes }} = 0\). The only way for a fraction to equal zero is when the numerator is zero. This means there is no way for event A to occur. A probability of zero means that the event is impossible.
What does it mean if we say the probability of an event is one? \(P(A)=\dfrac{ \text {number of ways for A to occur}}{\text { total number of outcomes }} = 1\). The only way for a fraction to equal one is if the numerator and denominator are the same. The number of ways for A to occur is the same as the number of outcomes. There are no outcomes where A does not occur. A probability of 1 means that the event always happens.
- \(P(A) = 0\) means that A is impossible
- \(P(A) = 1\) means that A is certain