13.1: The Basics of Probability Theory
The probability of a specified event is the chance or likelihood that it will occur. There are several ways of viewing probability. One would be experimental in nature, where we repeatedly conduct an experiment. Suppose we flipped a coin over and over and over again and it came up heads about half of the time; we would expect that in the future whenever we flipped the coin it would turn up heads about half of the time. When a weather reporter says “there is a 10% chance of rain tomorrow,” she is basing that on prior evidence; that out of all days with similar weather patterns, it has rained on 1 out of 10 of those days.
Another view would be subjective in nature, in other words an educated guess. If someone asked you the probability that the Seattle Mariners would win their next baseball game, it would be impossible to conduct an experiment where the same two teams played each other repeatedly, each time with the same starting lineup and starting pitchers, each starting at the same time of day on the same field under the precisely the same conditions. Since there are so many variables to take into account, someone familiar with baseball and with the two teams involved might make an educated guess that there is a 75% chance they will win the game; that is, if the same two teams were to play each other repeatedly under identical conditions, the Mariners would win about three out of every four games. But this is just a guess, with no way to verify its accuracy, and depending upon how educated the educated guesser is, a subjective probability may not be worth very much.
We will return to the experimental and subjective probabilities from time to time, but in this course we will mostly be concerned with theoretical probability, which is defined as follows: Suppose there is a situation with \(n\) equally likely possible outcomes and that m of those \(n\) outcomes correspond to a particular event; then the probability of that event is defined as \(\frac{m}{n}\).
If you roll a die, pick a card from deck of playing cards, or randomly select a person and observe their hair color, we are executing an experiment or procedure. In probability, we look at the likelihood of different outcomes. We begin with some terminology.
The result of an experiment is called an outcome .
An event is any particular outcome or group of outcomes.
The sample space is the set of all possible simple events.
If we roll a standard 6-sided die, describe the sample space and some simple events.
The sample space is the set of all possible simple events: \(\{1,2,3,4,5,6\}\)
Some examples of events:
We roll a 1: outcome \(\{1\}\)
We roll a 5: outcome \(\{5\}\)
We roll a number bigger than 4: outcomes \(\{5,6\}\)
We roll an even number: outcomes \(\{2,4,6\}\)
Two dice are rolled. Write the sample space.
Solution
We assume one of the dice is red, and the other green. We have the following 36 possibilities.
|
Green |
||||||
|
Red |
1 |
2 |
3 |
4 |
5 |
6 |
|
1 |
(1, 1) |
(1, 2) |
(1, 3) |
(1, 4) |
(1, 5) |
(1, 6) |
|
2 |
(2, 1) |
(2, 2) |
(2, 3) |
(2, 4) |
(2, 5) |
(2, 6) |
|
3 |
(3, 1) |
(3, 2) |
(3, 3) |
(3, 4) |
(3, 5) |
(3, 6) |
|
4 |
(4, 1) |
(4, 2) |
(4, 3) |
(4, 4) |
(4, 5) |
(4, 6) |
|
5 |
(5, 1) |
(5, 2) |
(5, 3) |
(5, 4) |
(5, 5) |
(5, 6) |
|
6 |
(6, 1) |
(6, 2) |
(6, 3) |
(6, 4) |
(6, 5) |
(6, 6) |
The entry (2, 5), for example, indicates that the red die shows a 2 and the green shows a 5. This is different than the entry (5, 2) which indicates that the red die shows a 5 and the green die shows a 2.
A family has three children. Write a sample space.
Solution
The sample space consists of eight possibilities.
{ BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG }
The possibility BGB, for example, indicates that the first born is a boy, the second born a girl, and the third a boy.
We illustrate these possibilities with a tree diagram.
Given that all outcomes are equally likely, we can compute the probability of an event \(E\) using this formula:
\(P(E)=\frac{\text { Number of outcomes corresponding to the event } \mathrm{E}}{\text { Total number of equally likely outcomes }}\)
If we roll a 6-sided die, calculate
- P(rolling a 1)
- P(rolling a number bigger than 4)
Solution
Recall that the sample space is \(\{1,2,3,4,5,6\}\)
- There is one outcome corresponding to “rolling a 1”, so the probability is \(\frac{1}{6}\)
- There are two outcomes bigger than a 4, so the probability is \(\frac{2}{6}=\frac{1}{3}\)
Probabilities are essentially fractions, and should be reduced to lowest terms.
Let's say you have a bag with 20 cherries, 14 sweet and 6 sour. If you pick a cherry at random, what is the probability that it will be sweet?
Solution
There are 20 possible cherries that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable (sweet), so the probability that the cherry will be sweet is \(\frac{14}{20}=\frac{7}{10}\).
There is one potential complication to this example, however. It must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other. This wouldn't be true if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.
At some random moment, you look at your clock and note the minutes reading.
- What is probability the minutes reading is 15?
- What is the probability the minutes reading is 15 or less?
- Answer
-
There are 60 possible readings, from 00 to 59.
- \(\frac{1}{60}\)
- \(\frac{16}{60}\) (counting 00 through 15)
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.
Compute the probability of randomly drawing one card from a deck and getting an Ace.
Solution
There are 52 cards in the deck and 4 Aces so
\[P(A c e)=\dfrac{4}{52}=\frac{1}{13} \approx 0.0769 \nonumber\]
We can also think of probabilities as percents: There is a 7.69% chance that a randomly selected card will be an Ace.
Notice that the smallest possible probability is 0 – if there are no outcomes that correspond with the event. The largest possible probability is 1 – if all possible outcomes correspond with the event.
An impossible event has a probability of 0.
A certain event has a probability of 1.
The probability of any event must be \(0 \leq P(E) \leq 1\).
In the course of this chapter, if you compute a probability and get an answer that is negative or greater than 1, you have made a mistake and should check your work .