# Computing the Probability of an Event

### Learning Objectives

The learning outcomes for this section include:- Describe a sample space and simple and compound events in it using standard notation
- Calculate the probability of an event using standard notation
- Calculate the probability of two independent events using standard notation
- Recognize when two events are mutually exclusive
- Calculate a conditional probability using standard notation

## Basic Concepts

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.### Events and Outcomes

- The result of an experiment is called an
**outcome**. - An
**event**is any particular outcome or group of outcomes. - A
**simple event**is an event that cannot be broken down further - The
**sample space**is the set of all possible simple events.

### example

If we roll a standard 6-sided die, describe the sample space and some simple events.Answer: The sample space is the set of all possible simple events: {1,2,3,4,5,6} Some examples of simple events:

- We roll a 1
- We roll a 5

- We roll a number bigger than 4
- We roll an even number

### Basic Probability

Given that all outcomes are equally likely, we can compute the probability of an event*E*using this formula:

[latex]P(E)=\frac{\text{Number of outcomes corresponding to the event E}}{\text{Total number of equally-likely outcomes}}[/latex]

### examples

If we roll a 6-sided die, calculate- P(rolling a 1)
- P(rolling a number bigger than 4)

Answer: 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 [latex]\frac{1}{6}[/latex]
- There are two outcomes bigger than a 4, so the probability is [latex]\frac{2}{6}=\frac{1}{3}[/latex]

Answer: 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 [latex]\frac{14}{20}=\frac{7}{10}[/latex]. 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.

### Try It Now

At some random moment, you look at your clock and note the minutes reading. a. What is probability the minutes reading is 15? b. What is the probability the minutes reading is 15 or less?### Cards

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.

### example

Compute the probability of randomly drawing one card from a deck and getting an Ace.Answer: There are 52 cards in the deck and 4 Aces so [latex]P(Ace)=\frac{4}{52}=\frac{1}{13}\approx 0.0769[/latex] 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.

This video demonstrates both this example and the previous cherry example on the page. https://youtu.be/EBqj_R3dzd4### Certain and Impossible events

- An impossible event has a probability of 0.
- A certain event has a probability of 1.
- The probability of any event must be [latex]0\le P(E)\le 1[/latex]

### Try it now

### Probability of two independent events

### example

Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin and a 6 on the die.Answer: We could list all possible outcomes: {H1,H2,h2,H4,H5,H6,T1,T2,T3,T4,T5,T6}. Notice there are [latex]2\cdot6=12[/latex] total outcomes. Out of these, only 1 is the desired outcome, so the probability is [latex]\frac{1}{12}[/latex].

*independent*events. Getting a certain outcome from rolling a die had no influence on the outcome from flipping the coin.

### Independent Events

Events A and B are**independent events**if the probability of Event B occurring is the same whether or not Event A occurs.

### example

Are these events independent?- A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.
- The two events (1) "It will rain tomorrow in Houston" and (2) "It will rain tomorrow in Galveston” (a city near Houston).
- You draw a card from a deck, then draw a second card without replacing the first.

Answer:

- The probability that a head comes up on the second toss is 1/2 regardless of whether or not a head came up on the first toss, so these events are independent.
- These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.
- The probability of the second card being red depends on whether the first card is red or not, so these events are not independent.

*P*(*A* and *B*) for independent events

If events *A*and

*B*are independent, then the probability of both

*A*and

*B*occurring is

[latex]P\left(A\text{ and }B\right)=P\left(A\right)\cdot{P}\left(B\right)[/latex]

where*P*(

*A*and

*B*) is the probability of events

*A*and

*B*both occurring,

*P*(

*A*) is the probability of event

*A*occurring, and

*P*(

*B*) is the probability of event

*B*occurring

### example

In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you randomly reach in and pull out a pair of socks and a tee shirt, what is the probability both are white?Answer: The probability of choosing a white pair of socks is [latex]\frac{6}{10}[/latex]. The probability of choosing a white tee shirt is [latex]\frac{3}{7}[/latex]. The probability of both being white is [latex]\frac{6}{10}\cdot\frac{3}{7}=\frac{18}{70}=\frac{9}{35}[/latex]

Examples of joint probabilities are discussed in this video. https://youtu.be/6F17WLp-EL8### Try it now

### example

Suppose we draw one card from a standard deck. What is the probability that we get a red card or a King?Answer: Half the cards are red, so [latex]P(\text{red})=\frac{26}{52}[/latex] There are four kings, so [latex]P(\text{King})=\frac{4}{52}[/latex] There are two red kings, so [latex]P(\text{Red and King})=\frac{2}{52}[/latex] We can then calculate

[latex]P(\text{Red or King})=P(\text{Red})+P(\text{King})-P(\text{Red and King})=\frac{26}{52}+\frac{4}{52}-\frac{2}{52}=\frac{28}{52}[/latex]

### Try It Now

In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you reach in and randomly grab a pair of socks and a tee shirt, what the probability at least one is white?### Example

The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:- Has a red car
*and*got a speeding ticket - Has a red car
*or*got a speeding ticket.

Speeding ticket | No speeding ticket | Total | |

Red car | 15 | 135 | 150 |

Not red car | 45 | 470 | 515 |

Total | 60 | 605 | 665 |

Answer: We can see that 15 people of the 665 surveyed had both a red car and got a speeding ticket, so the probability is [latex]\frac{15}{665}\approx0.0226[/latex]. Notice that having a red car and getting a speeding ticket are not independent events, so the probability of both of them occurring is not simply the product of probabilities of each one occurring. We could answer this question by simply adding up the numbers: 15 people with red cars and speeding tickets + 135 with red cars but no ticket + 45 with a ticket but no red car = 195 people. So the probability is [latex]\frac{195}{665}\approx0.2932[/latex]. We also could have found this probability by:

P(had a red car) + P(got a speeding ticket) – P(had a red car and got a speeding ticket)

= [latex]\frac{150}{665}+\frac{60}{665}-\frac{15}{665}=\frac{195}{665}[/latex].

This table example is detailed in the following explanatory video. https://youtu.be/HWrGoM1yRaU### Try it now

## Conditional Probability

In the previous section we computed the probabilities of events that were independent of each other. We saw that getting a certain outcome from rolling a die had no influence on the outcome from flipping a coin, even though we were computing a probability based on doing them at the same time. In this section, we will consider events that*are*dependent on each other, called

**conditional probabilities**.

### Conditional Probability

The probability the event*B*occurs, given that event

*A*has happened, is represented as

*P*(*B* | *A*)

*B*given

*A*”

### example

What is the probability that two cards drawn at random from a deck of playing cards will both be aces?Answer:
It might seem that you could use the formula for the probability of two independent events and simply multiply [latex]\frac{4}{52}\cdot\frac{4}{52}=\frac{1}{169}[/latex]. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck.
Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the **conditional probability** of drawing an ace. In this case the "condition" is that the first card is an ace. Symbolically, we write this as:
*P*(ace on second draw | an ace on the first draw).
The vertical bar "|" is read as "given," so the above expression is short for "The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw." What is this probability? After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is [latex]\frac{3}{51}=\frac{1}{17}[/latex].
Thus, the probability of both cards being aces is [latex]\frac{4}{52}\cdot\frac{3}{51}=\frac{12}{2652}=\frac{1}{221}[/latex].

### Conditional Probability Formula

If Events*A*and

*B*are not independent, then

*P*(

*A*and

*B*) =

*P*(

*A*) ·

*P*(

*B*|

*A*)

### example

If you pull 2 cards out of a deck, what is the probability that both are spades?Answer: The probability that the first card is a spade is [latex]\frac{13}{52}[/latex]. The probability that the second card is a spade, given the first was a spade, is [latex]\frac{12}{51}[/latex], since there is one less spade in the deck, and one less total cards. The probability that both cards are spades is [latex]\frac{13}{52}\cdot\frac{12}{51}=\frac{156}{2652}\approx0.0588[/latex]

### Try it now

### Example

A home pregnancy test was given to women, then pregnancy was verified through blood tests. The following table shows the home pregnancy test results. Find*P*(not pregnant | positive test result)*P*(positive test result | not pregnant)

Positive test | Negative test | Total | |

Pregnant | 70 | 4 | 74 |

Not Pregnant | 5 | 14 | 19 |

Total | 75 | 18 | 93 |

Answer:

- Since we know the test result was positive, we’re limited to the 75 women in the first column, of which 5 were not pregnant.
*P*(not pregnant | positive test result) = [latex]\frac{5}{75}\approx0.067[/latex]. - Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test.
*P*(positive test result | not pregnant) = [latex]\frac{5}{19}\approx0.263[/latex]

### Try it now

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