Calculate the probability of different types of events
Find the conditional probability of an event
Sample Space and Events
A sample space is a set of all possible outcomes of a random experiment. For example, if we were rolling a [latex]6[/latex]-sided die, the sample space would be the set [latex]{1, 2, 3, 4, 5, 6}[/latex].
Simple events are single outcomes within a sample space. For example, rolling a [latex]4[/latex] on a [latex]6[/latex]-sided die would be a simple event. We can represent simple events using standard notation, which in the example given would look like [latex]{4}[/latex].
Compound events are combinations of two or more simple events. For example, rolling an even number on a [latex]6[/latex]-sided die would be a compound event, as it is the combination of the simple events [latex]{2}[/latex], [latex]{4}[/latex], and [latex]{6}[/latex]. We can represent compound events using standard notation, which in the example given would look like [latex]{2, 4, 6}[/latex]
It’s important to note that compound events can also be represented using logical operators such as “and” and “or”. For example, the event “rolling a [latex]4[/latex] or a [latex]5[/latex]” on a fair die would be represented as [latex]{4,5}[/latex].
It’s also important to understand that sample space is the set of all possible outcomes, while events are the subset of outcomes that we are interested in.
Basic Probability
It can be confusing to distinguish between probability, likelihood, and chance. Probability is a measure of the likelihood of an event occurring, and it is calculated as the number of favorable outcomes divided by the total number of possible outcomes. In standard notation, probability is denoted as [latex]P(A)[/latex], where [latex]A[/latex] is an event. For example, if we were flipping a coin, the probability of getting heads would be [latex]1/2[/latex]. Likelihood is a measure of how well a particular hypothesis or model fits the data. It is used in statistical inference to determine the probability of a certain set of data given a certain hypothesis. For example, the likelihood of observing data [latex]X[/latex] given a hypothesis [latex]H[/latex] is denoted as [latex]L(X|H)[/latex]. Unlike probability, likelihood is not necessarily between [latex]0[/latex] and [latex]1[/latex]. Chance refers to the notion of randomness or uncertainty in an outcome. It is often used to describe the likelihood of an event occurring without any specific information about the event or the sample space. For example, “there’s a chance it will rain tomorrow” implies that the outcome of rain or no rain is uncertain, but it doesn’t specify the probability of either outcome. For this lesson, we will be working with only probability.If we roll a [latex]6[/latex]-sided die, calculate
[latex]P(\text{rolling a }1)[/latex]
[latex]P(\text{rolling a number bigger than }4)[/latex]
Recall that the sample space is [latex]{1,2,3,4,5,6}[/latex]
There is one outcome corresponding to “rolling a [latex]1[/latex],” so the probability is [latex]\frac{1}{6}[/latex]
There are two outcomes bigger than a [latex]4[/latex], so the probability is [latex]\frac{2}{6}=\frac{1}{3}[/latex]
Probabilities are essentially fractions and can be reduced to lower terms like fractions.
Watch the following video to see this example worked out.
Let’s say you have a bag with [latex]20[/latex] cherries, [latex]14[/latex] sweet and [latex]6[/latex] sour. If you pick a cherry at random, what is the probability that it will be sweet?
There are [latex]20[/latex] possible cherries that could be picked, so the number of possible outcomes is [latex]20[/latex]. Of these [latex]20[/latex] possible outcomes, [latex]14[/latex] 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.
Complementary Events
Now let us examine the probability that an event does not happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is [latex]P(\text{six}) =1/6[/latex]. Now consider the probability that we do not roll a six: there are [latex]5[/latex] outcomes that are not a six, so the answer is [latex]P(\text{not a six}) = \frac{5}{6}[/latex]. Notice that
[latex]P(\text{six})+P(\text{not a six})=\frac{1}{6}+\frac{5}{6}=\frac{6}{6}=1[/latex]
This is not a coincidence. Consider a generic situation with [latex]n[/latex] possible outcomes and an event [latex]E[/latex] that corresponds to [latex]m[/latex] of these outcomes. Then the remaining [latex]n - m[/latex] outcomes correspond to [latex]E[/latex] not happening, thus
A fair coin is tossed two times. The two events are [latex](1)[/latex] first toss is a head and [latex](2)[/latex] second toss is a head.
The two events [latex](1)[/latex] “It will rain tomorrow in Houston” and [latex](2)[/latex] “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.
The probability that a head comes up on the second toss is [latex]1/2[/latex] 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.
[latex]P(A \text{ and } B)[/latex] for Independent Events
When dealing with two independent events, the probability of both events occurring together can be calculated using the formula [latex]P(A \text{ and } B)= P(A) * P(B)[/latex]. Remember, independent events are events that don’t affect each other, meaning the outcome of one event doesn’t change the outcome of the other.
In your drawer you have [latex]10[/latex] pairs of socks, [latex]6[/latex] of which are white, and [latex]7[/latex] tee shirts, [latex]3[/latex] 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?
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]
[latex]P(A \text{ or } B)[/latex] for Independent Events
Two events that are not mutually exclusive can happen at the same time. The probability of two events happening together can be calculated using the formula [latex]P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)[/latex].
Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?
There are [latex]4[/latex] Queens and 4 Kings in the deck, hence [latex]8[/latex] outcomes corresponding to a Queen or King out of [latex]52[/latex] possible outcomes. Thus the probability of drawing a Queen or a King is:
[latex]P(\text{King or Queen})=\frac{8}{52}[/latex]
Note that in this case, there are no cards that are both a Queen and a King, so [latex]P(\text{King and Queen})=0[/latex]. Using our probability rule, we could have said:
[latex]P(\text{King or Queen})=P(\text{King})+P(\text{Queen})-P(\text{King and Queen})=\frac{4}{52}+\frac{4}{52}-0=\frac{8}{52}[/latex]
See more about this example and the previous one in the following video.
In your drawer, you have [latex]10[/latex] pairs of socks, [latex]6[/latex] of which are white, and [latex]7[/latex] tee shirts, [latex]3[/latex] of which are white. If you reach in and randomly grab a pair of socks and a tee shirt, what is the probability at least one is white?
The probability that at least one of them is white is [latex]77[/latex]%
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
[latex]15[/latex]
[latex]135[/latex]
[latex]150[/latex]
Not red car
[latex]45[/latex]
[latex]470[/latex]
[latex]515[/latex]
Total
[latex]60[/latex]
[latex]605[/latex]
[latex]665[/latex]
We can see that [latex]15[/latex] people of the [latex]665[/latex] 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: [latex]15[/latex] people with red cars and speeding tickets [latex]+ 135[/latex] with red cars but no ticket [latex]+ 45[/latex] with a ticket but no red car [latex]= 195[/latex] people. So the probability is [latex]\frac{195}{665}\approx0.2932[/latex].
We also could have found this probability by:
[latex]P(\text{ had a red car }) + P(\text{ got a speeding ticket }) – P(\text{ had a red car and got a speeding ticket })[/latex]
Remember to work through each example in the text and in the EXAMPLE boxes with a pencil on paper, pausing as frequently as needed to digest the process. Watch the videos by working them out on paper, pausing the video as frequently as you need to make sense of the demonstration. Don’t be afraid to ask for help — hard work and willingness to learn translate into success!
A conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted using standard notation as [latex]P(A|B)[/latex], which is read as “the probability of [latex]A[/latex] given [latex]B[/latex].” For example, if we were flipping a coin and rolling a die, the probability of getting heads and rolling a [latex]4[/latex] is [latex]P(\text{heads and } 4) = 1/12[/latex], since there is only [latex]1[/latex] favorable outcome (getting heads and rolling a [latex]4[/latex]) out of a total of [latex]12[/latex] possible outcomes (getting heads or tails and rolling any number from [latex]1[/latex] to [latex]6[/latex]).
To calculate a conditional probability, we use the formula [latex]P(A \text{ and } B) = P(A) · P(B | A)[/latex]. Using the example above, we can find the conditional probability of getting heads given that we rolled a [latex]4[/latex] as:
If you pull [latex]2[/latex] cards out of a deck, what is the probability that both are spades?
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]
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 speeding ticket given they have a red car
has a red car given they have a speeding ticket
Speeding ticket
No speeding ticket
Total
Red car
[latex]15[/latex]
[latex]135[/latex]
[latex]150[/latex]
Not red car
[latex]45[/latex]
[latex]470[/latex]
[latex]515[/latex]
Total
[latex]60[/latex]
[latex]605[/latex]
[latex]665[/latex]
Since we know the person has a red car, we are only considering the [latex]150[/latex] people in the first row of the table. Of those, [latex]15[/latex] have a speeding ticket, so [latex]P(\text {ticket } | \text{ red car }) = \frac{15}{150}=\frac{1}{10}=0.1[/latex]
Since we know the person has a speeding ticket, we are only considering the [latex]60[/latex] people in the first column of the table. Of those, [latex]15[/latex] have a red car, so [latex]P(\text{ red car } | \text {ticket }) = \frac{15}{60}=\frac{1}{4}=0.25[/latex].
Notice from the last example that [latex]P(B | A)[/latex] is not equal to [latex]P(A | B)[/latex].
These kinds of conditional probabilities are what insurance companies use to determine your insurance rates. They look at the conditional probability of you having accident, given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood.
View more about conditional probability in the following video.