Concepts of Probability and Data Interpretation

The probability of something occurring, like the probability that it will rain today, is a common use of data to provide a prediction. The probability is the likelihood of occurrence and is expressed as a ratio or a percent.

When we looked at Scatter Plots, we began making a prediction. The question was whether we could predict what might occur given the information known on the scatter plot. We did this by examining the plotted points for a pattern, positive, negative, or no correlation between the points. If a pattern occurred we could state a prediction of an unknown x-value's y-value within a range of values.

Probability is a prediction method of what might occur based on the given data. The weather people predict the weather based on current conditions and the weather that is heading towards the area. A 50% chance of rain simply means that 50% of the area might have rain today. It does not mean that you will have rain today. Another way of saying this is that your area has a 1 in 2, 1/2, chance of having rain, the ratio, that when divided provides the decimal .50 which means 50% when multiplied by 100.

Relative Frequency - the ratio of the actual number of favorable events to the total possible number of events; often taken as an estimate of probability.
Conditional Probability - The probability of an event A, assuming that B has already occurred.

We will examine two types of probability in this section: relative frequency and conditional probability.

Relative frequency or theoretical probability is the number of favorable events to the total possible outcomes. In other words, the probability of an event occurring. As we saw above this is expressed as a ratio first. P(E) stands for the probability of an event occurring.

P(E) = number of favorable outcomes / total number of outcomes

Probability Example 1

image of chocolate chip cookie You would like the chocolate chip cookie on the dessert platter. On the platter is one chocolate chip cookie and two sugar cookies.

You and your two friends are heading for the cookie platter. As you walk to the cookie platter you have a 1 in 3, 1/3, 33% probability (chance) of getting the chocolate chip cookie.

One of your friends reaches the platter first and takes a sugar cookie. You now have a 1 in 2, 1/2, 50% probability of getting a chocolate chip cookie.

Either you must pick first or your friend needs to take the sugar cookie for you to have the chocolate chip cookie. Your second friend takes the sugar cookie.

Yeah, the chocolate chip cookie is yours. Yum!!

Notice that the probability of your success changes as the number of items remaining changes(the number of outcomes changes).

Think of this method as part to whole, part/whole, the number of outcomes you want (the part)/ the total outcomes that could happen (the whole).

You wanted 1 chocolate chip cookie (one favorable outcome), the whole that you could get your 1 part from was 3 cookies (the total outcomes). So initially you had a 1/3 chance or probability of getting the correct cookie. As conditions changed, your probability went up to a 1/2 chance as there were only two (the whole that was left) cookies left to choose from. The favorable event was you getting to eat the chocolate chip cookie. Your probability increased because the total number of items decreased.

Conditional probability is the probability that of an event happening assuming a second condition has already taken place. This is what happened after the first sugar cookie was taken on the platter.
Dependent events is what conditional probability is about. Dependent events mean that something else happens first and then the second event. In the video below a dependent event is created when the stones are not replaced in the bag.
Independent events are events that do not depend on what has happened prior. Examples of independent events are shown in the video when the stones are replaced in the bag prior to the next event occurring.

Venn Diagrams and Probability

Now let's examine another method to illustrate probability.

Probability Example 2

There are 49 students in the media center. Geometry is taken by 28 of the students and 10 of the students take Geometry and Chemistry. What is the probability of a randomly chosen student in the media center taking Geometry?

This information may be represented pictorially using a Venn Diagram to help organize your thoughts. A Venn Diagram contains intersecting circles and a sample space outside the circles for an event that does not belong in the circles (any students not taking Geometry of Chemistry).

Here is the initial Venn Diagram thought process.

You know that 18 students are taking Geometry only because 10 are taking both Geometry and Chemistry. There are 28 students in Geometry class, so 28 - 10 = 18 taking Geometry only.

image of Venn Diagram with G18 on left side C? on right and 10 where they meet in the middle

Take note here: Check out the percents that you just calculated. What do they add to?

36.73% + 42.86% + 20.41% = 100%

If this is true you have accounted for all of the students in the media center.

Image of VennDiagram with G18 on left and C21 on right and 10 where they meet in the middle

Thought question.

What would happen if there were 51 students in the media center and the extra 2 students did not take Geometry or Chemistry? Use the diagram below to help formulate your answer. Note: A Venn Diagram with items outside the categories being compared use a rectangular box indicating that the number outside the circles are still included in the universe of numbers being considered.

image of venn diagram with G18 on left, C21 on right and 10 where they meet in the middle. the number 2 is displayed at bottom right of box representing 2 students who did not take either course

(Solution: All of the probabilities need to recalculated with 51 as the divisor and another probability is calculated for those students not belonging either to the Geometry or Chemistry group.

18/51 = .3529 35.29% probability of the students taking only geometry

21/51 = .4118 41.18% of the students taking Chemistry only

10/51 = .1961 19.61% taking both

2/51 = .0392 3.92%

Note that when you add the percentages up, they add up to 100%, work is correct. Notice that the rounding is consistent at the same decimal place each time.)

Now, what if we only knew the percentages of each of the events. What is the probability of a student taking Chemistry if we know the student is taking Geometry.

Note that I have changed the percentages. These are the percentages of students that take Geometry, take Chemistry, take both courses, and take neither course.

28 / 51 = .5490 = 54.90% probability of the students taking Geometry

31 / 51 = .6079 = 60.79% of the students taking Chemistry

10 / 51 = .1961 = 19.61% of students taking both courses

2 / 51 = .0392 = 3.92%

These no longer add up to 100% because the intersection, ∩, is covered two additional times with the percentage of students that take Geometry and Chemistry.

Geometry 54.90% - 19.61% = 35.29%

Chemistry 60.79% - 19.61% = 41.18%

image of venn diagram with G54.90% on left C 60.79% on right and 19.61% where they meet in the middle. Unrepresented students listed at 3.92%

Now we will handle the conditional probability. The probability of an event occurring assuming another event has already occurred is the probability of both events divided by the probability of the individual event. So the probability of students taking both Geometry and Chemistry divided by the probability of the student taking Geometry is the probability of a student taking Chemistry given that they are already taking Geometry.

P(Geometry and Chemistry) / P(Geometry) = P(Chemistry | Geometry)

Note: The straight up bar line, | , means given, a dependent event, not divide.

.1961 / .5490 = .3572 = 35.72%

So there is a 35.72% probability that a Geometry student is also taking Chemistry.

Now you try. Let me put it in words like you may see on a test and have to figure it out.

Example

The probability that a student takes Chemistry is .6079. The probability that a student takes Geometry and Chemistry is .1961. What is the probability that a student takes Geometry given that the student is taking Chemistry?

(Solution: Both/Single to be eliminated = P(Geometry&Chemistry)/P(Chemistry)

= .1961/.6079 = .3226 = 32.36%)

Bell Curve and Probability

In the last lesson, we looked at the normal distribution bell curve in terms of shape. We looked at data to view normal or skew spread of data. Now let's look at what we can predict with a normal spread of data.

As with any plotting of data there is a mean (the average). Remember we find the average by adding up all the numbers and dividing by the number of numbers. The median is the middle number or halfway between the two middle numbers.

With the bell curve, also called normal distribution curve, the mean is important. It is the midway point of the data to create a symmetric bell curve.

The second most important item is the standard deviation, the spread of the data. Below is a normal distribution bell curve with probability percentages. Watch the video to understand the parts of this curve.

Image of Normal Bell Curve Distribution with Probability percentages

Note that the easiest way to get the mean, median, and standard deviation of a set of data is to put the data into the list of your graphing utility and then follow the one variable stats directions to give you the mean, median, and standard deviation for the data set.

We can predict the probability of data being a specific distance, a standard deviation, away from the mean. How do we get the data to create the distribution curve? Watch the video below.

Lastly here is how to work a word problem given only the mean and the standard deviation.

Review

Problem solving is the process of interpreting information given in work or data format to draw conclusions. In the module we have covered ratios and percentages and their use with analyzing data. We have also interpreted graphs using information learned in the Heart of Algebra.

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