Section 5.4: Two-Way Tables and Conditional Probability

This section is a “mashup” of sections 5.2 and 5.3. We will focus on using two-way tables to find conditional probability. We will also verify that the two methods we have seen to compute $P(E\text{ AND }F)$, one from section 5.2 and the other using the multiplication rule in 5.3, are equivalent.

Conditional Probability from Two-way Tables

Recall the definition of conditional probability from section 5.3. When we compute a conditional probability, we are restricting the sample space.

When we have a two-way table and we wish to compute the conditional probability $P(F|E)$, we shrink our view to the row or column containing the data where event E happens. Then, we use the total number of ways E happens as the denominator, and the number of ways where E and F both happen as the numerator.

$$\begin{array}{cc}P\left(F\right|E)& =\frac{\text{number of ways }F\text{ happens with }E}{\text{Number of ways }E\text{ happens}}=\frac{P\left(E\text{ AND }F\right)}{P\left(E\right)}\end{array}$$

Tip!

Solution:a)We want to find $P\left(A|\text{Taking English}\right).$ Since we are assuming that the student is taking English, we restrict the table to the row of students taking English, highlighted in red in figure 5.28. There are 18 of these students, and 8 of them are earning an A. So, the probability of earning an A, given a student is taking English, is $P\left(A|\text{Taking English}\right)=\frac{8}{18}\approx 0.4444$.

Figure 5.28: The given condition of “Taking English” is highlighted.

b)Here we wish to find $P\left(A|\text{Not taking English}\right).$ The row for students not taking English is highlighted in figure 5.29. There are 22 students that are not taking English, with 8 of these students earning an A. So, $P\left(A|\text{Not taking English}\right)=\frac{8}{22}\approx 0.3636$.

Figure 5.29: The given condition of “Not Taking English” is highlighted.

c)Here we are looking for $P\left(A\text{ or }B\right|\text{Taking English}).$ There are 18 students taking English, and 8 are earning an A while 6 are earning a B. We will add these together using the addition rule for disjoint events to get $P\left(A\text{ or }B\right|\text{Taking English})=\frac{8+6}{18}=\frac{14}{18}\approx 0.7778$.

Figure 5.30: The given condition of “Taking English” is highlighted.

d)We want to find $P\left(\text{Taking English}|\text{Earning an }A\right).$ Here we are assuming the student is earning an A, so we restrict the table to the column of students who are earning an A, shown in figure 5.31. Of the 16 students earning an A, 8 of them are taking English. So,

$$\begin{array}{cc}P\left(\text{Taking English}\right|\text{Earning an }A)& =\frac{8}{16}=0.5.\end{array}$$

Figure 5.31: The given condition of “Earning an A” is highlighted.

The difference between parts (a) and (d) can be confusing. The probability of earning an A given a student is taking English sounds very similar to the probability of taking English given the student is earning an A. Be very careful to identify what is assumed in the statement, then focus on that assumption’s row or column. It is always the total of the assumption’s row or column that goes into the denominator of the conditional probability.

$$\begin{array}{cc}P\left(\text{Earning an }A|\text{Taking English}\right)& =\frac{8}{18}\ne P\left(\text{Taking English}\right|\text{Earning an }A)=\frac{8}{16}\end{array}$$

Finally, what does the professor conclude? Is there a connection between student grade and taking English? One way to use probability to answer the question is as follows: The probability of earning an A if you are taking English 0.4444. The probability of earning an A if you are not taking English is 0.3636. So, it appears you are more likely to be earning an A if you are also taking English. Similarly, the probability of earning a C is less likely if you are taking English and more likely if you are not taking English. So, taking English may improve your chances of earning a higher grade in Statistics. In unit 3 we will study how to determine when two variables such as these are connected.

Solution:a)We want to find $P\left(\text{Major Factor}\right|\text{White})$, so we highlight the column for White participants, shown in figure 5.32. The total number of White participants is 2997, and there are 120 of these who think race or ethnicity should be a Major Factor in determining college admissions. So, $P\left(\text{Major Factor}\right|\text{White})=\frac{120}{2997}\approx 0.0400$.

Figure 5.32: The given condition of “White” is highlighted.

b)Here we want to find $P\left(\text{Major Factor}\right|\text{Black})$, so we highlight the column for Black participants, shown in figure 5.33. The total number of Black participants is 1518, and there are 273 of these who think race or ethnicity should be a Major Factor in determining college admissions. So, $P\left(\text{Major Factor}\right|\text{Black})=\frac{273}{1518}\approx 0.1798$.

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Figure 5.33: The given condition of “Black” is highlighted.

c)We want to find $P\left(\text{Not }a\text{ Factor}\right|\text{White})$, so we highlight the column for White participants, shown in figure 5.34. The total number of White participants is 2997, and there are 2338 of these who think race or ethnicity should not be a factor in determining college admissions. So, $P\left(\text{Not }a\text{ Factor}\right|\text{White})=\frac{2338}{2997}\approx 0.7801$.

Figure 5.34: The given condition of “White” is highlighted.

d)We want to find $P\left(\text{Not }a\text{ Factor}\right|\text{non White})$, so we highlight the columns for non-White participants, which includes all the columns except the column for White, shown in figure 5.35. The total number of non-White participants is 3424, and there are 2189 of these who think race or ethnicity should not be a factor in determining college admissions. So, $P\left(\text{Not }a\text{ Factor}\right|\text{non White})=\frac{2189}{3424}\approx 0.6393$.

Figure 5.35: The given condition of “non-White” including Black, Hispanic, and Asian is highlighted.

e)To answer this question, we need to compare the probability of thinking race ethnicity should be a Major factor when individuals are White versus when they are non-White. We already found $\left(\text{Major Factor}\right|\text{White})\approx 0.0400.$ Next, we find $P\left(\text{Major Factor}\right|\text{non White})$. We have once again grouped together all the non-White columns, shown in figure 5.36. The total number of non-White individuals who think race or ethnicity should be a Major factor in college admissions is 489. So, $P\left(\text{Major Factor}\right|\text{non White})=\frac{489}{3424}\approx 0.1428.$

Figure 5.36: The given condition of “non-White” including Black, Hispanic, and Asian is highlighted.

So, 14.28% of non-White individuals think that race or ethnicity should be a Major factor in college admissions. On the other hand, only 4.0% of White individuals think that race or ethnicity should be a Major factor. If we divide the conditional probabilities,

$$\begin{array}{cc}\frac{P\left(\text{Major Factor}\right|\text{non White})}{P\left(\text{Major Factor}\right|\text{White})}& =\frac{0.1428}{0.0400}\approx 3.6\end{array}$$

This tells us that a non-White person is about 3.6 times more likely to think race or ethnicity should be a Major factor in college admissions than a White person.

Part (e) of the last example is an example of computing relative risk. In this case, what is the “risk” of believing race or ethnicity should be a Major factor in college admissions if you are non-White versus White? We found you are 3.6 times more likely if you are non-White than if you are White. The term “risk” comes from examples in which we compare the risk factors (such as race or job type) for things like heart attacks, lung disease, mental health issues, etc.

Solution:a)Since we are given that the occupant used a seatbelt, we restrict our attention to the first row, shown in figure 5.37. There were 11388 fatalities out 43027 occupants that were using seatbelts. So, $P\left(\text{Fatal}\right|\text{Seatbelt})=\frac{11388}{43027}\approx 0.2647.$ Thus about 26.5% of accidents involving occupants wearing seatbelts proved to be fatal.

Figure 5.37: The given condition of “Seatbelt Restraint Used” is highlighted.

b)Since we are given that the occupant did not use a seatbelt restraint, we restrict our attention to the second row, shown in figure 5.38. There were 10076 fatalities out of 14867 occupants that were not using seatbelts. So, $P\left(\text{Fatal}\right|\text{No Seatbelt})=\frac{10076}{14867}\approx 0.6777.$ Thus about 67.8% of accidents involving occupants not wearing seatbelts were fatalities.

Figure 5.38: The given condition of “No Restraint Used” is highlighted.

c)Comparing the two answers, 26.5% of accidents were fatal for those wearing seatbelts, while 67.8% of accidents were fatal for those not wearing seatbelts. We can compute the ratio

$$\begin{array}{cc}\text{Relative risk of Fatality given No Seatbelt or Seatbelt}& =\frac{P\left(\text{Fatality}\right|\text{No Seatbelt})}{P\left(\text{Fatality}\right|\text{Seatbelt})}\\ & =\frac{0.6777}{0.2647}\\ & \approx 2.56\end{array}$$

This means that a person is over 2.5 times more likely to die in a highway traffic accident if they do not use a seatbelt than if they do.

Usually, when we calculate relative risk, we use the larger probability in the numerator so that the risk is a number that is larger than 1. That way, we can phrase the conclusion in the form “You are (some number greater than 1) times more likely to ... if you ... than if you ....”

Probability Statements Involving “AND”

We have seen the word “AND” appear to be used in two different ways in section 5.2 and section 5.3. Because of that, interpreting the word “AND” may seem a little tricky. In section 5.2, we wrote $P(E\text{ AND }F)$ and we counted the number of times both E and F simultaneously occurred in the data, then divided by the total. When we used “AND” in section 5.3, we used the multiplication rule, $P\left(E\text{ AND }F\right)=P\left(E\right)\times P(F|E)$. It turns out that these are the same. Example 4 will show how the two interpretations are equivalent.

Solution:a)There are 8 students out of 40 that are earning an A AND taking English, so $P\left(\text{Earning }A\text{AND }\text{Taking English}\right)=\frac{8}{40}=0.2$. The 8 students can be seen as the overlap of the Taking English Row and the Earning an A column, shown in figure 5.39.

Figure 5.39: The intersection of the Taking English row and the Earning an A column.

b)To use the multiplication rule, we must find

$$\begin{array}{cc}P\left(\text{Earning }A\text{AND }\text{Taking English}\right)& =P\left(\text{Earning }A\right)\times P\left(\text{Taking English}\right|\text{Earning }A)\end{array}$$

The probability of earning an A is the number earning an A, divided by the total. So, $P\left(\text{Earning }A\right)=\frac{16}{40}=0.4$.

Figure 5.40: The probability of earning an A.

The conditional probability of $P\left(\text{Taking English}\right|\text{Earning an }A)$ is found by restricting the sample space to the column containing the A grades.

Figure 5.41: The probability of earning an A given the student is Taking English.

So, $P\left(\text{Taking English}\right|\text{Earning }A)=\frac{8}{16}=0.5$

Putting the parts together,

$$\begin{array}{cc}P\left(\text{Earning }A\text{ AND }\text{Taking English}\right)& =P\left(\text{Earning }A\right)\times P\left(\text{Taking English}\right|\text{Earning }A)\\ & =\frac{16}{40}\times \frac{8}{16}\\ & =0.4\times 0.5\\ & =0.2\end{array}$$

c)The answers to both (a) and (b) are the same, $P\left(\text{Earning }A\text{ AND }\text{Taking English}\right)=0.2$. However, if you have a two-way table, it is much easier to use the data in its organized form to find the probability than it is to use the multiplication rule. The use of the multiplication rule in this example is only to assure the reader that the word “AND” is not being used in different ways.

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