3.4 Contingency Tables
SLO 3
Describe and apply probability concepts and distributions.
Contingency tables are where a probability concept meets raw data: you sort a sample by two variables at once, then read P(A), P(A AND B), P(A OR B) and P(A | B) straight off the counts, and test independence by checking whether the joint count matches the product of the margins.
Learning Objectives
By the end of this section, you will be able to:
- read a contingency table, and identify its joint counts, its marginal (row and column) totals, and its grand total;
- compute \(P(A)\), \(P(A \text{ AND } B)\), \(P(A \text{ OR } B)\), and \(P(A \mid B)\) directly from the counts in a table;
- explain why a conditional probability uses a row or column total as its denominator instead of the grand total;
- fill in the missing entries of a partially completed contingency table using the fact that rows, columns, and totals must agree;
- test whether two events are independent by comparing \(P(A \text{ AND } B)\) against \(P(A)P(B)\) using table counts;
- read a probability contingency table, in which the entries are probabilities that sum to 1 rather than counts.
Sections 3.1 through 3.3 gave you the vocabulary and the two rules. What they did not give you is a place to put the numbers. Every problem so far arrived with its probabilities already computed and handed to you in the sentence; real data almost never does.
A contingency table is that place. It is a grid that sorts a sample by two variables at once — one variable across the rows, the other across the columns — and reports how many members of the sample fall into each combination. Once the data sits in that grid, every probability in this chapter becomes a matter of reading two numbers off the table and dividing. Conditional probabilities in particular become almost mechanical, which is why this display is worth learning as a tool rather than as a topic. You will meet contingency tables again much later, when the question turns from "what is the probability" to "is this pattern real"; the reading skill you build here is the same one you will need then.
3.4.1 What a Contingency Table Shows
A contingency table displays the members of a sample classified simultaneously by two variables. Each row corresponds to one value of the first variable, each column to one value of the second, and the entry where a row and a column meet reports how many sample members have both of those values.
The two variables may be dependent — contingent — on one another, and the table is how you find out.
The name is worth unpacking. A table of one variable tells you how a sample splits. A contingency table tells you whether the split on one variable changes depending on where you are in the other, and that is exactly the question independence asks.
Definition 3.4.1 — A contingency table is one sample partitioned by two variables at once; the six interior counts are what falls out.
In a contingency table:
- a joint count is an interior cell — the number of sample members in one specific row and one specific column at the same time;
- a marginal total is a row total or a column total — the number of sample members in that row, or that column, regardless of the other variable;
- the grand total is the size of the whole sample. It sits in the lower-right corner, and it equals both the sum of the row totals and the sum of the column totals.
Those three kinds of number are the only ingredients any probability in this section needs, and which one you put in the denominator is the entire game:
- \(P(A)\) — a single event. Marginal total for \(A\), over the grand total.
- \(P(A \text{ AND } B)\) — both at once. The joint count in the cell where \(A\)'s row meets \(B\)'s column, over the grand total.
- \(P(A \text{ OR } B)\) — at least one. Add the two marginal totals and subtract the joint count once (the addition rule from section 3.3), then divide by the grand total.
- \(P(A \mid B)\) — \(A\), restricted to \(B\). The joint count, over \(B\)'s marginal total. The grand total does not appear at all.
Every one of these four probabilities can use the same interior cell as its numerator. What distinguishes them is the bottom of the fraction. A grand total in the denominator means "out of everybody"; a row or column total means "out of this group only." If you ever lose track of which probability you are computing, read your own denominator back to yourself.
Imagine a lecture hall of 755 people and you ask everyone who got a speeding ticket last year to stand. Once they are standing, you stop looking at the seated crowd entirely — the question "how many of these people use a phone while driving?" is now being asked of the standing group and nobody else. The 755 is gone; the standing count replaced it. That replacement is conditional probability, and in a contingency table it is literally the act of covering the rest of the table with your hand and reading one row.
Definition 3.4.2 — Joint count, marginal total, grand total: the same table, three different regions, three different denominators.
Table 3.4.1 shows the number of athletes who stretch before exercising and how many had injuries within the past year.
| Injury in last year | No injury in last year | Total | |
|---|---|---|---|
| Stretches | 55 | 295 | 350 |
| Does not stretch | 231 | 219 | 450 |
| Total | 286 | 514 | 800 |
a. What is \(P(\text{athlete stretches before exercising})\)?
b. What is \(P(\text{athlete stretches before exercising} \mid \text{no injury in the last year})\)?
Solution
Part a — one event, so use a marginal total over the grand total. Stretching is a row, and its row total is 350. The whole sample is 800.
$$P(\text{stretches}) = \frac{350}{800} = 0.4375$$Part b — "given" replaces the denominator. The condition is no injury in the last year, which is a column with a marginal total of 514. Restricted to that column, the athletes who also stretch are the 295 in the interior cell.
$$P(\text{stretches} \mid \text{no injury}) = \frac{295}{514} \approx 0.5739$$Notice the numerator 295 and the denominator 514 both come out of the same column; the 800 never appears.
Answer: a. \(\frac{350}{800} = 0.4375\); b. \(\frac{295}{514} \approx 0.5739\).
Suppose a study of speeding violations and drivers who use cell phones produced the following data.
| Speeding violation in the last year | No speeding violation in the last year | Total | |
|---|---|---|---|
| Uses cell phone while driving | 25 | 280 | 305 |
| Does not use cell phone while driving | 45 | 405 | 450 |
| Total | 70 | 685 | 755 |
The total number of people in the sample is 755. The row totals are 305 and 450, and the column totals are 70 and 685. Notice that \(305 + 450 = 755\) and \(70 + 685 = 755\) — the two marginal breakdowns of the same sample must agree.
Calculate the following probabilities using the table.
a. Find \(P(\text{driver is a cell phone user})\).
b. Find \(P(\text{driver had no violation in the last year})\).
c. Find \(P(\text{driver had no violation in the last year AND was a cell phone user})\).
d. Find \(P(\text{driver is a cell phone user OR driver had no violation in the last year})\).
e. Find \(P(\text{driver is a cell phone user} \mid \text{driver had a violation in the last year})\).
f. Find \(P(\text{driver had no violation last year} \mid \text{driver was not a cell phone user})\).
Solution
Part a — a marginal total over the grand total. Cell phone users form a row whose total is 305.
$$P(\text{cell phone user}) = \frac{305}{755}$$Part b — the other variable, so a column total this time. Drivers with no violation form a column whose total is 685.
$$P(\text{no violation}) = \frac{685}{755}$$Part c — "AND" is one interior cell. Find where the cell phone user row meets the no violation column: 280 drivers are in both groups at once.
$$P(\text{no violation AND cell phone user}) = \frac{280}{755}$$Part d — "OR" uses the addition rule, and the overlap is part c. Adding 305 and 685 counts the 280 drivers who are in both groups twice, so subtract one copy of them.
$$P(\text{cell phone user OR no violation}) = \frac{305}{755} + \frac{685}{755} - \frac{280}{755} = \frac{710}{755}$$You can check this against the table directly: the only drivers not counted are the 45 who neither use a phone nor went without a violation, and \(755 - 45 = 710\).
Part e — "given" shrinks the sample space to one column. The condition is had a violation, a column totalling 70. Inside that column, 25 drivers are cell phone users.
$$P(\text{cell phone user} \mid \text{violation}) = \frac{25}{70}$$The sample space is reduced to the 70 drivers who had a violation.
Part f — "given" again, this time a row. The condition is does not use a cell phone, a row totalling 450. Inside that row, 405 drivers had no violation.
$$P(\text{no violation} \mid \text{not a cell phone user}) = \frac{405}{450}$$The sample space is reduced to the 450 drivers who were not cell phone users.
Answer: a. \(\frac{305}{755}\); b. \(\frac{685}{755}\); c. \(\frac{280}{755}\); d. \(\frac{710}{755}\); e. \(\frac{25}{70}\); f. \(\frac{405}{450}\).
3.4.2 Completing a Table and Testing Independence
Two events \(A\) and \(B\) recorded in a contingency table are independent exactly when
$$P(A \text{ AND } B) = P(A) \, P(B)$$The two sides of the test are often close — 0.18 against 0.153 in the worked example below — and "close" is not "equal." Independence is an exact statement about the population the table describes; there is no tolerance band here. Carry your fractions or decimals far enough that a real difference cannot hide in a rounding.
Read all three probabilities off the table — the joint count over the grand total on the left, the two marginal totals over the grand total on the right — and compare. If the two sides match, the events are independent. If they do not, the events are dependent.
Equivalently, you may compare \(P(A \mid B)\) with \(P(A)\): independence says conditioning on \(B\) leaves \(A\)'s probability alone. The two tests always agree, so use whichever numbers the table hands you more cheaply.
A contingency table is heavily constrained: every row must sum to its row total, every column to its column total, and both sets of totals to the same grand total. That redundancy pays off twice. If a table arrives with holes in it, the constraints will usually fill them for you and let you check your own arithmetic before you compute a single probability; and once the table is complete, its counts already contain everything the independence test needs — no extra data required.
Definition 3.4.3 — The independence test is a height comparison: the joint probability against the product of the margins.
Table 3.4.3 shows a random sample of 200 cyclists and the routes they prefer. Let \(O =\) older and \(H =\) hilly path.
| Age Group | Lake Path | Hilly Path | Wooded Path | Total |
|---|---|---|---|---|
| Younger | 45 | 38 | 27 | 110 |
| Older | 26 | 52 | 12 | 90 |
| Total | 71 | 90 | 39 | 200 |
a. Out of the older group, what is the probability that the cyclist prefers a hilly path?
b. Are the events "being older" and "preferring the hilly path" independent events?
Solution
Part a — "out of the older group" is a conditional. The phrase names the sample space: the 90 older cyclists. Inside that row, 52 prefer the hilly path.
$$P(H \mid O) = \frac{52}{90} \approx 0.5778$$Part b — run the test. Read all three probabilities off the table.
$$P(O \text{ AND } H) = \frac{52}{200} = 0.26$$ $$P(O) \, P(H) = \left(\frac{90}{200}\right)\left(\frac{90}{200}\right) = (0.45)(0.45) = 0.2025$$Since \(0.26 \neq 0.2025\), the events are not independent. The comparison in part a says the same thing in different words: older cyclists prefer the hilly path 57.78% of the time, while cyclists overall prefer it only \(\frac{90}{200} = 45\%\) of the time. Knowing a cyclist is older changes the odds, which is precisely dependence.
Answer: a. \(\frac{52}{90} \approx 0.5778\); b. no — \(0.26 \neq 0.2025\), so the events are dependent.
Table 3.4.4 shows a random sample of 100 hikers and the areas of hiking they prefer.
| Sex | The Coastline | Near Lakes and Streams | On Mountain Peaks | Total |
|---|---|---|---|---|
| Women | 18 | 16 | 45 | |
| Men | 14 | 55 | ||
| Total | 41 |
a. Complete the table.
b. Are the events "being a woman" and "preferring the coastline" independent events? Let \(F =\) being a woman and let \(C =\) preferring the coastline.
c. Find the probability that a person is a man given that the person prefers hiking near lakes and streams. Let \(M =\) being a man, and let \(L =\) prefers hiking near lakes and streams. What word tells you this is a conditional? Is the sample space for this problem all 100 hikers?
d. Find the probability that a person is a woman or prefers hiking on mountain peaks. Let \(F =\) being a woman, and let \(P =\) prefers mountain peaks.
Solution
Part a — let the totals close the gaps. Work one constraint at a time.
- The women's row must total 45, and \(18 + 16 = 34\), so the mountain-peaks entry for women is \(45 - 34 = 11\).
- The grand total is 100, given in the problem, and it is confirmed by the row totals \(45 + 55 = 100\).
- The lakes-and-streams column totals 41 with 16 women in it, so the men's entry is \(41 - 16 = 25\).
- The men's row must total 55, and \(25 + 14 = 39\), so the coastline entry for men is \(55 - 39 = 16\).
- The remaining column totals follow: coastline \(18 + 16 = 34\), mountain peaks \(11 + 14 = 25\).
Check the arithmetic before going on: \(34 + 41 + 25 = 100\), which matches the grand total.
| Sex | The Coastline | Near Lakes and Streams | On Mountain Peaks | Total |
|---|---|---|---|---|
| Women | 18 | 16 | 11 | 45 |
| Men | 16 | 25 | 14 | 55 |
| Total | 34 | 41 | 25 | 100 |
Part b — compute both sides of the independence test.
$$P(F \text{ AND } C) = \frac{18}{100} = 0.18$$ $$P(F) \, P(C) = \left(\frac{45}{100}\right)\left(\frac{34}{100}\right) = (0.45)(0.34) = 0.153$$Since \(P(F \text{ AND } C) \neq P(F)P(C)\), the events \(F\) and \(C\) are not independent.
Part c — the word "given" is the signal.
- The word given tells you this is a conditional.
- The condition is prefers lakes and streams, a column totalling 41; the men inside that column number 25.
- No — the sample space for this problem is the 41 hikers who prefer lakes and streams, not all 100.
Part d — the addition rule, with the overlap read off the completed table.
$$P(F) = \frac{45}{100} \qquad P(P) = \frac{25}{100} \qquad P(F \text{ AND } P) = \frac{11}{100}$$ $$P(F \text{ OR } P) = \frac{45}{100} + \frac{25}{100} - \frac{11}{100} = \frac{59}{100}$$Answer: a. see the completed table above; b. not independent, since \(0.18 \neq 0.153\); c. \(\frac{25}{41}\), sample space is the 41 lakes-and-streams hikers; d. \(\frac{59}{100}\).
Let the tool build the counterfactual
Open rāSHio, choose Calculators → Probability, switch the mode to Conditional, and enter the two margins from Table 3.4.5 — \(P(F) = 0.45\) and \(P(C) = 0.34\) — with Assume independence ticked. The panel returns \(P(F \text{ AND } C) = 0.153\), because ticking that box is exactly the assumption that makes the AND equal the product. That 0.153 is not the answer; it is the number the table's real 0.18 gets measured against, and the fact that the two differ is the verdict. Read the panel as "what this cell would hold if the two variables ignored each other."
Figure 3.4.1 — rāSHio's Calculators → Probability panel in Conditional mode: with Assume independence ticked it returns the AND as the product of the margins, which is the value the independence test compares the table's own joint probability against.
3.4.3 Probability Contingency Tables
A probability contingency table is a contingency table whose interior entries are the joint probabilities \(P(\text{row AND column})\) rather than counts. Its row and column margins are the marginal probabilities \(P(\text{row})\) and \(P(\text{column})\), and its lower-right corner is 1 rather than a sample size.
In a count table, the lower-right corner is the sample size, and you would notice immediately if the rows and columns disagreed about it. In a probability table that corner must be exactly 1, because every member of the sample space is in exactly one interior cell. If your entries do not sum to 1, you have made an arithmetic error somewhere — find it before you compute anything else. This is the single cheapest error check in the section.
Every rule from the count version carries over unchanged; only the grand total is different. In particular \(P(A \mid B)\) is still an interior entry divided by a margin — just an entry and a margin made of probabilities.
Counts are not the only thing a contingency table can hold. In a great many applied problems probabilities are what you are handed directly, and the display works just as well with those in its cells.
Definition 3.4.4 — In a probability contingency table the six interior entries tile the unit interval exactly, which is why the corner is 1.
Andi has to buy a new car. They have two choices, car A and car B. Andi can afford only one car. The probability that they will buy car A is \(P(A) = 0.35\), and the probability that they will buy car B is \(P(B) = 0.65\). Find:
a. \(P(A \text{ AND } B)\)
b. \(P(A \text{ OR } B)\)
Solution
Part a — read the constraint in the problem. "Andi can afford only one car" means the two purchases cannot both happen, so the events are mutually exclusive and there is no overlap at all.
$$P(A \text{ AND } B) = 0$$Part b — the addition rule, with a zero overlap. Because \(P(A \text{ AND } B) = 0\), the subtraction term vanishes.
$$P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) = 0.35 + 0.65 - 0 = 1.00$$The answer of exactly 1 is telling you something real: buying A and buying B are not only mutually exclusive, they exhaust the possibilities as the problem has set them up. Andi buys one of the two cars with certainty.
Answer: a. 0; b. 1.00.
Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is \(\frac{1}{5}\) and the probability he is not caught is \(\frac{4}{5}\). If he goes out the second door, the probability he gets caught by Alissa is \(\frac{1}{4}\) and the probability he is not caught is \(\frac{3}{4}\). The probability that Alissa catches Muddy coming out of the third door is \(\frac{1}{2}\) and the probability she does not catch Muddy is \(\frac{1}{2}\). It is equally likely that Muddy will choose any of the three doors, so the probability of choosing each door is \(\frac{1}{3}\).
| Caught or Not | Door One | Door Two | Door Three | Total |
|---|---|---|---|---|
| Caught | \(\frac{1}{15}\) | \(\frac{1}{12}\) | \(\frac{1}{6}\) | |
| Not Caught | \(\frac{4}{15}\) | \(\frac{3}{12}\) | \(\frac{1}{6}\) | |
| Total | 1 |
The first entry \(\frac{1}{15} = \left(\frac{1}{5}\right)\left(\frac{1}{3}\right)\) is \(P(\text{Door One AND Caught})\). The entry \(\frac{4}{15} = \left(\frac{4}{5}\right)\left(\frac{1}{3}\right)\) is \(P(\text{Door One AND Not Caught})\). Verify the remaining entries.
a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.
b. What is the probability that Alissa does not catch Muddy?
c. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?
Solution
Step 1 — see where the interior entries come from. Each interior entry is a multiplication-rule product: the probability of choosing that door, times the probability of the outcome given that door.
$$P(\text{Door Two AND Caught}) = \left(\frac{1}{4}\right)\left(\frac{1}{3}\right) = \frac{1}{12} \qquad P(\text{Door Two AND Not Caught}) = \left(\frac{3}{4}\right)\left(\frac{1}{3}\right) = \frac{3}{12} = \frac{1}{4}$$ $$P(\text{Door Three AND Caught}) = \left(\frac{1}{2}\right)\left(\frac{1}{3}\right) = \frac{1}{6} \qquad P(\text{Door Three AND Not Caught}) = \left(\frac{1}{2}\right)\left(\frac{1}{3}\right) = \frac{1}{6}$$Step 2 — add across the rows, using a common denominator of 60.
$$P(\text{Caught}) = \frac{1}{15} + \frac{1}{12} + \frac{1}{6} = \frac{4}{60} + \frac{5}{60} + \frac{10}{60} = \frac{19}{60}$$ $$P(\text{Not Caught}) = \frac{4}{15} + \frac{1}{4} + \frac{1}{6} = \frac{16}{60} + \frac{15}{60} + \frac{10}{60} = \frac{41}{60}$$Step 3 — add down the columns. Each column is one door, and each door is chosen with probability \(\frac{1}{3}\), so every column total should be \(\frac{1}{3}\) — a built-in check.
$$\frac{1}{15} + \frac{4}{15} = \frac{5}{15} = \frac{1}{3} \qquad \frac{1}{12} + \frac{3}{12} = \frac{4}{12} = \frac{1}{3} \qquad \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$Step 4 — check the corner. Both ways of totalling agree and land on 1: \(\frac{19}{60} + \frac{41}{60} = \frac{60}{60} = 1\), and \(\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\).
| Caught or Not | Door One | Door Two | Door Three | Total |
|---|---|---|---|---|
| Caught | \(\frac{1}{15}\) | \(\frac{1}{12}\) | \(\frac{1}{6}\) | \(\frac{19}{60}\) |
| Not Caught | \(\frac{4}{15}\) | \(\frac{1}{4}\) | \(\frac{1}{6}\) | \(\frac{41}{60}\) |
| Total | \(\frac{1}{3}\) | \(\frac{1}{3}\) | \(\frac{1}{3}\) | 1 |
Part b — read the margin. The Not Caught row total is the answer.
$$P(\text{Not Caught}) = \frac{41}{60}$$Part c — a conditional, so the denominator is the "Caught" margin. Restricted to the caught outcomes, the doors One and Two contribute \(\frac{1}{15} + \frac{1}{12}\).
$$P(\text{Door One OR Door Two} \mid \text{Caught}) = \frac{\frac{1}{15} + \frac{1}{12}}{\frac{19}{60}} = \frac{\frac{4}{60} + \frac{5}{60}}{\frac{19}{60}} = \frac{\frac{9}{60}}{\frac{19}{60}} = \frac{9}{19}$$The 60ths cancel, which is why a common denominator was worth setting up in step 2.
Answer: a. see the completed table above, corner \(= 1\); b. \(\frac{41}{60}\); c. \(\frac{9}{19}\).
3.4.4 Tables of Rates and Measured Totals
Nothing in the machinery of the last three subsections required the interior entries to be headcounts. They only had to be additive — quantities that can be summed across a row or a column and still mean something.
Crime rates per 100,000 inhabitants are additive in exactly that sense, and so are dollar amounts, hours logged, and tonnes shipped. When the entries are rates rather than counts, the grand total is no longer a sample size, and no individual "person" corresponds to a probability. What you are computing is the share of the total that a given row, column, or cell accounts for. That share behaves algebraically like a probability, and every formula in this section still applies.
A grand total of 4,520.7 is not 4,520 people; it is the sum of four years' worth of rates per 100,000 inhabitants. "Find \(P(\text{Year 2 AND Robbery})\)" is asking what fraction of the total recorded crime rate falls in that one cell. The arithmetic is identical to the count case, but the sentence you write to interpret the answer is not — say "share of the total rate," not "chance that a person."
Table 3.4.8 relates the weights and heights of a group of individuals participating in an observational study.
| Weight/Height | Tall | Medium | Short | Totals |
|---|---|---|---|---|
| Overweight | 18 | 28 | 14 | |
| Typical Weight Range | 20 | 51 | 28 | |
| Underweight | 12 | 25 | 9 | |
| Totals |
a. Find the total for each row and column.
b. Find the probability that a randomly chosen individual from this group is Tall.
c. Find the probability that a randomly chosen individual from this group is Overweight and Tall.
d. Find the probability that a randomly chosen individual from this group is Tall given that the individual is Overweight.
e. Find the probability that a randomly chosen individual from this group is Overweight given that the individual is Tall.
f. Find the probability a randomly chosen individual from this group is Tall and Underweight.
g. Are the events Overweight and Tall independent?
Solution
Part a — sum each row and each column, then confirm the two grand totals agree.
Row totals: \(18 + 28 + 14 = 60\); \(20 + 51 + 28 = 99\); \(12 + 25 + 9 = 46\).
Column totals: \(18 + 20 + 12 = 50\); \(28 + 51 + 25 = 104\); \(14 + 28 + 9 = 51\).
Check: \(60 + 99 + 46 = 205\) and \(50 + 104 + 51 = 205\). The grand total is 205.
| Weight/Height | Tall | Medium | Short | Totals |
|---|---|---|---|---|
| Overweight | 18 | 28 | 14 | 60 |
| Typical Weight Range | 20 | 51 | 28 | 99 |
| Underweight | 12 | 25 | 9 | 46 |
| Totals | 50 | 104 | 51 | 205 |
Part b — a column total over the grand total.
$$P(\text{Tall}) = \frac{50}{205} \approx 0.2439$$Part c — one interior cell over the grand total.
$$P(\text{Overweight AND Tall}) = \frac{18}{205} \approx 0.0878$$Part d — condition on Overweight, so the denominator is that row's total of 60.
$$P(\text{Tall} \mid \text{Overweight}) = \frac{18}{60} = 0.3$$Part e — condition on Tall instead, so the denominator becomes that column's total of 50. Same numerator, different question.
$$P(\text{Overweight} \mid \text{Tall}) = \frac{18}{50} = 0.36$$Part f — another interior cell over the grand total.
$$P(\text{Tall AND Underweight}) = \frac{12}{205} \approx 0.0585$$Part g — run the independence test.
$$P(\text{Overweight AND Tall}) = \frac{18}{205} \approx 0.0878$$ $$P(\text{Overweight}) \, P(\text{Tall}) = \left(\frac{60}{205}\right)\left(\frac{50}{205}\right) \approx (0.2927)(0.2439) \approx 0.0714$$The two values differ, so Overweight and Tall are not independent. Parts d and e say the same thing: \(P(\text{Tall} \mid \text{Overweight}) = 0.3\) while \(P(\text{Tall}) \approx 0.2439\), so knowing an individual is overweight raises the chance they are tall.
Answer: a. row totals 60, 99, 46; column totals 50, 104, 51; grand total 205. b. \(\frac{50}{205}\); c. \(\frac{18}{205}\); d. \(\frac{18}{60} = 0.3\); e. \(\frac{18}{50} = 0.36\); f. \(\frac{12}{205}\); g. not independent.
Table 3.4.10 contains the number of crimes per 100,000 inhabitants in the United States over the span of several years. Total each column and each row; the total of all the data is 4,520.7.
| Year | Robbery | Burglary | Vandalism | Vehicle | Total |
|---|---|---|---|---|---|
| 1 | 145.7 | 732.1 | 29.7 | 314.7 | |
| 2 | 133.1 | 717.7 | 29.1 | 259.2 | |
| 3 | 119.3 | 701.0 | 27.7 | 239.1 | |
| 4 | 113.7 | 702.2 | 26.8 | 229.6 | |
| Total |
a. Find \(P(\text{Year 2 AND Robbery})\).
b. Find \(P(\text{Year 3 AND Burglary})\).
c. Find \(P(\text{Year 3 OR Burglary})\).
d. Find \(P(\text{Year 4} \mid \text{Vandalism})\).
e. Find \(P(\text{Vehicle} \mid \text{Year 1})\).
Solution
Step 1 — total the table first. Every part below needs one of these margins, so compute them once.
Row totals: \(145.7 + 732.1 + 29.7 + 314.7 = 1{,}222.2\); \(133.1 + 717.7 + 29.1 + 259.2 = 1{,}139.1\); \(119.3 + 701.0 + 27.7 + 239.1 = 1{,}087.1\); \(113.7 + 702.2 + 26.8 + 229.6 = 1{,}072.3\).
Column totals: Robbery \(511.8\); Burglary \(2{,}853.0\); Vandalism \(113.3\); Vehicle \(1{,}042.6\).
Both add to the stated grand total: \(1{,}222.2 + 1{,}139.1 + 1{,}087.1 + 1{,}072.3 = 4{,}520.7\) and \(511.8 + 2{,}853.0 + 113.3 + 1{,}042.6 = 4{,}520.7\).
| Year | Robbery | Burglary | Vandalism | Vehicle | Total |
|---|---|---|---|---|---|
| 1 | 145.7 | 732.1 | 29.7 | 314.7 | 1,222.2 |
| 2 | 133.1 | 717.7 | 29.1 | 259.2 | 1,139.1 |
| 3 | 119.3 | 701.0 | 27.7 | 239.1 | 1,087.1 |
| 4 | 113.7 | 702.2 | 26.8 | 229.6 | 1,072.3 |
| Total | 511.8 | 2,853.0 | 113.3 | 1,042.6 | 4,520.7 |
Part a — one interior cell over the grand total.
$$P(\text{Year 2 AND Robbery}) = \frac{133.1}{4{,}520.7} \approx 0.0294$$Part b — same shape, different cell.
$$P(\text{Year 3 AND Burglary}) = \frac{701.0}{4{,}520.7} \approx 0.1551$$Part c — the addition rule, subtracting the overlap from part b.
$$P(\text{Year 3 OR Burglary}) = \frac{1{,}087.1 + 2{,}853.0 - 701.0}{4{,}520.7} = \frac{3{,}239.1}{4{,}520.7} \approx 0.7165$$Part d — condition on Vandalism, so the denominator is that column's total.
$$P(\text{Year 4} \mid \text{Vandalism}) = \frac{26.8}{113.3} \approx 0.2365$$Part e — condition on Year 1, so the denominator is that row's total.
$$P(\text{Vehicle} \mid \text{Year 1}) = \frac{314.7}{1{,}222.2} \approx 0.2575$$Answer: a. 0.0294; b. 0.1551; c. 0.7165; d. 0.2365; e. 0.2575.
Problem Set 3.4
Problems 3.4.1 through 3.4.4: Table 3.4.12 shows a random sample of musicians and how they learned to play their instruments.
| Gender | Self-taught | Studied in School | Private Instruction | Total |
|---|---|---|---|---|
| Woman | 12 | 38 | 22 | 72 |
| Man | 19 | 24 | 15 | 58 |
| Total | 31 | 62 | 37 | 130 |
Problem 1. Find \(P(\text{musician is a woman})\).
Problem 1 Solution
Step 1 — identify what kind of probability this is. "Musician is a woman" names one event only, so this is a marginal probability: a row total over the grand total.
Step 2 — read the two numbers off Table 3.4.12. The Woman row totals 72, and the sample is 130 musicians.
$$P(\text{woman}) = \frac{72}{130} \approx 0.5538$$Answer: \(\frac{72}{130} \approx 0.5538\).
Problem 2. Find \(P(\text{musician is a man AND had private instruction})\).
Problem 2 Solution
Step 1 — "AND" points at a single interior cell. The problem asks for musicians who are in the Man row and the Private Instruction column at the same time, so no addition or subtraction is needed — just find where that row meets that column.
Step 2 — read the cell and divide by the grand total. The Man / Private Instruction cell holds 15 musicians.
$$P(\text{man AND private instruction}) = \frac{15}{130} \approx 0.1154$$Answer: \(\frac{15}{130} \approx 0.1154\).
Problem 3. Find \(P(\text{musician is a woman OR is self-taught})\).
Problem 3 Solution
Step 1 — "OR" calls for the addition rule. Let \(W\) be the event that the musician is a woman and \(T\) the event that the musician is self-taught. The addition rule is \(P(W \text{ OR } T) = P(W) + P(T) - P(W \text{ AND } T)\).
Step 2 — read all three pieces off Table 3.4.12. The Woman row totals 72, the Self-taught column totals 31, and the cell where they meet holds 12.
$$P(W \text{ OR } T) = \frac{72}{130} + \frac{31}{130} - \frac{12}{130} = \frac{91}{130} = 0.7$$Step 3 — check the subtraction was needed. The 12 self-taught women appear in both the row total and the column total, so adding 72 and 31 counts them twice; removing one copy leaves 91 distinct musicians.
Answer: \(\frac{91}{130} = 0.7\).
Problem 4. Are the events "being a woman musician" and "learning music in school" mutually exclusive events?
Problem 4 Solution
Step 1 — recall the test. Two events are mutually exclusive when they cannot both happen on the same trial, that is, when \(P(A \text{ AND } B) = 0\).
Step 2 — find the joint count. The cell where the Woman row meets the Studied in School column holds 38 musicians, so
$$P(\text{woman AND studied in school}) = \frac{38}{130} \approx 0.2923 \neq 0$$Step 3 — interpret. Thirty-eight of the sampled musicians are women who learned in school, so the two descriptions plainly apply to the same people.
Answer: no — the events are not mutually exclusive, because \(P(\text{woman AND studied in school}) = \frac{38}{130} \neq 0\).
Problems 3.4.5 through 3.4.12: Table 3.4.13 shows the political party affiliation for various members of the U.S. Senate during two separate years when they are up for reelection.
| Up for reelection: | Democratic Party | Republican Party | Other | Total |
|---|---|---|---|---|
| Year A | 20 | 13 | 0 | |
| Year B | 10 | 24 | 0 | |
| Total |
Problem 5. What is the probability that a randomly selected senator has an "Other" affiliation?
Problem 5 Solution
Step 1 — complete the totals for Table 3.4.13 once, since every problem in this group needs them. Year A totals \(20 + 13 + 0 = 33\); Year B totals \(10 + 24 + 0 = 34\); the Democratic column totals \(20 + 10 = 30\); the Republican column totals \(13 + 24 = 37\); the Other column totals \(0 + 0 = 0\). The grand total is \(33 + 34 = 67\), which agrees with \(30 + 37 + 0 = 67\).
Step 2 — read the marginal probability. The Other column total is 0.
$$P(\text{Other}) = \frac{0}{67} = 0$$Answer: 0 — no senator in this table has an "Other" affiliation.
Problem 6. What is the probability that a randomly selected senator is up for reelection in Year B?
Problem 6 Solution
Step 1 — one event, so use a marginal total. Year B is a row, and from the totals computed in problem 3.4.5 that row holds 34 senators.
Step 2 — divide by the grand total of 67.
$$P(\text{Year B}) = \frac{34}{67} \approx 0.5075$$Answer: \(\frac{34}{67} \approx 0.5075\).
Problem 7. What is the probability that a randomly selected senator is a Democrat and up for reelection in Year B?
Problem 7 Solution
Step 1 — "and" is one interior cell. Find where the Year B row meets the Democratic Party column: 10 senators.
Step 2 — divide by the grand total.
$$P(\text{Democrat AND Year B}) = \frac{10}{67} \approx 0.1493$$Answer: \(\frac{10}{67} \approx 0.1493\).
Problem 8. What is the probability that a randomly selected senator is a Republican or is up for reelection in Year A?
Problem 8 Solution
Step 1 — "or" calls for the addition rule. Let \(R\) be Republican and \(A\) be up for reelection in Year A.
Step 2 — read the three pieces. The Republican column totals 37, the Year A row totals 33, and their intersection holds 13.
$$P(R \text{ OR } A) = \frac{37}{67} + \frac{33}{67} - \frac{13}{67} = \frac{57}{67} \approx 0.8507$$Step 3 — sanity check against the table. The only senators excluded are the 10 Democrats in Year B, and \(67 - 10 = 57\), which matches the numerator.
Answer: \(\frac{57}{67} \approx 0.8507\).
Problem 9. Suppose that a member of the U.S. Senate is randomly selected. Given that the randomly selected senator is up for reelection in Year B, what is the probability that this senator is a Democrat?
Problem 9 Solution
Step 1 — "given that" replaces the sample space. The condition is up for reelection in Year B, a row totalling 34, so 34 becomes the denominator and the 67 disappears.
Step 2 — the numerator is the joint count inside that row. Of the 34 Year B senators, 10 are Democrats.
$$P(\text{Democrat} \mid \text{Year B}) = \frac{10}{34} \approx 0.2941$$Answer: \(\frac{10}{34} \approx 0.2941\).
Problem 10. Suppose that a member of the U.S. Senate is randomly selected. What is the probability that the senator is up for reelection in Year A, knowing that this senator is a Republican?
Problem 10 Solution
Step 1 — "knowing that" is the same signal as "given". The condition is this senator is a Republican, a column totalling 37.
Step 2 — the numerator is the Republicans who are also in Year A. That cell holds 13.
$$P(\text{Year A} \mid \text{Republican}) = \frac{13}{37} \approx 0.3514$$Answer: \(\frac{13}{37} \approx 0.3514\).
Problem 11. The events "Republican" and "Up for reelection in Year B" are
a) mutually exclusive.
b) independent.
c) both mutually exclusive and independent.
d) neither mutually exclusive nor independent.
Problem 11 Solution
Step 1 — test mutual exclusivity first. Let \(R\) be Republican and \(B\) be up for reelection in Year B. Their intersection holds 24 senators, so
$$P(R \text{ AND } B) = \frac{24}{67} \approx 0.3582 \neq 0$$The events can and do happen together, so they are not mutually exclusive. Options a and c are out.
Step 2 — test independence. Compare the joint probability against the product of the marginals.
$$P(R) \, P(B) = \left(\frac{37}{67}\right)\left(\frac{34}{67}\right) = \frac{1{,}258}{4{,}489} \approx 0.2802$$Since \(0.3582 \neq 0.2802\), the events are not independent either. Option b is out.
Step 3 — pick the remaining option.
Answer: d — neither mutually exclusive nor independent.
Problem 12. The events "Other" and "Up for reelection in Year B" are
a) mutually exclusive.
b) independent.
c) both mutually exclusive and independent.
d) neither mutually exclusive nor independent.
Problem 12 Solution
Step 1 — test mutual exclusivity. Let \(O\) be an "Other" affiliation and \(B\) be up for reelection in Year B. The Other / Year B cell holds 0 senators, so
$$P(O \text{ AND } B) = \frac{0}{67} = 0$$The events cannot both happen, so they are mutually exclusive.
Step 2 — test independence with the same two numbers. From problem 3.4.5, \(P(O) = 0\), so the product of the marginals is
$$P(O) \, P(B) = (0)\left(\frac{34}{67}\right) = 0 = P(O \text{ AND } B)$$The independence equation is satisfied, so the events are also independent.
Step 3 — read what happened. Both tests pass at once only because \(P(O) = 0\): an impossible event is trivially independent of everything, since nothing can change odds that are already zero. This is the degenerate case that problem 3.4.11 does not have, and it is why "mutually exclusive" and "independent" are not opposites.
Answer: c — both mutually exclusive and independent.
Problem 13. Table 3.4.14 gives the number of participants in a recent National Health Interview Survey who had been treated for cancer in the previous 12 months. The results are sorted by age, race (Black or White), and sex. We are interested in possible relationships between age, race, and sex. Do not include "all others" for parts f and g.
| Race and sex | 15–24 | 25–40 | 41–65 | Over 65 | TOTALS |
|---|---|---|---|---|---|
| White, male | 1,165 | 2,036 | 3,703 | 8,395 | |
| White, female | 1,076 | 2,242 | 4,060 | 9,129 | |
| Black, male | 142 | 194 | 384 | 824 | |
| Black, female | 131 | 290 | 486 | 1,061 | |
| All others | |||||
| TOTALS | 2,792 | 5,279 | 9,354 | 21,081 |
a) Fill in the column for cancer treatment for individuals over age 65.
b) Fill in the row for all other races.
c) Find the probability that a randomly selected individual was a White male.
d) Find the probability that a randomly selected individual was a Black female.
e) Find the probability that a randomly selected individual was Black.
f) Find the probability that a randomly selected individual was male.
g) Out of the individuals over age 65, find the probability that a randomly selected individual was a Black or White male.
Problem 13 Solution
Step 1 (parts a and b) — fill the holes using the row and column constraints. Each row must reach its stated total, and each column must reach its stated total.
Over-65 entries, from the row totals:
$$8{,}395 - (1{,}165 + 2{,}036 + 3{,}703) = 1{,}491 \quad \text{(White, male)}$$ $$9{,}129 - (1{,}076 + 2{,}242 + 4{,}060) = 1{,}751 \quad \text{(White, female)}$$ $$824 - (142 + 194 + 384) = 104 \quad \text{(Black, male)}$$ $$1{,}061 - (131 + 290 + 486) = 154 \quad \text{(Black, female)}$$All-others entries, from the column totals:
$$2{,}792 - (1{,}165 + 1{,}076 + 142 + 131) = 278 \quad \text{(ages 15–24)}$$ $$5{,}279 - (2{,}036 + 2{,}242 + 194 + 290) = 517 \quad \text{(ages 25–40)}$$ $$9{,}354 - (3{,}703 + 4{,}060 + 384 + 486) = 721 \quad \text{(ages 41–65)}$$The all-others row total is \(21{,}081 - (8{,}395 + 9{,}129 + 824 + 1{,}061) = 1{,}672\), so its over-65 entry is \(1{,}672 - (278 + 517 + 721) = 156\). The over-65 column then totals \(1{,}491 + 1{,}751 + 104 + 154 + 156 = 3{,}656\), which also equals \(21{,}081 - (2{,}792 + 5{,}279 + 9{,}354)\) — the two routes agree, so the completed table is consistent.
| Race and sex | 15–24 | 25–40 | 41–65 | Over 65 | TOTALS |
|---|---|---|---|---|---|
| White, male | 1,165 | 2,036 | 3,703 | 1,491 | 8,395 |
| White, female | 1,076 | 2,242 | 4,060 | 1,751 | 9,129 |
| Black, male | 142 | 194 | 384 | 104 | 824 |
| Black, female | 131 | 290 | 486 | 154 | 1,061 |
| All others | 278 | 517 | 721 | 156 | 1,672 |
| TOTALS | 2,792 | 5,279 | 9,354 | 3,656 | 21,081 |
Step 2 (part c) — a row total over the grand total.
$$P(\text{White male}) = \frac{8{,}395}{21{,}081} \approx 0.3982$$Step 3 (part d) — same shape, different row.
$$P(\text{Black female}) = \frac{1{,}061}{21{,}081} \approx 0.0503$$Step 4 (part e) — combine the two Black rows, which are mutually exclusive.
$$P(\text{Black}) = \frac{824 + 1{,}061}{21{,}081} = \frac{1{,}885}{21{,}081} \approx 0.0894$$Step 5 (part f) — add the two male rows, leaving out "all others" as instructed. The all-others row does not record sex, so it cannot contribute to a male count.
$$P(\text{male}) = \frac{8{,}395 + 824}{21{,}081} = \frac{9{,}219}{21{,}081} \approx 0.4373$$Step 6 (part g) — a conditional, so the denominator is the over-65 column. Restricted to the 3,656 individuals over 65, the Black males and White males number \(104 + 1{,}491 = 1{,}595\).
$$P(\text{Black or White male} \mid \text{over 65}) = \frac{1{,}595}{3{,}656} \approx 0.4363$$Answer: a. over-65 column: 1,491; 1,751; 104; 154; 156. b. all-others row: 278; 517; 721; 156; total 1,672. c. \(\frac{8{,}395}{21{,}081} \approx 0.3982\); d. \(\frac{1{,}061}{21{,}081} \approx 0.0503\); e. \(\frac{1{,}885}{21{,}081} \approx 0.0894\); f. \(\frac{9{,}219}{21{,}081} \approx 0.4373\); g. \(\frac{1{,}595}{3{,}656} \approx 0.4363\).
Problems 3.4.14 and 3.4.15: Table 3.4.16 shows hit information for four well-known baseball players. Suppose that one hit from the table is randomly selected.
| NAME | Single | Double | Triple | Home Run | TOTAL HITS |
|---|---|---|---|---|---|
| Babe Ruth | 1,517 | 506 | 136 | 714 | 2,873 |
| Jackie Robinson | 1,054 | 273 | 54 | 137 | 1,518 |
| Ty Cobb | 3,035 | 724 | 295 | 117 | 4,189 |
| Hank Aaron | 2,294 | 624 | 98 | 755 | 3,771 |
| TOTAL | 7,918 | 2,127 | 583 | 1,723 | 12,351 |
Problem 14. Find \(P(\text{hit was made by Babe Ruth})\).
a) \(\frac{1{,}518}{2{,}873}\)
b) \(\frac{2{,}873}{12{,}351}\)
c) \(\frac{583}{12{,}351}\)
d) \(\frac{4{,}189}{12{,}351}\)
Problem 14 Solution
Step 1 — identify the shape. One event only, so this is a row total over the grand total. No conditioning phrase appears.
Step 2 — read Table 3.4.16. Babe Ruth's TOTAL HITS is 2,873, and the whole table totals 12,351 hits.
$$P(\text{hit was made by Babe Ruth}) = \frac{2{,}873}{12{,}351} \approx 0.2326$$Step 3 — match against the options. Option a uses Jackie Robinson's total as the numerator and Babe Ruth's as the denominator; option c is the triples column; option d is Ty Cobb's total. Only option b has the right pair.
Answer: b — \(\frac{2{,}873}{12{,}351}\).
Problem 15. Find \(P(\text{hit was made by Ty Cobb} \mid \text{the hit was a Home Run})\).
a) \(\frac{4{,}189}{12{,}351}\)
b) \(\frac{1{,}163}{1{,}723}\)
c) \(\frac{117}{1{,}723}\)
d) \(\frac{117}{4{,}189}\)
Problem 15 Solution
Step 1 — the bar means the sample space shrinks. The condition is the hit was a Home Run, a column totalling 1,723. That column total, not 12,351, is the denominator.
Step 2 — the numerator is the joint count. Ty Cobb's Home Run cell holds 117.
$$P(\text{Ty Cobb} \mid \text{Home Run}) = \frac{117}{1{,}723} \approx 0.0679$$Step 3 — match against the options. Option a ignores the condition entirely, option b is the complement's count over the home-run total, and option d divides by Ty Cobb's own total — that would answer "given the hit was Ty Cobb's, was it a home run?", the reversed question.
Answer: c — \(\frac{117}{1{,}723}\).
Problem 16. Table 3.4.17 identifies a group of children by one of four hair colors, and by type of hair.
| Hair Type | Brown | Blond | Black | Red | Totals |
|---|---|---|---|---|---|
| Wavy | 20 | 15 | 3 | 43 | |
| Straight | 80 | 15 | 12 | ||
| Totals | 20 | 215 |
a) Complete the table.
b) What is the probability that a randomly selected child will have wavy hair?
c) What is the probability that a randomly selected child will have either brown or blond hair?
d) What is the probability that a randomly selected child will have wavy brown hair?
e) What is the probability that a randomly selected child will have red hair, given that they have straight hair?
f) If \(B\) is the event of a child having brown hair, find the probability of the complement of \(B\).
g) In words, what does the complement of \(B\) represent?
Problem 16 Solution
Step 1 (part a) — fill the table from the constraints. The Wavy row totals 43 and already shows \(20 + 15 + 3 = 38\), so the wavy blond entry is \(43 - 38 = 5\). The Blond column totals 20, which confirms the straight blond entry of 15. The grand total is 215, so the Straight row totals \(215 - 43 = 172\), making the straight black entry \(172 - (80 + 15 + 12) = 65\). The column totals follow: brown \(20 + 80 = 100\), black \(15 + 65 = 80\), red \(3 + 12 = 15\). Check: \(100 + 20 + 80 + 15 = 215\).
| Hair Type | Brown | Blond | Black | Red | Totals |
|---|---|---|---|---|---|
| Wavy | 20 | 5 | 15 | 3 | 43 |
| Straight | 80 | 15 | 65 | 12 | 172 |
| Totals | 100 | 20 | 80 | 15 | 215 |
Step 2 (part b) — a row total over the grand total.
$$P(\text{wavy}) = \frac{43}{215} = 0.2$$Step 3 (part c) — brown and blond are different columns, so they are mutually exclusive and simply add.
$$P(\text{brown OR blond}) = \frac{100 + 20}{215} = \frac{120}{215} \approx 0.5581$$Step 4 (part d) — "wavy brown hair" is one interior cell.
$$P(\text{wavy AND brown}) = \frac{20}{215} \approx 0.0930$$Step 5 (part e) — "given that they have straight hair" makes the Straight row total the denominator.
$$P(\text{red} \mid \text{straight}) = \frac{12}{172} \approx 0.0698$$Step 6 (parts f and g) — use the complement rule. With \(B\) the event of brown hair, \(P(B) = \frac{100}{215}\), so
$$P(B') = 1 - \frac{100}{215} = \frac{115}{215} \approx 0.5349$$In words, \(B'\) is the event that a randomly selected child does not have brown hair — that is, the child has blond, black, or red hair.
Answer: a. see the completed table above; b. \(\frac{43}{215} = 0.2\); c. \(\frac{120}{215} \approx 0.5581\); d. \(\frac{20}{215} \approx 0.0930\); e. \(\frac{12}{172} \approx 0.0698\); f. \(\frac{115}{215} \approx 0.5349\); g. the child does not have brown hair (their hair is blond, black, or red).
Problem 17. In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published. The factual data were compiled into Table 3.4.19. For the following, suppose that you randomly select one player from the 49ers or Cowboys.
| Shirt # | ≤ 210 | 211–250 | 251–290 | > 290 |
|---|---|---|---|---|
| 1–33 | 21 | 5 | 0 | 0 |
| 34–66 | 6 | 18 | 7 | 4 |
| 66–99 | 6 | 12 | 22 | 5 |
a) Find the probability that his shirt number is from 1 to 33.
b) Find the probability that he weighs at most 210 pounds.
c) Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds.
d) Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds.
e) Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds.
Problem 17 Solution
Step 1 — total Table 3.4.19 before doing anything else. The table ships without margins, and every part needs them.
Row totals: \(21 + 5 + 0 + 0 = 26\) for shirts 1–33; \(6 + 18 + 7 + 4 = 35\) for 34–66; \(6 + 12 + 22 + 5 = 45\) for 66–99.
Column totals: \(\le 210\) gives \(21 + 6 + 6 = 33\); 211–250 gives \(5 + 18 + 12 = 35\); 251–290 gives \(0 + 7 + 22 = 29\); \(> 290\) gives \(0 + 4 + 5 = 9\).
Grand total: \(26 + 35 + 45 = 106\), which matches \(33 + 35 + 29 + 9 = 106\).
Step 2 (part a) — a row total over the grand total.
$$P(\text{shirt 1–33}) = \frac{26}{106} \approx 0.2453$$Step 3 (part b) — a column total over the grand total.
$$P(\text{at most 210 pounds}) = \frac{33}{106} \approx 0.3113$$Step 4 (part c) — "AND" is the cell where that row meets that column.
$$P(\text{shirt 1–33 AND} \le 210) = \frac{21}{106} \approx 0.1981$$Step 5 (part d) — "OR" is the addition rule, subtracting the part c overlap once.
$$P(\text{shirt 1–33 OR} \le 210) = \frac{26}{106} + \frac{33}{106} - \frac{21}{106} = \frac{38}{106} \approx 0.3585$$Step 6 (part e) — "GIVEN" swaps the denominator to the conditioning group. Restricted to the 33 players weighing at most 210 pounds, 21 wear shirts 1–33.
$$P(\text{shirt 1–33} \mid \le 210) = \frac{21}{33} \approx 0.6364$$Notice how much larger this is than the unconditional \(\frac{26}{106} \approx 0.2453\) from part a — the lighter players are heavily concentrated in the low shirt numbers, which is exactly the kind of dependence a contingency table is built to reveal.
Answer: a. \(\frac{26}{106} \approx 0.2453\); b. \(\frac{33}{106} \approx 0.3113\); c. \(\frac{21}{106} \approx 0.1981\); d. \(\frac{38}{106} \approx 0.3585\); e. \(\frac{21}{33} \approx 0.6364\).
Key Terms
contingency table — a display of a sample classified by two variables at once, with rows for one variable, columns for the other, and counts in the interior cells.
joint count — an interior cell of a contingency table; the number of sample members in one row and one column simultaneously.
marginal total — a row total or column total in a contingency table; the count for one value of one variable, regardless of the other.
grand total — the lower-right entry of a contingency table; the size of the whole sample, equal to both the sum of the row totals and the sum of the column totals.
probability contingency table — a contingency table whose entries are joint probabilities rather than counts, so that its grand total is 1.