Two-way table probability practice problems

Check the table before you use it. The rows give $36 + 84 = 120$ and $24 + 96 = 120$, the columns give $36 + 24 = 60$ and $84 + 96 = 180$, and the margins agree with the grand total: $120 + 120 = 240$ and $60 + 180 = 240$.

Part (a) asks for a marginal probability, a column total over the grand total. The rented-shoes column holds 60 visits, so $P(\text{rented}) = \frac{60}{240} = 0.25$.

Part (b) asks for a joint probability, a single cell over the grand total. The bouldered-and-rented cell holds 36 visits, so $P(\text{bouldered} \cap \text{rented}) = \frac{36}{240} = 0.15$.

Part (c) is also a joint probability. The roped-and-own-shoes cell holds 96 visits, so $P(\text{roped} \cap \text{own}) = \frac{96}{240} = 0.4$.

All three parts divide by 240 because none of them restricted you to a smaller group. Nothing here says 'given'.

Confirm the margins: $36 + 44 = 80$ and $84 + 36 = 120$ across the rows, $36 + 84 = 120$ and $44 + 36 = 80$ down the columns, and $80 + 120 = 200 = 120 + 80$.

Part (a): the condition is 'it was electronics', so throw away the textiles row and work inside the electronics row, which holds 80 items.

Of those 80 electronics, the repaired cell holds 36, so $P(\text{repaired} \mid \text{electronics}) = \frac{36}{80} = 0.45$.

Part (b): the condition is now 'it was repaired', so work inside the repaired column, which holds 120 items.

Of those 120 repairs, the electronics cell holds 36, so $P(\text{electronics} \mid \text{repaired}) = \frac{36}{120} = 0.30$.

Part (c): both fractions start from the same 36 items. Part (a) divides by the row total 80 and part (b) divides by the column total 120, which is the whole reason the two answers differ.

A conditional distribution across a row divides every cell in that row by that row's own total, and the results must add to 1.

Part (a): the weekday row totals 300. Commute is $\frac{180}{300} = 0.60$, errand is $\frac{90}{300} = 0.30$, and leisure is $\frac{30}{300} = 0.10$.

Check the weekday row: $0.60 + 0.30 + 0.10 = 1.00$.

Part (b): the weekend row totals 200. Commute is $\frac{40}{200} = 0.20$, errand is $\frac{60}{200} = 0.30$, and leisure is $\frac{100}{200} = 0.50$.

Check the weekend row: $0.20 + 0.30 + 0.50 = 1.00$.

Part (c): compare category by category. Commuting falls from 60% of weekday trips to 20% of weekend trips, leisure climbs from 10% to 50%, and errands hold at 30% in both.

Because the two conditional distributions sit far apart on two of the three purposes, day type and trip purpose are associated in this data.

List what you were given: 400 trees in all, 150 recent plantings, 100 damaged trees, and 24 trees that are both recent and damaged.

Finish the recent row. If 24 of the 150 recent trees are damaged, then $150 - 24 = 126$ recent trees are undamaged.

Get the other row total: $400 - 150 = 250$ trees were planted earlier.

Split that row using the damage column: $100 - 24 = 76$ damaged trees were planted earlier, so $250 - 76 = 174$ earlier trees are undamaged.

Assemble the table and check the second column and the grand total: $126 + 174 = 300$ and $100 + 300 = 400$.

Part (b): condition on each row in turn. $P(\text{damage} \mid \text{recent}) = \frac{24}{150} = 0.16$ and $P(\text{damage} \mid \text{earlier}) = \frac{76}{250} = 0.304$.

Part (c): now condition on the damage column, which holds 100 trees, of which 24 are recent plantings, so $P(\text{recent} \mid \text{damage}) = \frac{24}{100} = 0.24$.

Parts (b) and (c) both use the same 24 trees. The first divides by 150 recent trees, the second by 100 damaged trees.

Check the margins: $45 + 5 = 50$, $135 + 315 = 450$, $45 + 135 = 180$, $5 + 315 = 320$, and $50 + 450 = 500 = 180 + 320$.

Part (a): the condition is having run a marathon, so the denominator is that row total, 50.

$P(\text{GPS watch} \mid \text{marathon}) = \frac{45}{50} = 0.90$.

Part (b): the condition is now owning a GPS watch, so the denominator is that column total, 180.

$P(\text{marathon} \mid \text{GPS watch}) = \frac{45}{180} = 0.25$.

Part (c): the same 45 members sit on top of both fractions, but the groups underneath are very different sizes. Only 50 members have run a marathon while 180 own a watch, and $\frac{180}{50} = 3.6$.

That factor is exactly the gap between the two answers, since $\frac{0.90}{0.25} = 3.6$. In words, nearly every marathoner owns a watch, but only a quarter of watch owners have run a marathon.

Do not compare raw counts. The returning row beats the first-time row in every channel only because it holds 300 buyers against 200, so convert each row to proportions first.

First-time row, total 200: online $\frac{124}{200} = 0.62$, phone $\frac{28}{200} = 0.14$, box office $\frac{48}{200} = 0.24$. Check: $0.62 + 0.14 + 0.24 = 1.00$.

Returning row, total 300: online $\frac{177}{300} = 0.59$, phone $\frac{45}{300} = 0.15$, box office $\frac{78}{300} = 0.26$. Check: $0.59 + 0.15 + 0.26 = 1.00$.

Compare channel by channel: online 0.62 against 0.59, phone 0.14 against 0.15, box office 0.24 against 0.26.

The largest gap is 3 percentage points, on online buying. Knowing whether a buyer is new barely changes the channel you would predict.

Conclude that these data show little evidence of association. Identical conditional distributions would mean the two variables show no association in this sample, and these two are close to identical.

Check the margins first. Rows: $27 + 108 = 135$, $21 + 84 = 105$, $42 + 78 = 120$. Columns: $27 + 21 + 42 = 90$ and $108 + 84 + 78 = 270$. Grand total: $135 + 105 + 120 = 360 = 90 + 270$.

Part (a) is a marginal probability, so divide the advanced column total by the grand total: $P(\text{advanced}) = \frac{90}{360} = 0.25$.

Part (b): condition on each row in turn. $P(\text{advanced} \mid \text{life science}) = \frac{27}{135} = 0.20$.

$P(\text{advanced} \mid \text{physical science}) = \frac{21}{105} = 0.20$ and $P(\text{advanced} \mid \text{engineering}) = \frac{42}{120} = 0.35$.

Part (c) reverses the condition. Restrict to the 90 projects that advanced; 42 of them were engineering, so $P(\text{engineering} \mid \text{advanced}) = \frac{42}{90} \approx 0.467$.

Notice the contrast: engineering supplied $\frac{120}{360} \approx 0.333$ of all projects but about $0.467$ of the advancing ones.

Part (d): the advance rate is 0.20 in both science categories and 0.35 in engineering, a gap of 15 percentage points. Because the conditional probability of advancing changes when the category changes, category and advancing look associated.

Turn the percentages into counts out of 1000 riders. Pass holders: $0.40 \times 1000 = 400$. Riders without a pass: $1000 - 400 = 600$.

Split the pass holders with their own rate: $0.75 \times 400 = 300$ ride at least five days a week, so $400 - 300 = 100$ ride fewer.

Split the riders without a pass: $0.20 \times 600 = 120$ ride at least five days a week, so $600 - 120 = 480$ ride fewer.

Assemble the table and check the columns: $300 + 120 = 420$, $100 + 480 = 580$, and $420 + 580 = 1000$.

Part (b) is marginal: $P(\text{5 or more days}) = \frac{420}{1000} = 0.42$.

Part (c) reverses the condition given in the problem. Restrict to the 420 frequent riders, of whom 300 hold a pass: $P(\text{pass} \mid \text{5 or more days}) = \frac{300}{420} \approx 0.714$.

Part (d): independence would require $P(\text{5 or more days} \mid \text{pass}) = P(\text{5 or more days})$, that is $0.75 = 0.42$, which is false.

A second check agrees. The joint probability is $\frac{300}{1000} = 0.30$, while the product of the marginals is $0.40 \times 0.42 = 0.168$, and $0.30 \neq 0.168$.