A seed library packet holds 12 seeds: 5 heirloom and 7 hybrid. A gardener plants two seeds drawn at random one after the other, without replacement. First-stage branches: first seed heirloom (5/12) and first seed hybrid (7/12). After an heirloom seed is removed, 11 seeds remain and 4 are heirloom, so the second-stage branches are 4/11 heirloom and 7/11 hybrid. After a hybrid seed is removed, 11 seeds remain and 5 are heirloom, so the second-stage branches are 5/11 heirloom and 6/11 hybrid. (a) Find the probability both seeds are heirloom. (b) Find the probability exactly one seed is heirloom. (c) Find the probability at least one seed is heirloom.

(a) 5/33 ≈ 0.1515; (b) 35/66 ≈ 0.5303; (c) 15/22 ≈ 0.6818.

Tree diagram probability practice problems

By Jude Wallis · Updated July 31, 2026

These 8 problems multiply probabilities along a tree branch, add across the paths that qualify, handle a second stage that shifts after the first draw, and reverse the conditioning to get a first-stage probability from a second-stage result. Work each on paper, then check the steps.

AP Statistics: Unit 2 (topics 2.6 Conditional Probability, 2.7 Independent Events and Unions of Events). These problems cover Unit 2 topics 2.6 (conditional probability) and 2.7 (independent events and unions of events) in the Fall 2026 AP Statistics course. A tree diagram is a way to organize those calculations rather than a topic of its own. Unit 2 is 15 to 25% of the multiple-choice section.

What these problems build

These 8 problems build the two moves a tree diagram makes visual: multiply the branch probabilities along a path to get the probability of that whole path, then add the paths that satisfy the question. Each problem describes its tree in words with every branch probability stated, so sketch the tree before you compute. If you have never drawn one, start with how to use a tree diagram.

Use the branch sums as your arithmetic check. The first-stage branches add to 1, the second-stage branches leaving any single node add to 1, and the complete paths add to 1. The second-stage branch probabilities are the plain unconditional ones only when the stages are independent; otherwise they are conditional on the branch you are already on, a distinction set out in when to multiply probabilities. Two of these problems draw without replacement, so the second-stage probabilities change once the first item is gone, the distinction covered in with replacement vs without replacement.

Five problems run the tree backwards: you are given the second-stage outcome and asked for the probability of a particular first-stage branch. Divide the one path you care about by the sum of every path that ends in that same second-stage outcome. That reversal is worked through in the guide on which conditional you need, and for more drill see conditional probability practice and topic 2.6.

Problem 1

A pottery studio fires every mug twice. A mug survives the bisque firing with probability 0.90. Given that it survives the bisque firing, it survives the glaze firing with probability 0.80. A mug that cracks in the bisque firing is discarded and never reaches the glaze firing.

The tree has two first-stage branches: survives bisque (0.90) and cracks in bisque (0.10). Out of the survives-bisque node come two second-stage branches: survives glaze (0.80) and cracks in glaze (0.20).

(a) Find the probability a mug survives both firings. (b) Find the probability a mug survives the bisque firing but cracks in the glaze firing. (c) Show that the three possible outcomes have probabilities adding to 1.

Show the worked solution

Part (a): follow the path survives bisque, then survives glaze, and multiply the two branch probabilities.
$P(\text{survives both}) = 0.90 \times 0.80 = 0.72$ .
Part (b): follow the path survives bisque, then cracks in glaze. The second branch out of that node is $1 - 0.80 = 0.20$ .
$P(\text{survives bisque, cracks in glaze}) = 0.90 \times 0.20 = 0.18$ .
Part (c): the third outcome is cracking in the bisque firing, a single branch with probability $1 - 0.90 = 0.10$ .
Add all three complete paths: $0.72 + 0.18 + 0.10 = 1.00$ , so no outcome has been missed or counted twice.

(a) $0.72$ ; (b) $0.18$ ; (c) $0.72 + 0.18 + 0.10 = 1.00$ .

Problem 2

A commuter rides a ferry to work in the morning and a different ferry home in the evening. The morning ferry runs on time with probability 0.85 and the evening ferry runs on time with probability 0.75. The two crossings are independent.

First-stage branches: morning on time (0.85) and morning late (0.15). Out of each of those, the second-stage branches are evening on time (0.75) and evening late (0.25), unchanged because the crossings are independent.

(a) Find the probability both ferries run on time. (b) Find the probability exactly one runs on time. (c) Find the probability at least one runs on time.

Show the worked solution

Part (a): multiply along the on-time, on-time path. $P(\text{both on time}) = 0.85 \times 0.75 = 0.6375$ .
Part (b): exactly one on time covers two different paths, so compute each and add them.
Morning on time and evening late: $0.85 \times 0.25 = 0.2125$ .
Morning late and evening on time: $0.15 \times 0.75 = 0.1125$ .
Add: $P(\text{exactly one on time}) = 0.2125 + 0.1125 = 0.3250$ .
Part (c): the only path with no on-time ferry is late then late, $0.15 \times 0.25 = 0.0375$ , so $P(\text{at least one on time}) = 1 - 0.0375 = 0.9625$ .
Check all four paths: $0.6375 + 0.2125 + 0.1125 + 0.0375 = 1.0000$ .

(a) $0.6375$ ; (b) $0.3250$ ; (c) $0.9625$ .

Problem 3

A seed library packet holds 12 seeds: 5 heirloom and 7 hybrid. A gardener plants two seeds drawn at random one after the other, without replacement.

First-stage branches: first seed heirloom ( $\frac{5}{12}$ ) and first seed hybrid ( $\frac{7}{12}$ ). After an heirloom seed is removed, 11 seeds remain and 4 are heirloom, so the second-stage branches are $\frac{4}{11}$ heirloom and $\frac{7}{11}$ hybrid. After a hybrid seed is removed, 11 seeds remain and 5 are heirloom, so the second-stage branches are $\frac{5}{11}$ heirloom and $\frac{6}{11}$ hybrid.

(a) Find the probability both seeds are heirloom. (b) Find the probability exactly one seed is heirloom. (c) Find the probability at least one seed is heirloom.

Show the worked solution

Part (a): multiply along the heirloom, heirloom path, using the reduced second-stage probability. $P(\text{both heirloom}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}$ .
As a decimal, $\frac{20}{132} = \frac{5}{33} \approx 0.1515$ .
Part (b): exactly one heirloom is two paths. Heirloom then hybrid is $\frac{5}{12} \times \frac{7}{11} = \frac{35}{132}$ .
Hybrid then heirloom is $\frac{7}{12} \times \frac{5}{11} = \frac{35}{132}$ . The two match because both paths multiply the same four numbers in a different order.
Add them: $P(\text{exactly one heirloom}) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66} \approx 0.5303$ .
Part (c): the only path with no heirloom is hybrid then hybrid, $\frac{7}{12} \times \frac{6}{11} = \frac{42}{132}$ .
Subtract from 1: $P(\text{at least one heirloom}) = 1 - \frac{42}{132} = \frac{90}{132} = \frac{15}{22} \approx 0.6818$ .
Check all four paths: $\frac{20}{132} + \frac{35}{132} + \frac{35}{132} + \frac{42}{132} = \frac{132}{132} = 1$ .

(a) $\frac{5}{33} \approx 0.1515$ ; (b) $\frac{35}{66} \approx 0.5303$ ; (c) $\frac{15}{22} \approx 0.6818$ .

Problem 4

A juice bottler runs three filling lines. Line A fills 50% of the bottles, line B fills 30%, and line C fills 20%. A bottle comes out underfilled with probability 0.02 on line A, 0.05 on line B, and 0.10 on line C.

The tree has three first-stage branches (0.50, 0.30, 0.20), and each one splits into underfilled and correctly filled.

(a) Find the probability a randomly chosen bottle is underfilled. (b) Given that a bottle is underfilled, find the probability it came from line C. (c) Compare your answer in (b) with line C's share of production.

Show the worked solution

Part (a): find the underfilled path for each line, then add across all three first-stage branches.
Line A: $0.50 \times 0.02 = 0.010$ . Line B: $0.30 \times 0.05 = 0.015$ . Line C: $0.20 \times 0.10 = 0.020$ .
Add the three: $P(\text{underfilled}) = 0.010 + 0.015 + 0.020 = 0.045$ .
As a check, the correctly filled paths are $0.50 \times 0.98 = 0.490$ , $0.30 \times 0.95 = 0.285$ , and $0.20 \times 0.90 = 0.180$ , which sum to $0.955$ , and $0.045 + 0.955 = 1.000$ .
Part (b): run the tree backwards. The numerator is the single line C underfilled path, $0.020$ , and the denominator is every underfilled path, $0.045$ .
$P(\text{line C} \mid \text{underfilled}) = \frac{0.020}{0.045} = \frac{20}{45} = \frac{4}{9} \approx 0.4444$ .
Part (c): line C fills only 20% of bottles but produces about 44.4% of the underfilled ones, because its underfill rate is 5 times line A's and twice line B's.

(a) $P(\text{underfilled}) = 0.045$ ; (b) $P(\text{line C} \mid \text{underfilled}) = \frac{4}{9} \approx 0.4444$ ; (c) line C is 20% of production but about 44.4% of underfilled bottles.

Problem 5

In a coastal town in July, morning fog rolls in on 30% of days. On a foggy morning the whale-watching tour is cancelled with probability 0.60. On a clear morning it is cancelled with probability 0.05.

First-stage branches: foggy (0.30) and clear (0.70). Out of foggy: cancelled (0.60) and runs (0.40). Out of clear: cancelled (0.05) and runs (0.95).

(a) Find the probability the tour is cancelled on a randomly chosen July day. (b) Given that the tour was cancelled, find the probability the morning was foggy. (c) Given that the tour ran, find the probability the morning was foggy.

Show the worked solution

Part (a): multiply along each cancelled path and add. Foggy and cancelled is $0.30 \times 0.60 = 0.180$ ; clear and cancelled is $0.70 \times 0.05 = 0.035$ .
$P(\text{cancelled}) = 0.180 + 0.035 = 0.215$ .
Part (b): run the tree backwards. Divide the foggy cancelled path by the total cancelled probability: $P(\text{foggy} \mid \text{cancelled}) = \frac{0.180}{0.215} \approx 0.8372$ .
Part (c): first get the probability the tour ran. Foggy and ran is $0.30 \times 0.40 = 0.120$ ; clear and ran is $0.70 \times 0.95 = 0.665$ ; together $0.120 + 0.665 = 0.785$ .
That matches $1 - 0.215 = 0.785$ , and all four paths add correctly: $0.180 + 0.120 + 0.035 + 0.665 = 1.000$ .
Now divide: $P(\text{foggy} \mid \text{ran}) = \frac{0.120}{0.785} \approx 0.1529$ .
Read the contrast. Fog appears on 30% of July days overall, on about 83.7% of cancelled days, and on only about 15.3% of days the tour ran.

(a) $P(\text{cancelled}) = 0.215$ ; (b) $P(\text{foggy} \mid \text{cancelled}) = \frac{0.180}{0.215} \approx 0.8372$ ; (c) $P(\text{foggy} \mid \text{ran}) = \frac{0.120}{0.785} \approx 0.1529$ .

Problem 6

A theater props room keeps flashlights in two crates. Crate 1 holds 8 flashlights, 6 of which work. Crate 2 holds 12 flashlights, 3 of which work. A stagehand flips a fair coin to choose a crate, then pulls one flashlight at random from that crate.

First-stage branches: crate 1 ( $0.5$ ) and crate 2 ( $0.5$ ). Out of crate 1: works ( $\frac{6}{8}$ ) and dead ( $\frac{2}{8}$ ). Out of crate 2: works ( $\frac{3}{12}$ ) and dead ( $\frac{9}{12}$ ).

(a) Find the probability the flashlight pulled works. (b) Given that it works, find the probability it came from crate 1. (c) Given that it is dead, find the probability it came from crate 1.

Show the worked solution

Convert the second-stage fractions to decimals: $\frac{6}{8} = 0.75$ , $\frac{2}{8} = 0.25$ , $\frac{3}{12} = 0.25$ , and $\frac{9}{12} = 0.75$ .
Part (a): multiply along both working paths and add. Crate 1 and works is $0.5 \times 0.75 = 0.375$ ; crate 2 and works is $0.5 \times 0.25 = 0.125$ .
$P(\text{works}) = 0.375 + 0.125 = 0.500$ .
Get the dead paths the same way: crate 1 and dead is $0.5 \times 0.25 = 0.125$ , crate 2 and dead is $0.5 \times 0.75 = 0.375$ , so $P(\text{dead}) = 0.125 + 0.375 = 0.500$ .
Check all four paths: $0.375 + 0.125 + 0.125 + 0.375 = 1.000$ .
Part (b): divide the crate 1 working path by the total working probability, $P(\text{crate 1} \mid \text{works}) = \frac{0.375}{0.500} = 0.75$ .
Part (c): divide the crate 1 dead path by the total dead probability, $P(\text{crate 1} \mid \text{dead}) = \frac{0.125}{0.500} = 0.25$ .
The coin made the two crates equally likely before the draw. A working flashlight makes crate 1 three times as likely as crate 2, since $0.75 = 3 \times 0.25$ , and a dead one flips that judgement around.

(a) $P(\text{works}) = 0.500$ ; (b) $P(\text{crate 1} \mid \text{works}) = 0.75$ ; (c) $P(\text{crate 1} \mid \text{dead}) = 0.25$ .

Problem 7

A tree nursery screens seedlings for a root fungus. In the nursery's stock, 1% of seedlings actually carry the fungus. The screen flags 98% of the seedlings that carry it, and it also flags 5% of healthy seedlings.

First-stage branches: carries the fungus (0.01) and healthy (0.99). Out of carries: flagged (0.98) and not flagged (0.02). Out of healthy: flagged (0.05) and not flagged (0.95).

(a) Using a hypothetical 10,000 seedlings, find how many get flagged. (b) Find the probability a seedling gets flagged. (c) Given that a seedling is flagged, find the probability it actually carries the fungus. (d) Explain why the answer to (c) sits so far below 0.98.

Show the worked solution

Turn the branch probabilities into counts out of 10,000 seedlings. Carriers number $0.01 \times 10000 = 100$ , so healthy seedlings number $10000 - 100 = 9900$ .
Split the carriers: $0.98 \times 100 = 98$ are flagged and $100 - 98 = 2$ are missed.
Split the healthy seedlings: $0.05 \times 9900 = 495$ are flagged and $9900 - 495 = 9405$ are not.
Part (a): the flagged seedlings number $98 + 495 = 593$ . Check the whole count: $98 + 2 + 495 + 9405 = 10000$ .
Part (b): $P(\text{flagged}) = \frac{593}{10000} = 0.0593$ .
Part (c): restrict to the 593 flagged seedlings, of which 98 are true carriers, so $P(\text{carrier} \mid \text{flagged}) = \frac{98}{593} \approx 0.1653$ .
Part (d): the 98 true flags are swamped by 495 false ones. Flagging 5% of a group of 9900 healthy seedlings produces far more flags than catching 98% of a group of 100 carriers, so most flags come from healthy stock.
The two numbers describe the same 98 seedlings over different denominators: $P(\text{flagged} \mid \text{carrier}) = \frac{98}{100} = 0.98$ and $P(\text{carrier} \mid \text{flagged}) = \frac{98}{593} \approx 0.1653$ .

(a) 593 of the 10,000 seedlings get flagged; (b) $P(\text{flagged}) = 0.0593$ ; (c) $P(\text{carrier} \mid \text{flagged}) = \frac{98}{593} \approx 0.1653$ ; (d) the 495 false flags from 9900 healthy seedlings outnumber the 98 true flags from 100 carriers.

Problem 8

A tool library rack holds 15 drills: 6 are cordless and 9 are corded. Two borrowers arrive in turn and each takes one drill at random, without replacement.

First-stage branches: first drill cordless ( $\frac{6}{15}$ ) and first drill corded ( $\frac{9}{15}$ ). After a cordless drill leaves, 14 remain and 5 are cordless, so the second-stage branches are $\frac{5}{14}$ cordless and $\frac{9}{14}$ corded. After a corded drill leaves, 14 remain and 6 are cordless, so the second-stage branches are $\frac{6}{14}$ cordless and $\frac{8}{14}$ corded.

(a) Find the probability the second borrower gets a cordless drill. (b) Given that the second borrower got a cordless drill, find the probability the first borrower also got one. (c) Compare your answer to (a) with the probability that the first borrower gets a cordless drill.

Show the worked solution

Part (a): two paths end with a cordless second drill, so compute both and add.
Cordless then cordless: $\frac{6}{15} \times \frac{5}{14} = \frac{30}{210}$ .
Corded then cordless: $\frac{9}{15} \times \frac{6}{14} = \frac{54}{210}$ .
Add them: $P(\text{second cordless}) = \frac{30}{210} + \frac{54}{210} = \frac{84}{210} = \frac{2}{5} = 0.4$ .
Part (b): run the tree backwards. Divide the cordless-then-cordless path by the part (a) total, $P(\text{first cordless} \mid \text{second cordless}) = \frac{30/210}{84/210} = \frac{30}{84} = \frac{5}{14} \approx 0.3571$ .
Part (c): the first borrower gets a cordless drill with probability $\frac{6}{15} = 0.4$ , exactly the part (a) answer.
Before you know anything about the first draw, the second borrower is just as likely to get a cordless drill as the first was. Removing one unknown drill does not shift the second probability, even though it does shift each individual second-stage branch.

(a) $P(\text{second cordless}) = \frac{2}{5} = 0.4$ ; (b) $P(\text{first cordless} \mid \text{second cordless}) = \frac{5}{14} \approx 0.3571$ ; (c) it equals $\frac{6}{15} = 0.4$ , the probability for the first borrower.