8.3 Probability Using Tree Diagrams and Combinations
8.3 Probability Using Tree Diagrams and Combinations
Learning Objectives
By the end of this section, you will be able to:
- Use probability tree diagrams to compute probabilities of sequences of events.
- Use combinations to compute probabilities when order does not matter.
- Apply the complement rule to "at least one" problems.
8.3.1 Tree Diagrams for Sequential Events
When an experiment unfolds in stages — first one card is drawn, then another; first one marble, then a second — a tree diagram organizes every possible path through the stages and the probability along each branch.
Procedure: Reading a Probability Tree
- Each stage of the experiment becomes one level of the tree.
- The probability written on each branch is the conditional probability of that branch given everything earlier on the path.
- To find the probability of a specific path, multiply the branch probabilities along it.
- To find the probability of an event that includes several paths, add the path probabilities.
Context Pause: Why we multiply along a branch
The branch probabilities are conditional on the events earlier in the path. Their product is the joint probability that all those events happen in sequence — the multiplication axiom from Chapter 7. Tree diagrams are just a visual bookkeeping system for the multiplication axiom, with addition over disjoint paths layered on top.
Insight Note: When sampling without replacement, denominators shrink
Drawing two apples from a basket of 8 (5 red, 3 yellow) without replacement: the first draw has 8 in the denominator; the second has 7, because one apple is gone. If you ever forget whether to subtract from the denominator, ask: "Is the apple I just drew still in the basket?" If no, the denominator goes down by one.
Example 8.3.1
Two apples are chosen from a basket containing 5 red and 3 yellow. Find the probability that both apples are red by drawing a tree diagram.
Step 1 — Draw the tree. Stage 1 splits into Red (\(5/8\)) and Yellow (\(3/8\)). Each Stage-1 branch splits again, with denominators reduced by 1 because the first apple is not returned. `` Step 2 — Identify the path "both red". That is the R → R path. Step 3 — Multiply along the path: \[P(\text{both red}) \;=\; \frac{5}{8} \times \frac{4}{7} \;=\; \frac{20}{56} \;=\; \frac{5}{14}.\] Answer: \(5/14\).Solution
Stage 1 Stage 2 ┌── R: 4/7 R 5/8┤ ┌────┴── Y: 3/7 Start ────────┤ └────┬── R: 5/7 Y 3/8┤ └── Y: 2/7 ``
Example 8.3.2
A basket contains 6 red and 4 blue marbles. Three marbles are drawn at random without replacement. Find \(P(\text{all three red})\) using sequential probabilities (no combinations).
Step 1 — First draw. \(6/10\) red. Step 2 — Second draw, given a red on draw 1. \(5/9\) red. Step 3 — Third draw, given two reds. \(4/8\) red. Step 4 — Multiply: \[P(\text{all red}) \;=\; \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} \;=\; \frac{120}{720} \;=\; \frac{1}{6}.\] Answer: \(1/6\).Solution
Try It Now 8.3.1
Two apples are drawn without replacement from a basket of 5 red and 3 yellow apples. Find \(P(\text{one red and one yellow})\).
This event has two paths through the tree: \(R \to Y\) and \(Y \to R\). Add the two products: \[P = \frac{5}{8}\cdot\frac{3}{7} + \frac{3}{8}\cdot\frac{5}{7} = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}.\] Answer: \(15/28\).Solution
8.3.2 Counting with Combinations
When order doesn't matter — picking a committee, a hand of cards, a playlist — counting outcomes with combinations \(C(n, r)\) is faster than enumerating tree paths.
Definition 8.6: Probability via Combinations
If the sample space consists of equally likely unordered selections of \(r\) objects from a population of \(n\), and the event \(E\) corresponds to a set of those selections, then
\[P(E) \;=\; \frac{\text{number of favorable combinations}}{C(n, r)}.\]
The number of favorable combinations is found by multiplying combinations chosen separately from each subgroup.
Example 8.3.3
A jar contains 5 red, 4 white, and 3 blue marbles (12 total). Three marbles are drawn at random. Find \(P(\text{all three red})\).
Step 1 — Total ways to draw 3 from 12: \[C(12, 3) = \frac{12!}{3!\, 9!} = 220.\] Step 2 — Ways to draw 3 reds from 5: \[C(5, 3) = 10.\] Step 3 — Probability: \[P(\text{all red}) \;=\; \frac{C(5,3)}{C(12,3)} \;=\; \frac{10}{220} \;=\; \frac{1}{22}.\] Answer: \(1/22\).Solution
Example 8.3.4
From the same jar (5 red, 4 white, 3 blue), find \(P(\text{2 white and 1 blue})\).
Step 1 — Choose 2 of 4 whites: \(C(4, 2) = 6\). Step 2 — Choose 1 of 3 blues: \(C(3, 1) = 3\). Step 3 — Multiply for favorable count: \(6 \times 3 = 18\). Step 4 — Divide by total: \[P(\text{2 white, 1 blue}) \;=\; \frac{18}{220} \;=\; \frac{9}{110}.\] Answer: \(9/110\).Solution
Try It Now 8.3.2
From the same jar of 5 red, 4 white, and 3 blue marbles, find \(P(\text{none white})\) when 3 marbles are drawn.
"None white" means all 3 are chosen from the 8 non-white marbles: \[P = \frac{C(8, 3)}{C(12, 3)} = \frac{56}{220} = \frac{14}{55}.\] Answer: \(14/55\).Solution
8.3.3 "At Least One" via the Complement
Whenever an event reads at least one or no fewer than \(k\), the complement is usually easier — none or fewer than \(k\). Compute the complement and subtract from 1.
Example 8.3.5
From a jar of 5 red, 4 white, and 3 blue marbles, three are drawn. Find \(P(\text{at least one red})\).
Step 1 — Use the complement. "At least one red" is the complement of "no reds at all". The "no reds" event picks 3 from the 7 non-red marbles: \[P(\text{no reds}) = \frac{C(7, 3)}{C(12, 3)} = \frac{35}{220} = \frac{7}{44}.\] Step 2 — Subtract from 1: \[P(\text{at least one red}) = 1 - \frac{7}{44} = \frac{37}{44}.\] Answer: \(37/44\).Solution
Insight Note: The complement is often a one-liner
Direct counting of "at least one red" requires summing several cases (1 red, 2 reds, 3 reds). The complement collapses all of that into a single combination ratio. When you see "at least one," try the complement first.
8.3.4 The Birthday Problem
The Birthday Problem asks: in a group of \(k\) people, what's the probability that at least two share a birthday? It's a classic illustration of "at least one" via the complement.
Example 8.3.6
If there are 5 people in a room, find \(P(\text{at least 2 share a birthday})\). (Assume 365 equally likely birthdays and ignore leap years.)
Step 1 — Use the complement "all 5 birthdays are distinct". For the second person to differ from the first: \(364/365\). For the third to differ from the first two: \(363/365\). And so on: \[P(\text{all distinct}) = \frac{365}{365}\cdot\frac{364}{365}\cdot\frac{363}{365}\cdot\frac{362}{365}\cdot\frac{361}{365} \approx 0.9729.\] Step 2 — Subtract from 1: \[P(\text{at least 2 share}) = 1 - 0.9729 \approx 0.0271.\] Answer: about \(0.027\), or roughly \(2.7\%\).Solution
Context Pause: Why the answer is "more than you'd guess"
With only 5 people the probability is small (about \(2.7\%\)), but the answer climbs surprisingly fast. With 23 people the probability of a shared birthday already exceeds \(50\%\). The reason is combinatorial: the number of pairs in a group of \(n\) is \(C(n, 2)\), which grows much faster than \(n\) itself. With 23 people there are \(C(23, 2) = 253\) pairs — each one a chance for a match.
Try It Now 8.3.3
If there are 4 people in a room, find \(P(\text{at least two share a birthday})\) (365-day calendar).
\(P(\text{all 4 distinct}) = \dfrac{365 \cdot 364 \cdot 363 \cdot 362}{365^{4}} \approx 0.9836\). \[P(\text{at least two share}) \approx 1 - 0.9836 = 0.0164.\] Answer: about \(1.6\%\).Solution
Problem Set 8.3
Two apples are chosen from a basket containing five red and three yellow apples. Draw a tree diagram and find the following probabilities.
8.3.1 \(P(\text{both red})\)
8.3.2 \(P(\text{one red, one yellow})\)
8.3.3 \(P(\text{both yellow})\)
8.3.4 \(P(\text{first red and second yellow})\)
A basket contains six red and four blue marbles. Three marbles are drawn at random. Find the following probabilities using sequential branch products (do not use combinations).
8.3.5 \(P(\text{all three red})\)
8.3.6 \(P(\text{two red, one blue})\)
8.3.7 \(P(\text{one red, two blue})\)
8.3.8 \(P(\text{first red, second blue, third red})\)
Three marbles are drawn from a jar containing five red, four white, and three blue marbles. Find the following probabilities using combinations.
8.3.9 \(P(\text{all three red})\)
8.3.10 \(P(\text{two white and one blue})\)
8.3.11 \(P(\text{none white})\)
8.3.12 \(P(\text{at least one red})\)
A committee of four is selected from a total of 4 freshmen, 5 sophomores, and 6 juniors. Find the probabilities of the following events.
8.3.13 At least three freshmen.
8.3.14 No sophomores.
8.3.15 All four of the same class.
8.3.16 Not all four from the same class.
8.3.17 Exactly three of the same class.
8.3.18 More juniors than freshmen and sophomores combined.
Five cards are drawn from a standard 52-card deck. Find the probabilities of the following events.
8.3.19 Two hearts, two spades, and one club.
8.3.20 A flush of any suit (all five cards of a single suit).
8.3.21 A full house of nines and tens (three nines and two tens).
8.3.22 Any full house.
8.3.23 A pair of nines and a pair of tens (and the fifth card is not a nine or a ten).
8.3.24 Any two pairs (two cards of one value, two more cards of another value, and the fifth card does not match either pair).
Jorge has 6 rock songs, 7 rap songs, and 4 country songs that he likes for exercising. He randomly selects 6 songs to make a workout playlist. Find the following probabilities.
8.3.25 \(P(\text{playlist has 2 songs of each type})\)
8.3.26 \(P(\text{playlist has no country songs})\)
8.3.27 \(P(\text{playlist has 3 rock, 2 rap, and 1 country song})\)
8.3.28 \(P(\text{playlist has 3 or 4 rock songs and the rest are rap})\)
A project is staffed by 12 people: 5 engineers, 4 salespeople, and 3 customer service representatives. A committee of 5 is selected at random to present to senior management. Find the probabilities of the following events.
8.3.29 The committee has 2 engineers, 2 salespeople, and 1 customer service representative.
8.3.30 The committee contains 3 engineers and 2 salespeople.
8.3.31 The committee has no engineers.
8.3.32 The committee has all salespeople.
Birthday problems (assume 365 equally likely birthdays and ignore leap years).
8.3.33 If there are 5 people in a room, what is the probability that no two share a birthday?
8.3.34 If there are 5 people in a room, find the probability that at least 2 share a birthday.
Step 1 — First red: \(P = 5/8\). Step 2 — Second red, given the first was red: one red apple is gone, so \(P = 4/7\). Step 3 — Multiply: \[P(\text{both red}) = \frac{5}{8}\cdot\frac{4}{7} = \frac{20}{56} = \frac{5}{14}.\] Answer: \(5/14\).Problem 8.3.1 Solution
Step 1 — Two paths give one red and one yellow: R then Y, or Y then R. Step 2 — Path probabilities: \(\frac{5}{8}\cdot\frac{3}{7} = \frac{15}{56}\) and \(\frac{3}{8}\cdot\frac{5}{7} = \frac{15}{56}\). Step 3 — Add the disjoint paths: \[P = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}.\] Answer: \(15/28\).Problem 8.3.2 Solution
Step 1 — Path Y → Y: \(P = \frac{3}{8}\cdot\frac{2}{7} = \frac{6}{56} = \frac{3}{28}\). Answer: \(3/28\).Problem 8.3.3 Solution
Step 1 — Single path R then Y: \(P = \frac{5}{8}\cdot\frac{3}{7} = \frac{15}{56}\). Answer: \(15/56\).Problem 8.3.4 Solution
Step 1 — Sequential branches: \(\frac{6}{10}\cdot\frac{5}{9}\cdot\frac{4}{8}\). Step 2 — Multiply: \[P = \frac{6 \cdot 5 \cdot 4}{10 \cdot 9 \cdot 8} = \frac{120}{720} = \frac{1}{6}.\] Answer: \(1/6\).Problem 8.3.5 Solution
Step 1 — Three orderings give 2 red and 1 blue: RRB, RBR, BRR. Each path uses the same six numerators in some order, so each has the same probability: \[\frac{6}{10}\cdot\frac{5}{9}\cdot\frac{4}{8} = \frac{120}{720}.\] Step 2 — Sum the three disjoint paths: \[P = 3 \cdot \frac{120}{720} = \frac{360}{720} = \frac{1}{2}.\] Answer: \(1/2\).Problem 8.3.6 Solution
Step 1 — Three orderings give 1 red and 2 blue: RBB, BRB, BBR. Each has probability \(\frac{6 \cdot 4 \cdot 3}{10 \cdot 9 \cdot 8} = \frac{72}{720}\). Step 2 — Sum: \[P = 3 \cdot \frac{72}{720} = \frac{216}{720} = \frac{3}{10}.\] Answer: \(3/10\).Problem 8.3.7 Solution
Step 1 — Single ordered path R, B, R: \[P = \frac{6}{10}\cdot\frac{4}{9}\cdot\frac{5}{8} = \frac{120}{720} = \frac{1}{6}.\] Answer: \(1/6\).Problem 8.3.8 Solution
Step 1 — Total ways: \(C(12, 3) = 220\). Step 2 — Favorable: \(C(5, 3) = 10\). Step 3 — Probability: \[P = \frac{10}{220} = \frac{1}{22}.\] Answer: \(1/22\).Problem 8.3.9 Solution
Step 1 — Choose 2 of 4 whites and 1 of 3 blues: \(C(4,2)\cdot C(3,1) = 6 \cdot 3 = 18\). Step 2 — Probability: \[P = \frac{18}{220} = \frac{9}{110}.\] Answer: \(9/110\).Problem 8.3.10 Solution
Step 1 — Pick all 3 from the 8 non-white marbles: \(C(8,3) = 56\). Step 2 — Probability: \[P = \frac{56}{220} = \frac{14}{55}.\] Answer: \(14/55\).Problem 8.3.11 Solution
Step 1 — Use the complement "no reds at all": pick all 3 from the 7 non-red marbles. \(C(7,3) = 35\), so \[P(\text{no reds}) = \frac{35}{220} = \frac{7}{44}.\] Step 2 — Subtract from 1: \[P(\text{at least one red}) = 1 - \frac{7}{44} = \frac{37}{44}.\] Answer: \(37/44\).Problem 8.3.12 Solution
Step 1 — Total ways: \(C(15, 4) = 1365\). Step 2 — Favorable. "At least three freshmen" = (3 freshmen + 1 other) + (4 freshmen): \[C(4,3)\cdot C(11,1) + C(4,4) = 4 \cdot 11 + 1 = 45.\] Step 3 — Probability: \[P = \frac{45}{1365} = \frac{3}{91}.\] Answer: \(3/91\).Problem 8.3.13 Solution
Step 1 — Pick 4 from non-sophomores (10 of them): \(C(10, 4) = 210\). Step 2 — Probability: \[P = \frac{210}{1365} = \frac{2}{13}.\] Answer: \(2/13\).Problem 8.3.14 Solution
Step 1 — Sum the all-same-class counts: \(C(4,4) + C(5,4) + C(6,4) = 1 + 5 + 15 = 21\). Step 2 — Probability: \[P = \frac{21}{1365} = \frac{1}{65}.\] Answer: \(1/65\).Problem 8.3.15 Solution
Step 1 — Use the complement of 8.3.15: \[P = 1 - \frac{1}{65} = \frac{64}{65}.\] Answer: \(64/65\).Problem 8.3.16 Solution
Step 1 — For each class, count "exactly 3 from that class + 1 from the other 11": Step 2 — Sum: \(44 + 100 + 180 = 324\). Step 3 — Probability: \[P = \frac{324}{1365} = \frac{108}{455}.\] Answer: \(108/455\).Problem 8.3.17 Solution
Step 1 — Translate the condition. Let \(J\) be the number of juniors. We need \(J > 4 - J\), so \(J > 2\), i.e. \(J \in \{3, 4\}\). Step 2 — Count. Total = \(195\). Step 3 — Probability: \[P = \frac{195}{1365} = \frac{1}{7}.\] Answer: \(1/7\).Problem 8.3.18 Solution
Step 1 — Total 5-card hands: \(C(52, 5) = 2{,}598{,}960\). Step 2 — Favorable: \(C(13, 2)\cdot C(13, 2)\cdot C(13, 1) = 78 \cdot 78 \cdot 13 = 79{,}092\). Step 3 — Probability: \[P = \frac{79{,}092}{2{,}598{,}960} \approx 0.0304.\] Answer: \(\dfrac{79{,}092}{2{,}598{,}960} \approx 3.04\%\).Problem 8.3.19 Solution
Step 1 — Choose suit, then 5 cards of that suit: \(4 \cdot C(13, 5) = 4 \cdot 1287 = 5148\). Step 2 — Probability (includes straight flushes): \[P = \frac{5148}{2{,}598{,}960} \approx 0.00198.\] Answer: about \(0.198\%\) (≈ 1 in 505).Problem 8.3.20 Solution
Step 1 — Choose 3 of the 4 nines and 2 of the 4 tens: \(C(4,3)\cdot C(4,2) = 4 \cdot 6 = 24\). Step 2 — Probability: \[P = \frac{24}{2{,}598{,}960} = \frac{1}{108{,}290}.\] Answer: \(\dfrac{1}{108{,}290}\).Problem 8.3.21 Solution
Step 1 — Pick the rank for the triple (13), then 3 of its 4 cards (\(C(4,3)\)). Pick the rank for the pair from the remaining 12, then 2 of its 4 cards (\(C(4,2)\)): \[13 \cdot 4 \cdot 12 \cdot 6 = 3744.\] Step 2 — Probability: \[P = \frac{3744}{2{,}598{,}960} \approx 0.00144.\] Answer: about \(0.144\%\) (≈ 1 in 694).Problem 8.3.22 Solution
Step 1 — Two of 4 nines, two of 4 tens, fifth card from the 44 non-nines/non-tens: \[C(4,2)\cdot C(4,2)\cdot 44 = 6 \cdot 6 \cdot 44 = 1584.\] Step 2 — Probability: \[P = \frac{1584}{2{,}598{,}960} \approx 0.000610.\] Answer: about \(0.061\%\) (≈ 1 in 1640).Problem 8.3.23 Solution
Step 1 — Pick the two pair ranks: \(C(13, 2) = 78\). From each, pick 2 of 4 cards: \(C(4,2)^{2} = 36\). Pick the fifth card from the remaining \(11 \cdot 4 = 44\) cards. Step 2 — Multiply: \[78 \cdot 36 \cdot 44 = 123{,}552.\] Step 3 — Probability: \[P = \frac{123{,}552}{2{,}598{,}960} \approx 0.0475.\] Answer: about \(4.75\%\).Problem 8.3.24 Solution
Step 1 — Total playlists: \(C(17, 6) = 12{,}376\). Step 2 — Favorable: 2 of 6 rock, 2 of 7 rap, 2 of 4 country: \[C(6,2)\cdot C(7,2)\cdot C(4,2) = 15 \cdot 21 \cdot 6 = 1890.\] Step 3 — Probability: \[P = \frac{1890}{12{,}376} = \frac{135}{884}.\] Answer: \(135/884 \approx 0.153\).Problem 8.3.25 Solution
Step 1 — Pick 6 from the 13 non-country songs: \(C(13, 6) = 1716\). Step 2 — Probability: \[P = \frac{1716}{12{,}376} = \frac{33}{238}.\] Answer: \(33/238 \approx 0.139\).Problem 8.3.26 Solution
Step 1 — Favorable: \(C(6,3)\cdot C(7,2)\cdot C(4,1) = 20 \cdot 21 \cdot 4 = 1680\). Step 2 — Probability: \[P = \frac{1680}{12{,}376} = \frac{30}{221}.\] Answer: \(30/221 \approx 0.136\).Problem 8.3.27 Solution
Step 1 — Two cases (no country songs): Step 2 — Sum: \(700 + 315 = 1015\). Step 3 — Probability: \[P = \frac{1015}{12{,}376} = \frac{145}{1768}.\] Answer: \(145/1768 \approx 0.082\).Problem 8.3.28 Solution
Step 1 — Total committees: \(C(12, 5) = 792\). Step 2 — Favorable: \(C(5,2)\cdot C(4,2)\cdot C(3,1) = 10 \cdot 6 \cdot 3 = 180\). Step 3 — Probability: \[P = \frac{180}{792} = \frac{5}{22}.\] Answer: \(5/22\).Problem 8.3.29 Solution
Step 1 — Favorable: \(C(5,3)\cdot C(4,2) = 10 \cdot 6 = 60\). Step 2 — Probability: \[P = \frac{60}{792} = \frac{5}{66}.\] Answer: \(5/66\).Problem 8.3.30 Solution
Step 1 — Pick all 5 from the 7 non-engineers: \(C(7, 5) = 21\). Step 2 — Probability: \[P = \frac{21}{792} = \frac{7}{264}.\] Answer: \(7/264\).Problem 8.3.31 Solution
Step 1 — Check feasibility: the committee needs 5 salespeople, but there are only 4. Step 2 — Conclude: \[P(\text{all salespeople}) = 0.\] Answer: \(0\).Problem 8.3.32 Solution
Step 1 — Sequential distinct birthdays: \[P(\text{all distinct}) = \frac{365 \cdot 364 \cdot 363 \cdot 362 \cdot 361}{365^{5}}.\] Step 2 — Evaluate: the product equals approximately \(0.9729\). Answer: approximately \(0.9729\) (about \(97.3\%\)).Problem 8.3.33 Solution
Step 1 — Use the complement of 8.3.33: \[P(\text{at least 2 share}) = 1 - 0.9729 \approx 0.0271.\] Answer: about \(0.0271\) (about \(2.7\%\)).Problem 8.3.34 Solution