8.1 Sample Spaces and Probability
8.1 Sample Spaces and Probability
Learning Objectives
By the end of this section, you will be able to:
- Write the sample space for a chance experiment.
- Compute probabilities by counting equally likely outcomes.
8.1.1 Sample Spaces
Probability begins with chance experiments — actions whose outcomes we cannot predict in advance, such as tossing a coin, rolling a die, or drawing a card. Before we can talk about how likely something is, we have to be precise about everything that could happen. That complete list is called the sample space.
Definition 8.1: Sample Space
The sample space of a chance experiment is the set of all distinct possible outcomes. We usually denote it by \(S\). Each individual outcome is called a simple event or sample point.
For example, when one die is rolled the sample space is
\[S = \{1, 2, 3, 4, 5, 6\}.\]
When a single coin is tossed the sample space is \(S = \{H, T\}\).
Context Pause: Why list the outcomes one by one?
Many probability mistakes come from miscounting the sample space. A famous example: if two coins are tossed, students sometimes claim the outcomes are "no heads, one head, two heads," giving three possibilities. But these three results are not equally likely, so dividing by 3 leads to a wrong answer. Listing the underlying outcomes — \(HH, HT, TH, TT\) — makes the equally likely structure visible.
Insight Note: Distinguish the coins (or dice) even when they look identical
Pretend one coin is a penny and the other is a nickel, even when both are quarters. The outcome "head on penny, tail on nickel" is genuinely different from "tail on penny, head on nickel" — they happen on different coins. Treating the coins as distinguishable doesn't change the physical experiment; it just makes our counting honest.
Example 8.1.1
Two coins — a penny and a nickel — are tossed. Write the sample space.
For each coin we record \(H\) or \(T\). Listing the penny's outcome first, the four equally likely outcomes are \[S = \{HH,\; HT,\; TH,\; TT\}.\] Here \(HT\) means a head on the penny and a tail on the nickel, while \(TH\) means a tail on the penny and a head on the nickel. Answer: \(S = \{HH, HT, TH, TT\}\), which has \(4\) outcomes.Solution
Example 8.1.2
A die is rolled and a coin is tossed. Write the sample space.
Pair each die face \(\{1, 2, 3, 4, 5, 6\}\) with each coin face \(\{H, T\}\). That gives \(6 \times 2 = 12\) ordered pairs: \[S = \{1H, 2H, 3H, 4H, 5H, 6H,\; 1T, 2T, 3T, 4T, 5T, 6T\}.\] Answer: A 12-element sample space.Solution
Try It Now 8.1.1
Three coins are tossed. Write the sample space.
For each of the three coins we record \(H\) or \(T\). Order the coins (first, second, third) and list the \(2 \cdot 2 \cdot 2 = 8\) outcomes: \[S = \{HHH,\; HHT,\; HTH,\; HTT,\; THH,\; THT,\; TTH,\; TTT\}.\]Solution
8.1.2 Events and Their Probability
A sample space lists every possible outcome. Often we care about a particular collection of outcomes — for instance, "the coin shows at least one head," or "the die shows an even number." Such a collection is called an event.
Definition 8.2: Event
An event \(E\) is any subset of the sample space \(S\). When we perform the experiment we say the event \(E\) occurred if the actual outcome is one of the elements of \(E\).
For example, in the two-coin experiment with \(S = \{HH, HT, TH, TT\}\), the event "exactly one head" is
\[E = \{HT, TH\}.\]
Definition 8.3: Probability of an Event (Equally Likely Outcomes)
If a sample space \(S\) has \(n(S)\) equally likely outcomes and an event \(E\) contains \(n(E)\) of those outcomes, the probability of \(E\) is
\[P(E) \;=\; \frac{n(E)}{n(S)} \;=\; \frac{\text{number of outcomes in } E}{\text{total number of outcomes in } S}.\]
Probabilities always satisfy \(0 \le P(E) \le 1\). The probability of the empty event \(\varnothing\) is \(0\) (it cannot happen) and the probability of the full sample space \(S\) is \(1\) (something must happen).
Context Pause: "Equally likely" is a modeling assumption
Definition 8.3 assumes every outcome in \(S\) is equally likely. That assumption is exactly right for fair coins, well-balanced dice, and well-shuffled cards — and it is the standard set-up for an introductory chapter. When outcomes are not equally likely (a loaded die, a biased coin, a card guessed from someone's poker face), we need a more general definition that we will encounter later. Always check the equally-likely assumption before using \(P(E) = n(E)/n(S)\).
Insight Note: Probability as proportion
Because every outcome carries equal weight, \(P(E)\) is just the fraction of the sample space occupied by \(E\). If \(E\) covers half the sample space, \(P(E) = 1/2\); if it covers all of it, \(P(E) = 1\). Thinking of probability as "share of the sample space" makes many calculations feel natural — you're just measuring sizes.
Example 8.1.3
Two coins are tossed. Find the probability of getting exactly one head.
The sample space \(S = \{HH, HT, TH, TT\}\) has \(n(S) = 4\) equally likely outcomes. The event \(E\) of exactly one head is \(E = \{HT, TH\}\), so \(n(E) = 2\). \[P(E) \;=\; \frac{n(E)}{n(S)} \;=\; \frac{2}{4} \;=\; \frac{1}{2}.\] Answer: \(P(\text{exactly one head}) = 1/2\).Solution
Example 8.1.4
A single card is drawn from a standard 52-card deck. Find:
a) \(P(\text{the card is a king})\)
b) \(P(\text{the card is a heart})\)
c) \(P(\text{the card is a face card})\) — a face card is a jack, queen, or king.
The sample space contains \(n(S) = 52\) equally likely cards. a) King. There are 4 kings (one in each suit), so \[P(\text{king}) \;=\; \frac{4}{52} \;=\; \frac{1}{13}.\] b) Heart. Each of the four suits has 13 cards, so there are 13 hearts: \[P(\text{heart}) \;=\; \frac{13}{52} \;=\; \frac{1}{4}.\] c) Face card. Each suit has 3 face cards (jack, queen, king), giving \(3 \times 4 = 12\) face cards: \[P(\text{face card}) \;=\; \frac{12}{52} \;=\; \frac{3}{13}.\]Solution
Try It Now 8.1.2
Two coins are tossed. Find the probability of getting at least one head.
\(S = \{HH, HT, TH, TT\}\), so \(n(S) = 4\). The event "at least one head" excludes only \(TT\), so \[E = \{HH, HT, TH\}, \qquad n(E) = 3.\] \[P(\text{at least one head}) \;=\; \frac{3}{4}.\]Solution
8.1.3 Computing Probabilities by Counting
When the sample space is small, we can list it. When it grows — pairs of dice, hands of cards, families of three children — we count outcomes systematically using the multiplication rule we will sharpen in later chapters. For now, the strategy is the same:
Procedure: Computing a Probability with Equally Likely Outcomes
- Identify the chance experiment and its sample space \(S\).
- Verify that the listed outcomes are equally likely. (If not, this section's formula does not apply.)
- Identify the event \(E\) — the subset of outcomes you care about.
- Count \(n(S)\) and \(n(E)\).
- Compute \(P(E) = n(E)/n(S)\).
Example 8.1.5
Two dice are rolled. Find the probability that the sum on the two dice is 7.
Step 1 — Sample space. Treat the dice as distinguishable. Each die has \(6\) faces, so \[n(S) = 6 \times 6 = 36.\] The sample space is the set of ordered pairs \((a, b)\) with \(a, b \in \{1, 2, 3, 4, 5, 6\}\). Step 2 — Event. The pairs that sum to \(7\) are \[E = \{(1,6),\, (2,5),\, (3,4),\, (4,3),\, (5,2),\, (6,1)\}, \qquad n(E) = 6.\] Step 3 — Probability. \[P(\text{sum is 7}) \;=\; \frac{6}{36} \;=\; \frac{1}{6}.\] Answer: \(1/6\).Solution
Example 8.1.6
Consider a family of three children. Find the probability that the family has exactly two boys and one girl. (Assume each birth is independently a boy or girl with equal likelihood.)
Step 1 — Sample space. List the eight equally likely birth-order outcomes (oldest to youngest): \[S = \{BBB,\, BBG,\, BGB,\, BGG,\, GBB,\, GBG,\, GGB,\, GGG\}, \qquad n(S) = 8.\] Step 2 — Event. "Exactly two boys and one girl" matches \(BBG, BGB, GBB\), so \(n(E) = 3\). Step 3 — Probability. \[P(\text{two boys and one girl}) \;=\; \frac{3}{8}.\]Solution
Try It Now 8.1.3
A jar contains 6 red, 7 white, and 7 blue marbles. One marble is chosen at random. Find \(P(\text{red})\).
The total number of marbles is \(6 + 7 + 7 = 20\), so \(n(S) = 20\). The red event has \(n(E) = 6\): \[P(\text{red}) \;=\; \frac{6}{20} \;=\; \frac{3}{10}.\]Solution
Problem Set 8.1
In problems 8.1.1–8.1.6, write a sample space for the given experiment.
8.1.1 A die is rolled.
8.1.2 A penny and a nickel are tossed.
8.1.3 A die is rolled, and a coin is tossed.
8.1.4 Three coins are tossed.
8.1.5 Two dice are rolled.
8.1.6 A jar contains four marbles numbered 1, 2, 3, and 4. Two marbles are drawn.
In problems 8.1.7–8.1.12, one card is randomly selected from a standard 52-card deck. Find the following probabilities.
8.1.7 \(P(\text{an ace})\)
8.1.8 \(P(\text{a red card})\)
8.1.9 \(P(\text{a club})\)
8.1.10 \(P(\text{a face card})\)
8.1.11 \(P(\text{a jack or a spade})\)
8.1.12 \(P(\text{a jack and a spade})\)
For problems 8.1.13–8.1.16: A jar contains 6 red, 7 white, and 7 blue marbles. If one marble is chosen at random, find the following probabilities.
8.1.13 \(P(\text{red})\)
8.1.14 \(P(\text{white})\)
8.1.15 \(P(\text{red or blue})\)
8.1.16 \(P(\text{red and blue})\)
For problems 8.1.17–8.1.22: Consider a family of three children. Find the following probabilities.
8.1.17 \(P(\text{two boys and a girl})\)
8.1.18 \(P(\text{at least one boy})\)
8.1.19 \(P(\text{children of both sexes})\)
8.1.20 \(P(\text{at most one girl})\)
8.1.21 \(P(\text{first and third children are male})\)
8.1.22 \(P(\text{all children are of the same gender})\)
For problems 8.1.23–8.1.27: Two dice are rolled. Find the following probabilities.
8.1.23 \(P(\text{the sum of the dice is 5})\)
8.1.24 \(P(\text{the sum of the dice is 8})\)
8.1.25 \(P(\text{the sum is 3 or 6})\)
8.1.26 \(P(\text{the sum is more than 10})\)
8.1.27 \(P(\text{the result is a double})\). Hint: a double means both dice show the same value.
For problems 8.1.28–8.1.31: A jar contains four marbles numbered 1, 2, 3, and 4. Two marbles are drawn randomly without replacement — after a marble is drawn it is not returned to the jar before the second is selected. Find the following probabilities.
8.1.28 \(P(\text{the sum of the numbers is 5})\)
8.1.29 \(P(\text{the sum of the numbers is odd})\)
8.1.30 \(P(\text{the sum of the numbers is 9})\)
8.1.31 \(P(\text{one of the numbers is 3})\)
For problems 8.1.32–8.1.33: A jar contains four marbles numbered 1, 2, 3, and 4. Two marbles are drawn randomly with replacement — after a marble is drawn it is returned to the jar before the second is selected. Find the following probabilities.
8.1.32 \(P(\text{the sum of the numbers is 5})\)
8.1.33 \(P(\text{the sum of the numbers is 2})\)
Step 1 — Identify all possible outcomes: A standard die has six faces showing the numbers 1 through 6. Step 2 — Write the sample space: \[S = \{1, 2, 3, 4, 5, 6\}.\] Answer: \(S = \{1, 2, 3, 4, 5, 6\}\) with \(n(S) = 6\).Problem 8.1.1 Solution
Step 1 — Identify each coin's outcomes: The penny shows H or T; the nickel shows H or T. Step 2 — Form the ordered pairs (penny first, nickel second): \[S = \{HH,\; HT,\; TH,\; TT\}.\] Answer: \(S = \{HH, HT, TH, TT\}\) with \(n(S) = 4\).Problem 8.1.2 Solution
Step 1 — Pair die faces with coin faces: The die has 6 outcomes and the coin has 2, giving \(6 \times 2 = 12\) outcomes. Step 2 — List the sample space: \[S = \{1H, 2H, 3H, 4H, 5H, 6H,\; 1T, 2T, 3T, 4T, 5T, 6T\}.\] Answer: A 12-element sample space.Problem 8.1.3 Solution
Step 1 — Each coin contributes two outcomes: With three coins there are \(2 \times 2 \times 2 = 8\) ordered outcomes. Step 2 — List them: \[S = \{HHH,\; HHT,\; HTH,\; HTT,\; THH,\; THT,\; TTH,\; TTT\}.\] Answer: \(n(S) = 8\).Problem 8.1.4 Solution
Step 1 — Treat the dice as distinguishable: List ordered pairs \((a, b)\) with \(a, b \in \{1, 2, 3, 4, 5, 6\}\). There are \(6 \times 6 = 36\) outcomes. Step 2 — Describe the sample space: \[S = \{(a, b) : a, b \in \{1, 2, 3, 4, 5, 6\}\},\] i.e. all 36 ordered pairs from \((1,1)\) through \((6,6)\). Answer: \(n(S) = 36\).Problem 8.1.5 Solution
Step 1 — Two distinct marbles drawn in order: The first draw has 4 possibilities and the second has 3 (the first marble is not returned), giving \(4 \times 3 = 12\) ordered outcomes. Step 2 — List the sample space: \[S = \{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3)\}.\] Answer: \(n(S) = 12\).Problem 8.1.6 Solution
Step 1 — Count favorable outcomes: A standard deck has 4 aces (one per suit), so \(n(E) = 4\). Step 2 — Apply the probability formula: \[P(\text{ace}) \;=\; \frac{4}{52} \;=\; \frac{1}{13}.\] Answer: \(1/13\).Problem 8.1.7 Solution
Step 1 — Count red cards: Hearts and diamonds give \(13 + 13 = 26\) red cards. Step 2 — Compute the probability: \[P(\text{red}) \;=\; \frac{26}{52} \;=\; \frac{1}{2}.\] Answer: \(1/2\).Problem 8.1.8 Solution
Step 1 — Count clubs: There are 13 clubs in the deck. Step 2 — Compute the probability: \[P(\text{club}) \;=\; \frac{13}{52} \;=\; \frac{1}{4}.\] Answer: \(1/4\).Problem 8.1.9 Solution
Step 1 — Count face cards: Each suit contributes 3 face cards (jack, queen, king): \(3 \times 4 = 12\). Step 2 — Compute the probability: \[P(\text{face card}) \;=\; \frac{12}{52} \;=\; \frac{3}{13}.\] Answer: \(3/13\).Problem 8.1.10 Solution
Step 1 — Use the addition principle for "or": There are 4 jacks and 13 spades, but the jack of spades is counted in both groups. Subtract the overlap once: \[n(\text{jack or spade}) = 4 + 13 - 1 = 16.\] Step 2 — Compute the probability: \[P(\text{jack or spade}) \;=\; \frac{16}{52} \;=\; \frac{4}{13}.\] Answer: \(4/13\).Problem 8.1.11 Solution
Step 1 — Identify the overlap: Only one card is both a jack and a spade — the jack of spades. Step 2 — Compute the probability: \[P(\text{jack and spade}) \;=\; \frac{1}{52}.\] Answer: \(1/52\).Problem 8.1.12 Solution
Step 1 — Total marbles: \(6 + 7 + 7 = 20\), so \(n(S) = 20\). Step 2 — Probability of red: \[P(\text{red}) \;=\; \frac{6}{20} \;=\; \frac{3}{10}.\] Answer: \(3/10\).Problem 8.1.13 Solution
Step 1 — Count white marbles: There are \(7\) white marbles out of \(20\). Step 2 — Compute the probability: \[P(\text{white}) \;=\; \frac{7}{20}.\] Answer: \(7/20\).Problem 8.1.14 Solution
Step 1 — Combine the two disjoint events: Red and blue marbles never overlap (each marble is one color), so add: \[n(\text{red or blue}) = 6 + 7 = 13.\] Step 2 — Compute the probability: \[P(\text{red or blue}) \;=\; \frac{13}{20}.\] Answer: \(13/20\).Problem 8.1.15 Solution
Step 1 — Check for overlap: A single marble cannot be both red and blue. There are no outcomes in this event. Step 2 — Compute the probability: \[P(\text{red and blue}) \;=\; \frac{0}{20} \;=\; 0.\] Answer: \(0\).Problem 8.1.16 Solution
Step 1 — Sample space for three children: \[S = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG\}, \quad n(S) = 8.\] Step 2 — Identify the event: "Two boys and a girl" matches \(BBG, BGB, GBB\), so \(n(E) = 3\). Step 3 — Compute: \[P(\text{two boys, one girl}) \;=\; \frac{3}{8}.\] Answer: \(3/8\).Problem 8.1.17 Solution
Step 1 — Use the complement: "At least one boy" excludes only \(GGG\). So \(n(E) = 8 - 1 = 7\). Step 2 — Compute: \[P(\text{at least one boy}) \;=\; \frac{7}{8}.\] Answer: \(7/8\).Problem 8.1.18 Solution
Step 1 — Identify the complement: The event "children of both sexes" is everything except \(BBB\) and \(GGG\). Thus \(n(E) = 8 - 2 = 6\). Step 2 — Compute: \[P(\text{both sexes}) \;=\; \frac{6}{8} \;=\; \frac{3}{4}.\] Answer: \(3/4\).Problem 8.1.19 Solution
Step 1 — Enumerate "at most one girl": Either 0 girls (\(BBB\)) or exactly 1 girl (\(BBG, BGB, GBB\)). Total: \(n(E) = 4\). Step 2 — Compute: \[P(\text{at most one girl}) \;=\; \frac{4}{8} \;=\; \frac{1}{2}.\] Answer: \(1/2\).Problem 8.1.20 Solution
Step 1 — Fix the first and third positions to \(B\): The second child is free. Outcomes: \(BBB, BGB\). So \(n(E) = 2\). Step 2 — Compute: \[P(\text{first and third are male}) \;=\; \frac{2}{8} \;=\; \frac{1}{4}.\] Answer: \(1/4\).Problem 8.1.21 Solution
Step 1 — Identify same-gender outcomes: Only \(BBB\) and \(GGG\). So \(n(E) = 2\). Step 2 — Compute: \[P(\text{all same gender}) \;=\; \frac{2}{8} \;=\; \frac{1}{4}.\] Answer: \(1/4\).Problem 8.1.22 Solution
Step 1 — Sample space size: Two dice give \(n(S) = 36\). Step 2 — Pairs summing to 5: \((1,4), (2,3), (3,2), (4,1)\), so \(n(E) = 4\). Step 3 — Compute: \[P(\text{sum} = 5) \;=\; \frac{4}{36} \;=\; \frac{1}{9}.\] Answer: \(1/9\).Problem 8.1.23 Solution
Step 1 — Pairs summing to 8: \((2,6), (3,5), (4,4), (5,3), (6,2)\), so \(n(E) = 5\). Step 2 — Compute: \[P(\text{sum} = 8) \;=\; \frac{5}{36}.\] Answer: \(5/36\).Problem 8.1.24 Solution
Step 1 — Pairs summing to 3: \((1,2),(2,1)\), 2 outcomes. Step 2 — Pairs summing to 6: \((1,5),(2,4),(3,3),(4,2),(5,1)\), 5 outcomes. Step 3 — Add (the events are disjoint) and compute: \[P(\text{sum} = 3 \text{ or } 6) \;=\; \frac{2 + 5}{36} \;=\; \frac{7}{36}.\] Answer: \(7/36\).Problem 8.1.25 Solution
Step 1 — Pairs with sum > 10: Sum 11: \((5,6),(6,5)\); sum 12: \((6,6)\). Total \(n(E) = 3\). Step 2 — Compute: \[P(\text{sum} > 10) \;=\; \frac{3}{36} \;=\; \frac{1}{12}.\] Answer: \(1/12\).Problem 8.1.26 Solution
Step 1 — Identify doubles: \((1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\), so \(n(E) = 6\). Step 2 — Compute: \[P(\text{double}) \;=\; \frac{6}{36} \;=\; \frac{1}{6}.\] Answer: \(1/6\).Problem 8.1.27 Solution
Step 1 — Sample space without replacement: Two distinct ordered marbles from \(\{1,2,3,4\}\): \(n(S) = 4 \times 3 = 12\). Step 2 — Pairs summing to 5: \((1,4),(2,3),(3,2),(4,1)\), so \(n(E) = 4\). Step 3 — Compute: \[P(\text{sum} = 5) \;=\; \frac{4}{12} \;=\; \frac{1}{3}.\] Answer: \(1/3\).Problem 8.1.28 Solution
Step 1 — Enumerate ordered pairs with odd sum: A sum is odd when one marble is odd and the other is even. Listing all 12 outcomes and their sums: | Pair | Sum | |------|-----| | (1,2) | 3 (odd) | | (1,3) | 4 | | (1,4) | 5 (odd) | | (2,1) | 3 (odd) | | (2,3) | 5 (odd) | | (2,4) | 6 | | (3,1) | 4 | | (3,2) | 5 (odd) | | (3,4) | 7 (odd) | | (4,1) | 5 (odd) | | (4,2) | 6 | | (4,3) | 7 (odd) | That gives \(n(E) = 8\) odd-sum pairs. Step 2 — Compute: \[P(\text{sum is odd}) \;=\; \frac{8}{12} \;=\; \frac{2}{3}.\] Answer: \(2/3\).Problem 8.1.29 Solution
Step 1 — Maximum possible sum: Without replacement, the largest sum is \(3 + 4 = 7\). A sum of 9 is impossible. Step 2 — Compute: \[P(\text{sum} = 9) \;=\; \frac{0}{12} \;=\; 0.\] Answer: \(0\).Problem 8.1.30 Solution
Step 1 — Pairs containing a 3: \((1,3),(2,3),(3,1),(3,2),(3,4),(4,3)\), so \(n(E) = 6\). Step 2 — Compute: \[P(\text{one of the numbers is 3}) \;=\; \frac{6}{12} \;=\; \frac{1}{2}.\] Answer: \(1/2\).Problem 8.1.31 Solution
Step 1 — Sample space with replacement: \(n(S) = 4 \times 4 = 16\) ordered pairs. Step 2 — Pairs summing to 5: \((1,4),(2,3),(3,2),(4,1)\), so \(n(E) = 4\). Step 3 — Compute: \[P(\text{sum} = 5) \;=\; \frac{4}{16} \;=\; \frac{1}{4}.\] Answer: \(1/4\).Problem 8.1.32 Solution
Step 1 — Pairs summing to 2: With replacement only \((1,1)\) gives sum 2, so \(n(E) = 1\). Step 2 — Compute: \[P(\text{sum} = 2) \;=\; \frac{1}{16}.\] Answer: \(1/16\).Problem 8.1.33 Solution