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.
Source: Applied Finite Mathematics
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 \(S = \{H, T\}\).
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.
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.
Source: Applied Finite Mathematics
Two coins — a penny and a nickel — are tossed. Write the sample space.
Solution
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.
Source: Applied Finite Mathematics
A die is rolled and a coin is tossed. Write the sample space.
Solution
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.
Source: Applied Finite Mathematics
Three coins are tossed. Write the sample space.
Solution
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:
$$S = \{HHH,\; HHT,\; HTH,\; HTT,\; THH,\; THT,\; TTH,\; TTT\}.$$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.
Source: Applied Finite Mathematics
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
$$E = \{HT, TH\}.$$Source: Applied Finite Mathematics
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).
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 \(P(E) = n(E)/n(S)\).
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.
Source: Applied Finite Mathematics
Two coins are tossed. Find the probability of getting exactly one head.
Solution
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: \(P(\text{exactly one head}) = 1/2\).
Source: Applied Finite Mathematics
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.
Solution
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:
$$P(\text{face card}) \;=\; \frac{12}{52} \;=\; \frac{3}{13}.$$Source: Applied Finite Mathematics
Two coins are tossed. Find the probability of getting at least one head.
Solution
\(S = \{HH, HT, TH, TT\}\), so \(n(S) = 4\).
The event "at least one head" excludes only \(TT\), so
$$E = \{HH, HT, TH\}, \qquad n(E) = 3.$$ $$P(\text{at least one head}) \;=\; \frac{3}{4}.$$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 \(P(E) = n(E)/n(S)\).
Source: Applied Finite Mathematics
Two dice are rolled. Find the probability that the sum on the two dice is 7.
Solution
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: \(1/6\).
Source: Applied Finite Mathematics
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.)
Solution
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.
$$P(\text{two boys and one girl}) \;=\; \frac{3}{8}.$$Source: Applied Finite Mathematics
A jar contains 6 red, 7 white, and 7 blue marbles. One marble is chosen at random. Find \(P(\text{red})\).
Solution
The total number of marbles is \(6 + 7 + 7 = 20\), so \(n(S) = 20\). The red event has \(n(E) = 6\):
$$P(\text{red}) \;=\; \frac{6}{20} \;=\; \frac{3}{10}.$$Problem Set 8.1
Source: Applied Finite Mathematics
A card is drawn from a standard deck. Determine whether the events \(E\) and \(F\) below are mutually exclusive (i.e., whether \(E \cap F = \varnothing\)).
\(E = \{\text{the card is an ace}\}\), \(F = \{\text{the card is a heart}\}\).
Solution
The Ace of Hearts belongs to both events, so
$$E \cap F = \{\text{Ace of Hearts}\} \neq \varnothing.$$Answer: \(E\) and \(F\) are not mutually exclusive.
Source: Applied Finite Mathematics
Two dice are rolled. Determine whether the events below are mutually exclusive.
\(G = \{\text{the sum of the faces is 6}\}\), \(H = \{\text{one die shows a 4}\}\).
Solution
List both events as sets of ordered pairs:
$$G = \{(1,5),(2,4),(3,3),(4,2),(5,1)\}, \qquad H = \{(2,4),(4,2)\, \text{and any other pair with a 4}\}.$$Restricting \(H\) to pairs that intersect \(G\):
$$G \cap H = \{(2,4),(4,2)\} \neq \varnothing.$$Answer: \(G\) and \(H\) are not mutually exclusive.
Source: Applied Finite Mathematics
A family has three children. Determine whether the following events are mutually exclusive.
\(M = \{\text{the family has at least one boy}\}\), \(N = \{\text{the family has all girls}\}\).
Solution
Listing the events:
$$M = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB\}, \qquad N = \{GGG\}.$$These are disjoint sets:
$$M \cap N = \varnothing.$$Answer: \(M\) and \(N\) are mutually exclusive.
Source: Applied Finite Mathematics
A single die is rolled. Are the events \(E = \{\text{a multiple of 3 shows}\}\) and \(F = \{\text{a 2 shows}\}\) mutually exclusive?
Solution
\(E = \{3, 6\}\) and \(F = \{2\}\). Their intersection is empty, so
$$E \cap F = \varnothing.$$Answer: Yes, \(E\) and \(F\) are mutually exclusive.
Source: Applied Finite Mathematics
A die is rolled. Find the probability that the result is an even number or a number greater than four.
Solution
Step 1 — Identify the sample space and events.
$$S = \{1,2,3,4,5,6\}, \quad E = \{2, 4, 6\}, \quad F = \{5, 6\}.$$Step 2 — Naive addition is wrong.
\(P(E) = 3/6\) and \(P(F) = 2/6\). Adding gives \(5/6\), but that double-counts the outcome \(6\), which lies in both events.
Step 3 — Use \(E \cup F\) directly.
$$E \cup F = \{2, 4, 5, 6\}, \qquad n(E \cup F) = 4.$$So
$$P(E \cup F) \;=\; \frac{4}{6} \;=\; \frac{2}{3}.$$Step 4 — Equivalent form using the overlap.
Since \(E \cap F = \{6\}\) and \(P(E \cap F) = 1/6\),
$$P(E \cup F) \;=\; P(E) + P(F) - P(E \cap F) \;=\; \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6}.$$Answer: \(P(E \cup F) = 2/3\).
Source: Applied Finite Mathematics
In a club, 20% of the students take Finite Mathematics, 30% take Statistics, and 10% take both. What percent take Finite Mathematics or Statistics?
Solution
Let \(F = \{\text{takes Finite Math}\}\) and \(S = \{\text{takes Statistics}\}\). Then \(P(F) = 0.20\), \(P(S) = 0.30\), \(P(F \cap S) = 0.10\).
$$P(F \cup S) \;=\; 0.20 + 0.30 - 0.10 \;=\; 0.40.$$Answer: \(40\%\) of students take at least one of the two courses.
Source: Applied Finite Mathematics
If \(P(E) = 0.4\), \(P(F) = 0.5\), and \(P(E \cup F) = 0.7\), find \(P(E \cap F)\).
Solution
Solve the Addition Rule for the overlap:
$$P(E \cap F) = P(E) + P(F) - P(E \cup F) = 0.4 + 0.5 - 0.7 = 0.2.$$Answer: \(P(E \cap F) = 0.2\).
Source: Applied Finite Mathematics
The table below is the distribution of the 100 U.S. senators of the 114th Congress (January 2015) by political party and gender.
| Male (M) | Female (F) | Total | |
|---|---|---|---|
| Democrats (D) | 30 | 14 | 44 |
| Republicans (R) | 48 | 6 | 54 |
| Other (T) | 2 | 0 | 2 |
| Total | 80 | 20 | 100 |
A senator is selected at random. Find:
a) \(P(M \cap D)\) — male and Democrat.
b) \(P(M \cup D)\) — male or Democrat.
c) \(P(F \cap R)\).
d) \(P(F \cup R)\).
Solution
Part a — Intersection. The "male Democrat" cell shows 30:
$$P(M \cap D) \;=\; \frac{30}{100} \;=\; 0.30.$$Part b — Union via Addition Rule. \(P(M) = 80/100\), \(P(D) = 44/100\), and we just computed \(P(M \cap D) = 30/100\). So
$$P(M \cup D) = \frac{80}{100} + \frac{44}{100} - \frac{30}{100} = \frac{94}{100} = 0.94.$$Part c — Intersection. The "female Republican" cell shows 6:
$$P(F \cap R) \;=\; \frac{6}{100} \;=\; 0.06.$$Part d — Union. \(P(F) = 20/100\), \(P(R) = 54/100\), \(P(F \cap R) = 6/100\):
$$P(F \cup R) = \frac{20}{100} + \frac{54}{100} - \frac{6}{100} = \frac{68}{100} = 0.68.$$Source: Applied Finite Mathematics
Using the Senate table above, find \(P(F \cup T)\) — female or Other party.
Solution
\(P(F) = 20/100\), \(P(T) = 2/100\), and the "female Other" cell is 0, so \(P(F \cap T) = 0\). The events are mutually exclusive in the data:
$$P(F \cup T) = \frac{20}{100} + \frac{2}{100} - 0 = \frac{22}{100} = 0.22.$$Answer: \(0.22\).
Source: Applied Finite Mathematics
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.
Solution
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.
Stage 1 Stage 2
┌── R: 4/7
R 5/8┤
┌────┴── Y: 3/7
Start ────────┤
└────┬── R: 5/7
Y 3/8┤
└── Y: 2/7
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: \(5/14\).
Source: Applied Finite Mathematics
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).
Solution
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: \(1/6\).
Source: Applied Finite Mathematics
Two apples are drawn without replacement from a basket of 5 red and 3 yellow apples. Find \(P(\text{one red and one yellow})\).
Solution
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: \(15/28\).
Source: Applied Finite Mathematics
A jar contains 5 red, 4 white, and 3 blue marbles (12 total). Three marbles are drawn at random. Find \(P(\text{all three red})\).
Solution
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: \(1/22\).
Source: Applied Finite Mathematics
From the same jar (5 red, 4 white, 3 blue), find \(P(\text{2 white and 1 blue})\).
Solution
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: \(9/110\).
Source: Applied Finite Mathematics
From the same jar of 5 red, 4 white, and 3 blue marbles, find \(P(\text{none white})\) when 3 marbles are drawn.
Solution
"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: \(14/55\).
Source: Applied Finite Mathematics
From a jar of 5 red, 4 white, and 3 blue marbles, three are drawn. Find \(P(\text{at least one red})\).
Solution
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: \(37/44\).
Source: Applied Finite Mathematics
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.)
Solution
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 \(2.7\%\).
Source: Applied Finite Mathematics
If there are 4 people in a room, find \(P(\text{at least two share a birthday})\) (365-day calendar).
Solution
\(P(\text{all 4 distinct}) = \dfrac{365 \cdot 364 \cdot 363 \cdot 362}{365^{4}} \approx 0.9836\).
$$P(\text{at least two share}) \approx 1 - 0.9836 = 0.0164.$$Answer: about \(1.6\%\).
Source: Applied Finite Mathematics
A family has three children. Find the conditional probability of having two boys and a girl, given that the first born is a boy.
Solution
Step 1 — Full sample space:
$$S = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG\}, \quad n(S) = 8.$$Step 2 — Condition on "first born is a boy". That cuts the sample space to outcomes starting with \(B\):
$$F = \{BBB, BBG, BGB, BGG\}, \quad n(F) = 4.$$Step 3 — Count outcomes in \(F\) that also have two boys and one girl: \(BBG\) and \(BGB\).
Step 4 — Compute:
$$P(E \mid F) = \frac{2}{4} = \frac{1}{2}.$$Answer: \(1/2\).
Source: Applied Finite Mathematics
A six-sided die is rolled once. Let \(E = \{\text{the result is even}\}\) and \(T = \{\text{the result is greater than 3}\}\). Find:
a) \(P(E)\).
b) \(P(E \mid T)\).
Solution
Sample space: \(S = \{1, 2, 3, 4, 5, 6\}\). \(E = \{2, 4, 6\}\), \(T = \{4, 5, 6\}\).
Part a — Unconditional. \(P(E) = 3/6 = 1/2\).
Part b — Conditional. Given \(T\) holds, the only possible outcomes are \(\{4, 5, 6\}\). Of those, the even ones are \(\{4, 6\}\). Therefore
$$P(E \mid T) = \frac{2}{3}.$$Source: Applied Finite Mathematics
A fair coin is tossed twice. Find:
a) \(P(\text{two heads})\).
b) \(P(\text{two heads} \mid \text{at least one head})\).
Solution
Sample space: \(S = \{HH, HT, TH, TT\}\). Let \(E = \{\text{two heads}\} = \{HH\}\) and \(F = \{\text{at least one head}\} = \{HH, HT, TH\}\).
Part a — Unconditional. \(P(E) = 1/4\).
Part b — Conditional. Given \(F\), the reduced sample space is \(\{HH, HT, TH\}\) with \(n(F) = 3\). Of those, only \(HH\) gives two heads:
$$P(E \mid F) = \frac{1}{3}.$$Source: Applied Finite Mathematics
A single die is rolled. Find \(P(\text{the result is a 3} \mid \text{the result is odd})\).
Solution
The conditioning event "odd" is \(\{1, 3, 5\}\), with \(n = 3\). The event "result is 3" inside that reduced space is \(\{3\}\). Therefore
$$P = \frac{1}{3}.$$Answer: \(1/3\).
Source: Applied Finite Mathematics
Suppose \(P(A) = 0.3\), \(P(B) = 0.4\), and \(P(A \cap B) = 0.12\). Find \(P(A \mid B)\) and \(P(B \mid A)\).
Solution
Answer: \(P(A \mid B) = 0.30\), \(P(B \mid A) = 0.40\).
Source: Applied Finite Mathematics
If \(P(F) = 0.4\) and \(P(E \mid F) = 0.3\), find \(P(E \cap F)\).
Solution
Use the multiplication rule:
$$P(E \cap F) = P(F)\, P(E \mid F) = 0.4 \times 0.3 = 0.12.$$Answer: \(0.12\).
Source: Applied Finite Mathematics
The table below shows the distribution of the 100 U.S. senators of the 114th Congress (January 2015) by political party and gender.
| Male (M) | Female (F) | Total | |
|---|---|---|---|
| Democrats (D) | 30 | 14 | 44 |
| Republicans (R) | 48 | 6 | 54 |
| Other (T) | 2 | 0 | 2 |
| Total | 80 | 20 | 100 |
Find:
a) \(P(M \mid D)\).
b) \(P(D \mid M)\).
c) \(P(F \mid R)\).
d) \(P(R \mid F)\).
Solution
Part a — Condition on Democrats. Among the 44 Democrats, 30 are male:
$$P(M \mid D) = \frac{30}{44} = \frac{15}{22} \approx 0.682.$$Part b — Condition on Males. Among the 80 males, 30 are Democrats:
$$P(D \mid M) = \frac{30}{80} = \frac{3}{8} = 0.375.$$Part c — Condition on Republicans. Among the 54 Republicans, 6 are female:
$$P(F \mid R) = \frac{6}{54} = \frac{1}{9} \approx 0.111.$$Part d — Condition on Females. Among the 20 females, 6 are Republicans:
$$P(R \mid F) = \frac{6}{20} = \frac{3}{10} = 0.30.$$Source: Applied Finite Mathematics
Using the same Senate table, find \(P(M \mid F)\).
Solution
Among the 20 females, none are male:
$$P(M \mid F) = \frac{0}{20} = 0.$$Answer: \(0\) (Senators in this table are exclusively male or female; the events are mutually exclusive).
Source: Applied Finite Mathematics
A card is drawn from a standard deck. Find:
a) \(P(\text{king})\).
b) \(P(\text{king} \mid \text{face card})\).
Solution
Part a — Unconditional. \(P(\text{king}) = 4/52 = 1/13\).
Part b — Conditional. Among the 12 face cards there are 4 kings:
$$P(\text{king} \mid \text{face card}) = \frac{4}{12} = \frac{1}{3}.$$Note \(P(\text{king} \mid \text{face card}) \neq P(\text{king})\) — knowing the card is a face card changed the probability that it is a king. The two events depend on each other.
Source: Applied Finite Mathematics
A card is drawn from a standard deck. Find:
a) \(P(\text{king})\).
b) \(P(\text{king} \mid \text{red})\).
Solution
Part a — Unconditional. \(P(\text{king}) = 4/52 = 1/13\).
Part b — Conditional. Among the 26 red cards there are 2 kings:
$$P(\text{king} \mid \text{red}) = \frac{2}{26} = \frac{1}{13}.$$This time \(P(\text{king} \mid \text{red}) = P(\text{king})\) — knowing the card is red did not change the probability that it is a king. The two events are independent.
Source: Applied Finite Mathematics
Consider a two-child family. Let \(F = \{\text{the family has children of both sexes}\}\) and \(G = \{\text{the first born is a boy}\}\). Are \(F\) and \(G\) independent?
Solution
Step 1 — Sample space: \(S = \{BB, BG, GB, GG\}\), \(n(S) = 4\).
Step 2 — Compute probabilities.
- \(F = \{BG, GB\}\), so \(P(F) = 2/4 = 1/2\).
- \(G = \{BB, BG\}\), so \(P(G) = 2/4 = 1/2\).
- \(F \cap G = \{BG\}\), so \(P(F \cap G) = 1/4\).
Step 3 — Apply the test \(P(E)\,P(F) \stackrel{?}{=} P(E \cap F)\):
$$P(F)\, P(G) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} = P(F \cap G).$$The product equals the joint probability, so the events are independent.
Answer: Yes, \(F\) and \(G\) are independent.
Source: Applied Finite Mathematics
For the same two-child family let \(E = \{\text{at least one boy}\}\) and \(F = \{\text{children of both sexes}\}\). Are \(E\) and \(F\) independent?
Solution
\(P(E) = 3/4\) (every outcome but \(GG\)), \(P(F) = 1/2\), and \(E \cap F = \{BG, GB\}\), so \(P(E \cap F) = 1/2\). Then
$$P(E)\, P(F) = \frac{3}{4}\cdot\frac{1}{2} = \frac{3}{8} \neq \frac{1}{2} = P(E \cap F),$$so \(E\) and \(F\) are not independent.
Source: Applied Finite Mathematics
John's probability of passing Statistics is \(0.40\), and Linda's is \(0.70\). Assuming the two events are independent, find:
a) \(P(\text{both pass})\).
b) \(P(\text{at least one passes})\).
Solution
Part a — Both:
$$P(\text{both pass}) = 0.40 \times 0.70 = 0.28.$$Part b — At least one. Use the complement: the only way "at least one passes" fails is if both fail. The probability each fails is \(1 - p\); fails are independent too, so
$$P(\text{both fail}) = 0.60 \times 0.30 = 0.18.$$ $$P(\text{at least one passes}) = 1 - 0.18 = 0.82.$$Source: Applied Finite Mathematics
Jane's probabilities of making her two flight connections are \(0.80\) and \(0.90\), independent of each other. Find \(P(\text{at least one connection})\).
Solution
\(P(\text{misses both}) = (1 - 0.8)(1 - 0.9) = 0.2 \times 0.1 = 0.02\).
$$P(\text{at least one}) = 1 - 0.02 = 0.98.$$Answer: \(0.98\).
Source: Applied Finite Mathematics
The library checkout data below records, for one day, where each book was borrowed (Main vs. Branch) and its category (Fiction vs. Non-fiction).
| Main (M) | Branch (B) | Total | |
|---|---|---|---|
| Fiction (F) | 300 | 100 | 400 |
| Non-fiction (N) | 150 | 50 | 200 |
| Total | 450 | 150 | 600 |
Is the event "the book is fiction" independent of the event "the book was borrowed at the Main library"?
Solution
Step 1 — Compute marginals.
$$P(F) = \frac{400}{600} = \frac{2}{3}, \qquad P(M) = \frac{450}{600} = \frac{3}{4}.$$Step 2 — Compute the joint probability.
$$P(F \cap M) = \frac{300}{600} = \frac{1}{2}.$$Step 3 — Apply the test.
$$P(F)\, P(M) = \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}.$$The product equals \(P(F \cap M)\), so the events are independent.
Answer: Yes, fiction and Main library are independent (in this dataset).
Source: Applied Finite Mathematics
In problems 8.1.1–8.1.6, write a sample space for the given experiment.
Problem 1. A die is rolled.
Problem 2. A penny and a nickel are tossed.
Problem 3. A die is rolled, and a coin is tossed.
Problem 4. Three coins are tossed.
Problem 5. Two dice are rolled.
Problem 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.
Problem 7. \(P(\text{an ace})\)
Problem 8. \(P(\text{a red card})\)
Problem 9. \(P(\text{a club})\)
Problem 10. \(P(\text{a face card})\)
Problem 11. \(P(\text{a jack or a spade})\)
Problem 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.
Problem 13. \(P(\text{red})\)
Problem 14. \(P(\text{white})\)
Problem 15. \(P(\text{red or blue})\)
Problem 16. \(P(\text{red and blue})\)
For problems 8.1.17–8.1.22: Consider a family of three children. Find the following probabilities.
Problem 17. \(P(\text{two boys and a girl})\)
Problem 18. \(P(\text{at least one boy})\)
Problem 19. \(P(\text{children of both sexes})\)
Problem 20. \(P(\text{at most one girl})\)
Problem 21. \(P(\text{first and third children are male})\)
Problem 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.
Problem 23. \(P(\text{the sum of the dice is 5})\)
Problem 24. \(P(\text{the sum of the dice is 8})\)
Problem 25. \(P(\text{the sum is 3 or 6})\)
Problem 26. \(P(\text{the sum is more than 10})\)
Problem 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.
Problem 28. \(P(\text{the sum of the numbers is 5})\)
Problem 29. \(P(\text{the sum of the numbers is odd})\)
Problem 30. \(P(\text{the sum of the numbers is 9})\)
Problem 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.
Problem 32. \(P(\text{the sum of the numbers is 5})\)
Problem 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: 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
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)
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