8.5 Independent Events

8.5 Independent Events

Learning Objectives

By the end of this section, you will be able to:

In this section, you will learn to:
  1. Define independent events.
  2. Determine whether two events are independent or dependent.
  3. Use the multiplication rule for independent events to compute joint probabilities.

8.5.1 What "Independent" Means

In the previous section we learned that knowing event \(F\) has occurred can change the probability of event \(E\). Sometimes that extra information changes \(P(E)\); other times it leaves \(P(E)\) alone. The latter case is the heart of this section.

Example 8.5.1

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.

Example 8.5.2

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.

Context Pause: "Independent" is a strong claim about information

Saying that \(E\) and \(F\) are independent means that learning whether \(F\) happened gives you no information about whether \(E\) will happen. That's a very strong statement, and it is rarely true by accident. Independence usually arises from the experimental design — separate dice, separate coins, separate trials — or from an assumed model (a fair lottery, a memoryless service queue).

8.5.2 Tests for Independence

Definition 8.7: Independent Events

Two events \(E\) and \(F\) (with positive probabilities) are independent if any one of the following equivalent conditions holds:

  1. \(P(E \mid F) = P(E)\).
  2. \(P(F \mid E) = P(F)\).
  3. \(P(E \cap F) = P(E)\, P(F)\).

If \(E\) and \(F\) are not independent, they are dependent.

The third condition is usually the most useful in practice — it does not require computing a conditional probability — but all three are equivalent.

Insight Note: Independent ≠ Mutually Exclusive

Beginners often confuse the two. Mutually exclusive events cannot both happen; independent events can both happen, and one happening tells you nothing about the other. In fact, two events that are mutually exclusive (and have positive probability) are guaranteed to be dependent — knowing \(F\) happened tells you \(E\) cannot happen, which is plenty of information. The two notions describe different relationships.

Example 8.5.3

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.

Try It Now 8.5.1

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.

8.5.3 Multiplication Rule for Independent Events

The third condition in Definition 8.7 is itself a useful formula for computing joint probabilities.

Multiplication Rule for Independent Events

If \(E\) and \(F\) are independent,

\[P(E \cap F) \;=\; P(E)\, P(F).\]

More generally, the joint probability of any number of mutually independent events is the product of their individual probabilities.

Example 8.5.4

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

\[P(\text{both fail}) = 0.60 \times 0.30 = 0.18.\]

\[P(\text{at least one passes}) = 1 - 0.18 = 0.82.\]

Try It Now 8.5.2

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: \(0.98\).

8.5.4 Independence in Practice

Real-world independence claims must be checked against data. Two-way tables are the most direct test: compute \(P(E)\), \(P(F)\), and \(P(E \cap F)\) from the table and see whether \(P(E)\,P(F) = P(E \cap F)\).

Example 8.5.5

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).

Problem Set 8.5

For problems 8.5.1–8.5.4, use the library checkout table.

| | Main (M) | Branch (B) | Total | |--------------|---------:|-----------:|------:| | Fiction (F) | 300 | 100 | 400 | | Non-fiction (N) | 150 | 50 | 200 | | Total | 450 | 150 | 600 |

8.5.1 \(P(F)\)

8.5.2 \(P(M \mid F)\)

8.5.3 \(P(N \mid B)\)

8.5.4 Is the fact that a person checks out a fiction book independent of the main library? Use probabilities to justify your conclusion.

For a two-child family let the events be:

8.5.5 Find:

a) \(P(E)\)

b) \(P(F)\)

c) \(P(E \cap F)\)

d) Are \(E\) and \(F\) independent? Justify with probabilities.

8.5.6 Find:

a) \(P(F)\)

b) \(P(G)\)

c) \(P(F \cap G)\)

d) Are \(F\) and \(G\) independent? Justify with probabilities.

Solve each independence problem.

8.5.7 If \(P(E) = 0.6\), \(P(F) = 0.2\), and \(E\) and \(F\) are independent, find \(P(E \cap F)\).

8.5.8 If \(P(E) = 0.6\), \(P(F) = 0.2\), and \(E\) and \(F\) are independent, find \(P(E \cup F)\).

8.5.9 If \(P(E) = 0.9\), \(P(F \mid E) = 0.36\), and \(E\) and \(F\) are independent, find \(P(F)\).

8.5.10 If \(P(E) = 0.6\), \(P(E \cup F) = 0.8\), and \(E\) and \(F\) are independent, find \(P(F)\).

8.5.11 In a survey of 100 people, 40 were casual drinkers and 60 did not drink. Of the drinkers, 6 had minor headaches. Of the non-drinkers, 9 had minor headaches. Are the events "drinker" and "had a headache" independent?

8.5.12 80% of people wear seat belts and 5% quit smoking last year. If 4% of seat-belt wearers quit smoking, are the events "wears a seat belt" and "quit smoking" independent?

8.5.13 John's probability of passing Statistics is \(40\%\) and Linda's is \(70\%\). Assuming the events are independent, find:

a) \(P(\text{both pass})\)

b) \(P(\text{at least one passes})\)

8.5.14 Jane is flying home for the holidays and has to change planes twice. \(P(\text{makes 1st connection}) = 0.80\) and \(P(\text{makes 2nd connection}) = 0.90\). Assuming the events are independent, find:

a) \(P(\text{makes both connections})\)

b) \(P(\text{makes at least one connection})\)

For a three-child family let:

8.5.15 Find:

a) \(P(E)\)

b) \(P(F)\)

c) \(P(E \cap F)\)

d) Are \(E\) and \(F\) independent?

8.5.16 Find:

a) \(P(F)\)

b) \(P(G)\)

c) \(P(F \cap G)\)

d) Are \(F\) and \(G\) independent?

8.5.17 \(P(K \mid D) = 0.7\), \(P(D) = 0.25\), and \(P(K) = 0.7\).

a) Are events \(K\) and \(D\) independent? Justify with probabilities.

b) Find \(P(K \cap D)\).

8.5.18 \(P(R \mid S) = 0.4\), \(P(S) = 0.2\), and \(P(R) = 0.3\).

a) Are events \(R\) and \(S\) independent? Justify with probabilities.

b) Find \(P(R \cap S)\).

8.5.19 At a college, \(54\%\) of students are female, \(25\%\) of students are majoring in engineering, and \(15\%\) of female students are majoring in engineering. Let \(E = \{\text{majors in engineering}\}\) and \(F = \{\text{female}\}\).

a) Are \(E\) and \(F\) independent? Justify with probabilities.

b) Find \(P(E \cap F)\).

8.5.20 At a college, \(54\%\) of students are female, \(60\%\) of all students receive financial aid, and \(60\%\) of female students receive financial aid. Let \(A = \{\text{receives financial aid}\}\) and \(F = \{\text{female}\}\).

a) Are \(A\) and \(F\) independent? Justify with probabilities.

b) Find \(P(A \cap F)\).

Chapter 8 Review

8.5.21 Two dice are rolled. Find the probability that the sum of the dice is

a) four

b) five

8.5.22 A jar contains 3 red, 4 white, and 5 blue marbles. If a marble is chosen at random, find:

a) \(P(\text{red or blue})\)

b) \(P(\text{not blue})\)

8.5.23 A card is drawn from a standard deck. Find:

a) \(P(\text{a jack or a king})\)

b) \(P(\text{a jack or a spade})\)

8.5.24 A basket contains 3 red and 2 yellow apples. Two apples are chosen at random. Find:

a) \(P(\text{one red, one yellow})\)

b) \(P(\text{at least one red})\)

8.5.25 A basket contains 4 red, 3 white, and 3 blue marbles. Three marbles are chosen at random. Find:

a) \(P(\text{two red, one white})\)

b) \(P(\text{first red, second white, third blue})\)

c) \(P(\text{at least one red})\)

d) \(P(\text{none red})\)

8.5.26 For a family of four children find:

a) \(P(\text{all boys})\)

b) \(P(\text{1 boy and 3 girls})\)

8.5.27 For a family of three children find:

a) \(P(\text{children of both sexes} \mid \text{first born is a boy})\)

b) \(P(\text{all girls} \mid \text{children of both sexes})\)

8.5.28 Mrs. Rossetti is flying from San Francisco to New York. On her way to the airport she encounters heavy traffic and estimates a \(20\%\) chance she will be late and miss her flight. Even if she makes that flight, there is a \(10\%\) chance she will miss her connection in Chicago. What is the probability she makes it to New York as scheduled?

8.5.29 At a college, \(20\%\) of students take History, \(30\%\) take Math, and \(10\%\) take both. What percent take at least one of the two?

8.5.30 In a T-maze, a mouse may run right (R) or left (L). A mouse runs the maze three times. Let \(E = \{\text{runs right on the first trial}\}\) and \(F = \{\text{runs left two consecutive times}\}\). Determine whether \(E\) and \(F\) are independent.

8.5.31 At a college, \(20\%\) of students take advanced math, \(40\%\) take advanced English, and \(15\%\) take both. If a student is selected at random, find:

a) \(P(\text{takes English} \mid \text{takes math})\)

b) \(P(\text{takes math or English})\)

8.5.32 If there are 35 students in a class, what is the probability that at least two share a birthday? (365-day calendar.)

8.5.33 A student feels that her probability of passing Accounting is \(0.62\), of passing Mathematics is \(0.45\), and of passing Accounting or Mathematics is \(0.85\). Find the probability she passes both.

8.5.34 The U.S. Supreme Court has nine justices: 5 conservative and 4 liberal. The court will act on six major cases this year. What is the probability that the court favors the conservatives in at least four of the six cases?

8.5.35 Five cards are drawn from a standard deck. Find:

a) \(P(\text{four cards of a single suit})\)

b) \(P(\text{two cards of one suit, two of another, and one from a third suit})\)

c) \(P(\text{a pair (e.g. two aces and three other cards)})\)

d) \(P(\text{a straight flush (five in a row of a single suit, but not a royal flush)})\)

8.5.36 The following table shows a distribution of drink preferences by gender.

| | Coke (C) | Pepsi (P) | Seven Up (S) | Total | |---------|---------:|----------:|-------------:|------:| | Male (M)| 60 | 50 | 22 | 132 | | Female (F)| 50 | 40 | 18 | 108 | | Total| 110 | 90 | 40 | 240 |

Find:

a) \(P(F \mid S)\)

b) \(P(P \mid F)\)

c) \(P(C \mid M)\)

d) \(P(M \mid P \cup C)\)

e) Are the events \(F\) and \(S\) mutually exclusive?

f) Are the events \(F\) and \(S\) independent?

8.5.37 At a clothing outlet \(20\%\) of clothes are irregular, \(10\%\) have at least one button missing, and \(4\%\) are both irregular and have a button missing. If Martha found a dress with a missing button, what is the probability the dress is irregular?

8.5.38 A trade delegation has 4 Americans, 3 Japanese, and 2 Germans. Three people are chosen at random. Find:

a) \(P(\text{two Americans and one Japanese})\)

b) \(P(\text{at least one American})\)

c) \(P(\text{one of each nationality})\)

d) \(P(\text{no German})\)

8.5.39 A coin is tossed three times. Let \(E = \{\text{shows a head on the first toss}\}\) and \(F = \{\text{never turns up a tail}\}\). Are \(E\) and \(F\) independent?

8.5.40 If \(P(E) = 0.6\), \(P(F) = 0.4\), and \(E\) and \(F\) are mutually exclusive, find \(P(E \cap F)\).

8.5.41 If \(P(E) = 0.5\), \(P(F) = 0.3\), and \(E\) and \(F\) are independent, find \(P(E \cup F)\).

8.5.42 If \(P(F) = 0.9\), \(P(E \mid F) = 0.36\), and \(E\) and \(F\) are independent, find \(P(E)\).

8.5.43 If \(P(E) = 0.4\), \(P(E \cup F) = 0.9\), and \(E\) and \(F\) are independent, find \(P(F)\).

8.5.44 If \(P(E) = 0.4\) and \(P(F \mid E) = 0.5\), find \(P(E \cap F)\).

8.5.45 If \(P(E) = 0.6\) and \(P(E \cap F) = 0.3\), find \(P(F \mid E)\).

8.5.46 If \(P(E) = 0.3\), \(P(F) = 0.4\), and \(E\) and \(F\) are independent, find \(P(E \mid F)\).

Problem 8.5.1 Solution

\(P(F) = 400/600 = 2/3\).

Answer: \(2/3\).

Problem 8.5.2 Solution

Among the 400 fiction checkouts, 300 came from the Main library:

\[P(M \mid F) = \frac{300}{400} = \frac{3}{4}.\]

Answer: \(3/4\).

Problem 8.5.3 Solution

Among the 150 Branch checkouts, 50 are non-fiction:

\[P(N \mid B) = \frac{50}{150} = \frac{1}{3}.\]

Answer: \(1/3\).

Problem 8.5.4 Solution

Step 1 — Compute \(P(F)\) and \(P(F \mid M)\):

\[P(F) = \frac{400}{600} = \frac{2}{3}, \qquad P(F \mid M) = \frac{300}{450} = \frac{2}{3}.\]

Step 2 — Compare: \(P(F \mid M) = P(F)\), so the events are independent.

Answer: Yes — fiction-vs-Main are independent.

Problem 8.5.5 Solution

Sample space \(S = \{BB, BG, GB, GG\}\), \(n(S) = 4\).

a) \(E = \{BB, BG, GB\}\), \(P(E) = 3/4\).

b) \(F = \{BG, GB\}\), \(P(F) = 1/2\).

c) \(E \cap F = \{BG, GB\}\), \(P(E \cap F) = 1/2\).

d) \(P(E)\,P(F) = (3/4)(1/2) = 3/8 \neq 1/2 = P(E \cap F)\). Not independent.

Problem 8.5.6 Solution

a) \(F = \{BG, GB\}\), \(P(F) = 1/2\).

b) \(G = \{BB, BG\}\), \(P(G) = 1/2\).

c) \(F \cap G = \{BG\}\), \(P(F \cap G) = 1/4\).

d) \(P(F)\,P(G) = (1/2)(1/2) = 1/4 = P(F \cap G)\). Independent.

Problem 8.5.7 Solution

\(P(E \cap F) = P(E)\,P(F) = (0.6)(0.2) = 0.12\).

Answer: \(0.12\).

Problem 8.5.8 Solution

Use the Addition Rule with the independent product:

\[P(E \cup F) = P(E) + P(F) - P(E)\,P(F) = 0.6 + 0.2 - 0.12 = 0.68.\]

Answer: \(0.68\).

Problem 8.5.9 Solution

Independence implies \(P(F \mid E) = P(F)\), so \(P(F) = 0.36\).

Answer: \(0.36\).

Problem 8.5.10 Solution

\(P(E \cup F) = P(E) + P(F) - P(E)\,P(F)\). With \(P(E) = 0.6\) and \(P(E \cup F) = 0.8\):

\[0.8 = 0.6 + P(F) - 0.6\,P(F) = 0.6 + 0.4\,P(F).\]

\[P(F) = \frac{0.8 - 0.6}{0.4} = 0.5.\]

Answer: \(0.5\).

Problem 8.5.11 Solution

Step 1 — Compute marginals from the data.

  • Drinkers: \(P(D) = 40/100 = 0.40\).
  • Headaches: \(15/100 = 0.15\) (6 from drinkers + 9 from non-drinkers).
  • Both: \(P(D \cap H) = 6/100 = 0.06\).

Step 2 — Test: \(P(D)\,P(H) = (0.4)(0.15) = 0.06 = P(D \cap H)\). Independent.

Problem 8.5.12 Solution

\(P(Q) = 0.05\) and \(P(Q \mid S) = 0.04\). The two are not equal, so the events are not independent.

Problem 8.5.13 Solution

a) \(P(\text{both pass}) = 0.40 \times 0.70 = 0.28\).

b) \(P(\text{at least one passes}) = 1 - P(\text{both fail}) = 1 - (0.60)(0.30) = 1 - 0.18 = 0.82\).

Problem 8.5.14 Solution

a) \(P(\text{both connections}) = 0.80 \times 0.90 = 0.72\).

b) \(P(\text{misses both}) = (0.20)(0.10) = 0.02\). \(P(\text{at least one}) = 1 - 0.02 = 0.98\).

Problem 8.5.15 Solution

For a three-child family, \(n(S) = 8\).

a) \(E = \{\text{at least one boy}\} = S \setminus \{GGG\}\), so \(P(E) = 7/8\).

b) \(F = \{\text{both sexes}\} = S \setminus \{BBB, GGG\}\), so \(P(F) = 6/8 = 3/4\).

c) \(E \cap F = S \setminus \{BBB, GGG\} = \{BBG, BGB, BGG, GBB, GBG, GGB\}\) (since \(F\) already excludes \(BBB\) and \(GGG\), and every remaining outcome has at least one boy).

\[P(E \cap F) = 6/8 = 3/4.\]

d) \(P(E)\,P(F) = (7/8)(3/4) = 21/32\). \(P(E \cap F) = 3/4 = 24/32\). Not equal — not independent.

Problem 8.5.16 Solution

a) \(P(F) = 6/8 = 3/4\) (both-sexes outcomes).

b) \(P(G) = 4/8 = 1/2\) (first-born-boy outcomes).

c) \(F \cap G\): first born is boy AND family has both sexes. From \(\{BBB, BBG, BGB, BGG\}\) exclude \(BBB\): \(\{BBG, BGB, BGG\}\), so \(P(F \cap G) = 3/8\).

d) \(P(F)\,P(G) = (3/4)(1/2) = 3/8 = P(F \cap G)\). Independent.

Problem 8.5.17 Solution

a) \(P(K \mid D) = 0.7 = P(K)\). Independent.

b) \(P(K \cap D) = P(K \mid D)\,P(D) = (0.7)(0.25) = 0.175\).

Problem 8.5.18 Solution

a) \(P(R \mid S) = 0.4\) but \(P(R) = 0.3\). Not equal — not independent.

b) \(P(R \cap S) = P(R \mid S)\,P(S) = (0.4)(0.2) = 0.08\).

Problem 8.5.19 Solution

a) \(P(E \mid F) = 0.15\) but \(P(E) = 0.25\). Not equal — not independent.

b) \(P(E \cap F) = P(E \mid F)\,P(F) = (0.15)(0.54) = 0.081\).

Problem 8.5.20 Solution

a) \(P(A \mid F) = 0.60 = P(A)\). Independent.

b) \(P(A \cap F) = P(A \mid F)\,P(F) = (0.60)(0.54) = 0.324\).

Problem 8.5.21 Solution

a) Sum 4: \((1,3),(2,2),(3,1)\) — 3 outcomes. \(P = 3/36 = 1/12\).

b) Sum 5: \((1,4),(2,3),(3,2),(4,1)\) — 4 outcomes. \(P = 4/36 = 1/9\).

Problem 8.5.22 Solution

Total marbles \(= 3 + 4 + 5 = 12\).

a) \(P(\text{red or blue}) = (3 + 5)/12 = 8/12 = 2/3\).

b) \(P(\text{not blue}) = (12 - 5)/12 = 7/12\).

Problem 8.5.23 Solution

a) Jacks and kings are disjoint: \(P(\text{jack or king}) = 4/52 + 4/52 = 8/52 = 2/13\).

b) \(P(\text{jack or spade}) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13\).

Problem 8.5.24 Solution

\(C(5, 2) = 10\) total ways to draw 2 apples.

a) \(P(\text{1 R, 1 Y}) = C(3,1)\,C(2,1)/C(5,2) = 6/10 = 3/5\).

b) \(P(\text{at least 1 R}) = 1 - P(\text{no R}) = 1 - C(2,2)/10 = 1 - 1/10 = 9/10\).

Problem 8.5.25 Solution

\(C(10, 3) = 120\).

a) \(P(\text{2 R, 1 W}) = C(4,2)\,C(3,1)/120 = (6)(3)/120 = 18/120 = 3/20\).

b) Sequential ordered probability: \(P = (4/10)(3/9)(3/8) = 36/720 = 1/20\).

c) \(P(\text{no R}) = C(6, 3)/120 = 20/120 = 1/6\). \(P(\text{at least one R}) = 1 - 1/6 = 5/6\).

d) \(P(\text{none red}) = 1/6\) (from part c).

Problem 8.5.26 Solution

Family of 4: \(n(S) = 16\) equally likely outcomes.

a) \(P(\text{all boys}) = 1/16\).

b) \(P(\text{1 boy, 3 girls}) = C(4, 1)/16 = 4/16 = 1/4\).

Problem 8.5.27 Solution

a) "First born is boy" reduces \(S\) to \(\{BBB, BBG, BGB, BGG\}\) (4 outcomes). "Both sexes" within: \(\{BBG, BGB, BGG\}\) (3 outcomes). \(P = 3/4\).

b) "Both sexes" reduces \(S\) to 6 outcomes (all but \(BBB\) and \(GGG\)). "All girls" \(= \{GGG\}\) is not in this reduced space, so \(P = 0\).

Problem 8.5.28 Solution

\(P(\text{makes 1st flight}) = 0.80\). Given she made it, \(P(\text{makes 2nd flight} \mid \text{made 1st}) = 0.90\).

\[P(\text{makes both}) = (0.80)(0.90) = 0.72.\]

Answer: \(0.72\) (\(72\%\)).

Problem 8.5.29 Solution

\(P(H \cup M) = P(H) + P(M) - P(H \cap M) = 0.20 + 0.30 - 0.10 = 0.40\).

Answer: \(40\%\).

Problem 8.5.30 Solution

Each trial is L or R with probability \(1/2\). Three trials → \(n(S) = 8\) equally likely outcomes.

  • \(E = \{\text{first trial R}\} = \{RRR, RRL, RLR, RLL\}\), \(P(E) = 1/2\).
  • \(F = \{\text{at least two consecutive Ls}\} = \{LLR, LLL, RLL\}\), \(P(F) = 3/8\).
  • \(E \cap F = \{RLL\}\), so \(P(E \cap F) = 1/8\).

\(P(E)\,P(F) = (1/2)(3/8) = 3/16\), but \(P(E \cap F) = 1/8 = 2/16\). Not independent.

Problem 8.5.31 Solution

a) \(P(E \mid M) = P(E \cap M)/P(M) = 0.15/0.20 = 0.75\).

b) \(P(E \cup M) = 0.40 + 0.20 - 0.15 = 0.45\).

Problem 8.5.32 Solution

Use the complement "all 35 birthdays distinct":

\[P(\text{all distinct}) = \frac{365 \cdot 364 \cdots (365 - 34)}{365^{35}} \approx 0.1856.\]

\[P(\text{at least 2 share}) \approx 1 - 0.1856 = 0.8144.\]

Answer: about \(0.814\) (≈ \(81.4\%\)).

Problem 8.5.33 Solution

\[P(A \cap M) = P(A) + P(M) - P(A \cup M) = 0.62 + 0.45 - 0.85 = 0.22.\]

Answer: \(0.22\).

Problem 8.5.34 Solution

Model: each of 6 cases is decided independently with \(P(\text{conservative wins}) = 5/9\) (the proportion of conservative justices). Let \(X\) be the number of conservative-favored cases; \(X \sim \text{Binomial}(6, 5/9)\). Want \(P(X \ge 4)\):

\[P(X = k) = \binom{6}{k}\!\left(\tfrac{5}{9}\right)^{k}\!\left(\tfrac{4}{9}\right)^{6-k}.\]

  • \(P(X = 4) = 15 \cdot (5/9)^4 (4/9)^2 = 150{,}000/531{,}441\).
  • \(P(X = 5) = 6 \cdot (5/9)^5 (4/9) = 75{,}000/531{,}441\).
  • \(P(X = 6) = (5/9)^6 = 15{,}625/531{,}441\).

\[P(X \ge 4) = \frac{150{,}000 + 75{,}000 + 15{,}625}{531{,}441} = \frac{240{,}625}{531{,}441} \approx 0.4528.\]

Answer: about \(0.4528\) (≈ \(45.3\%\)).

Problem 8.5.35 Solution

\(C(52, 5) = 2{,}598{,}960\).

a) Exactly four cards of one suit (and the fifth of a different suit):

\[4 \cdot C(13, 4) \cdot 39 = 4 \cdot 715 \cdot 39 = 111{,}540.\]

\(P \approx 111{,}540/2{,}598{,}960 \approx 0.0429\).

b) 2-of-one-suit, 2-of-another, 1-of-a-third: number of suit-count assignments is \(\binom{4}{2}\cdot 2 = 12\) (pick the two "pair suits", then the singleton suit), and each gives \(C(13,2)^{2}\,C(13,1) = 78 \cdot 78 \cdot 13 = 79{,}092\) hands:

\[12 \cdot 79{,}092 = 949{,}104.\]

\(P \approx 949{,}104/2{,}598{,}960 \approx 0.3651\).

c) Exactly one pair (the standard "one pair" hand): \(13 \cdot C(4,2) \cdot C(12, 3) \cdot 4^{3} = 13 \cdot 6 \cdot 220 \cdot 64 = 1{,}098{,}240\).

\(P \approx 1{,}098{,}240/2{,}598{,}960 \approx 0.4226\).

d) Straight flush (excluding royal): there are 40 straight flushes total (10 high-cards × 4 suits), of which 4 are royal flushes. So non-royal straight flushes = 36.

\(P = 36/2{,}598{,}960 \approx 1.39 \times 10^{-5}\).

Problem 8.5.36 Solution

Marginals: \(M = 132\), \(F = 108\), \(C = 110\), \(P = 90\), \(S = 40\), grand total \(= 240\).

a) \(P(F \mid S) = 18/40 = 9/20\).

b) \(P(P \mid F) = 40/108 = 10/27 \approx 0.370\).

c) \(P(C \mid M) = 60/132 = 5/11 \approx 0.455\).

d) \(P \cup C\) has \(90 + 110 = 200\) people. Of these, the male count is \(50 + 60 = 110\). \(P(M \mid P \cup C) = 110/200 = 11/20\).

e) \(F \cap S = 18 \neq 0\), so not mutually exclusive.

f) \(P(F)\,P(S) = (108/240)(40/240) = (9/20)(1/6) = 9/120 = 3/40 = 0.075\). And \(P(F \cap S) = 18/240 = 3/40 = 0.075\). Equal — independent.

Problem 8.5.37 Solution

\(P(I \mid B) = P(I \cap B)/P(B) = 0.04/0.10 = 0.40\).

Answer: \(0.40\) (\(40\%\)).

Problem 8.5.38 Solution

\(C(9, 3) = 84\).

a) \(P(\text{2 A, 1 J}) = C(4,2)\,C(3,1)/84 = 18/84 = 3/14\).

b) \(P(\text{no Americans}) = C(5, 3)/84 = 10/84 = 5/42\). \(P(\text{at least one American}) = 1 - 5/42 = 37/42\).

c) \(P(\text{one of each}) = C(4,1)\,C(3,1)\,C(2,1)/84 = 24/84 = 2/7\).

d) \(P(\text{no Germans}) = C(7, 3)/84 = 35/84 = 5/12\).

Problem 8.5.39 Solution

\(F = \{HHH\}\), \(P(F) = 1/8\). \(E = \{\text{first toss is H}\} = \{HHH, HHT, HTH, HTT\}\), \(P(E) = 1/2\). \(E \cap F = \{HHH\}\), \(P(E \cap F) = 1/8\).

\(P(E)\,P(F) = (1/2)(1/8) = 1/16\), but \(P(E \cap F) = 1/8\). Not equal — not independent.

Problem 8.5.40 Solution

Mutually exclusive ⇒ \(P(E \cap F) = 0\).

Problem 8.5.41 Solution

\(P(E \cap F) = (0.5)(0.3) = 0.15\). \(P(E \cup F) = 0.5 + 0.3 - 0.15 = 0.65\).

Problem 8.5.42 Solution

Independent ⇒ \(P(E) = P(E \mid F) = 0.36\).

Problem 8.5.43 Solution

\(P(E \cup F) = P(E) + P(F) - P(E)\,P(F)\). With \(P(E) = 0.4\) and \(P(E \cup F) = 0.9\):

\[0.9 = 0.4 + P(F) - 0.4\,P(F) = 0.4 + 0.6\,P(F).\]

\[P(F) = \frac{0.5}{0.6} = \frac{5}{6} \approx 0.833.\]

Problem 8.5.44 Solution

\(P(E \cap F) = P(E)\,P(F \mid E) = (0.4)(0.5) = 0.20\).

Problem 8.5.45 Solution

\(P(F \mid E) = P(E \cap F)/P(E) = 0.3/0.6 = 0.5\).

Problem 8.5.46 Solution

Independent ⇒ \(P(E \mid F) = P(E) = 0.3\).