Introduction to Stats · Chapter 3 · Probability Topics

Independent and Mutually Exclusive Events

Two tests decide how a pair of events relate — does knowing one tell you anything about the other, and can they even happen at the same time?

bookSHelf · Introduction to Statistics · §3.2 · a self-paced section

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Outline — by the end of this section you will be able to

Objectives

Determine whether two events are independent by checking one of three equivalent conditions definition
Decide whether two events are mutually exclusive by computing their intersection definition
Distinguish independence from mutual exclusivity, which are often confused §3.2.3
Apply the multiplication rule for independent events to compute $P(A \text{ AND } B)$ mult. rule
Recognize when sampling with versus without replacement drives dependence replacement

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.1 — Independent Events

Sampling with vs. without replacement

With replacement

After each pick, the member goes back before the next draw. The same member can appear more than once, and the first result never changes the probabilities for the second pick. With-replacement draws are independent.

Without replacement

Once a member is picked, it is set aside. The next pick happens from a smaller population, so its probabilities are affected by what came before. Without-replacement draws are dependent.

Roll a fair die twice: whatever the first roll shows, the second roll still has six equally likely faces — the two outcomes are independent.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.1 — the formal test

What independent events are

Definition 3.2.1 — Independent Events

Two events $A$ and $B$ are independent if knowing that one occurred does not change the probability that the other occurs. Any one of these three equivalent statements confirms it:

P(A \mid B) = P(A) \quad \text{or} \quad P(B \mid A) = P(B) \quad \text{or} \quad P(A \text{ AND } B) = P(A) \cdot P(B)

If none of the three holds, the events are dependent — knowing one changes the chance of the other. The three are equivalent whenever $P(A)$ and $P(B)$ are both positive; pick whichever is easiest to check.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Insight Note — default to dependent when you are unsure

When in doubt, assume dependent

If a problem does not say whether two events are independent, assume they are dependent until you can show otherwise.

The independence test is a one-line check. Assuming independence when the events are actually dependent silently over-counts $P(A \text{ AND } B)$ and can push a probability above 1.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Context Pause — why "and" gets a product

Independence is what buys you the product.

When two events are independent, $P(A \text{ AND } B) = P(A) \cdot P(B)$ — the joint probability is just the product of the individual probabilities. Sections 3.3 and 3.4 generalize this to events that are not independent, where the AND has to be reconstructed from a conditional probability instead.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — two cards, two sampling plans

Try It Now 3.2.1

You draw two cards from a fair, well-shuffled 52-card deck. (a) You note the first card and put it back before drawing the second. Independent? (b) You set the first card aside before drawing the second. Independent?

(a) Replacing the card restores the full 52-card deck, so the first draw has no effect on the second — independent. (b) With only 51 cards left, the second draw's probabilities depend on what came out first — dependent.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.1 — a worked comparison

Example 3.2.1 — Cards drawn with and without replacement

With replacement: jack of spades, replace, ten of clubs, replace, jack of spades again. Every draw happens from a full 52-card deck — the three draws are independent.

Without replacement: seven of hearts (51 left), three of diamonds (50 left), jack of spades. Each draw's probabilities depend on what was already removed — the three draws are dependent.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — read the sampling plan from the cards

Try It Now 3.2.2

You pick three cards. (a) The cards are {jack of spades, seven of hearts, jack of spades}. With or without replacement? (b) The cards are {jack of spades, seven of hearts, five of diamonds}. With or without replacement?

(a) The jack of spades appears twice, and the deck has only one — it must have been replaced. (b) All three cards are distinct, which is consistent with either plan — cannot be determined from the cards alone.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.1 — reading the plan from a hand

Example 3.2.2 — Reading the sampling plan from the cards

(a) Four cards, not put back: $\{J\spadesuit,\; 5\heartsuit,\; 5\diamondsuit,\; K\clubsuit\}$ . (b) Four cards, each put back before the next pick: $\{J\spadesuit,\; 7\heartsuit,\; 6\heartsuit,\; K\spadesuit\}$ .

Both hands have no card repeated — a repeat is the only mark a hand can carry that proves replacement. Here the surrounding text settles it: (a) is without replacement; (b) is with replacement.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — which hands are even possible?

Try It Now 3.2.3

Sample four cards without replacement, then again with replacement. Which of these outcomes are possible under each plan? (a) $\{J\spadesuit, 5\heartsuit, 5\diamondsuit, K\clubsuit\}$ (b) $\{Q\spadesuit, 7\heartsuit, 6\heartsuit, K\spadesuit\}$ (c) $\{J\spadesuit, 7\heartsuit, 6\heartsuit, K\spadesuit\}$

All three hands are possible under both plans. Every card in every hand is distinct — none of the three hands carries a repeat, so none of them can distinguish with-replacement sampling from without.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.2 — Mutually Exclusive Events

Sharing no outcomes

Let $S = \{1,2,\dots,10\}$ , $A = \{1,2,3,4,5\}$ , $B = \{4,5,6,7,8\}$ , and $C = \{7,9\}$ .

$A \cap B = \{4,5\}$ , so $P(A \text{ AND } B) = \frac{2}{10} \neq 0$ — not mutually exclusive.

$A \cap C = \varnothing$ and $B \cap C = \varnothing$ — both pairs share no outcomes, so both pairs are mutually exclusive.

When two events are mutually exclusive, $A \cap B = \varnothing$ — the OR rule then collapses to plain addition, because there is no overlap left to subtract.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.2 — the formal test

What mutually exclusive events are

Definition 3.2.2 — Mutually Exclusive Events

Two events $A$ and $B$ are mutually exclusive if they cannot occur at the same time. Equivalently,

P(A \text{ AND } B) = 0

To be mutually exclusive, the AND has to be exactly zero — a small shared probability is enough to rule it out.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Context Pause — default to "not mutually exclusive"

When in doubt, assume two events are not mutually exclusive.

The test is one line of arithmetic — assuming mutual exclusivity when the events actually overlap silently under-counts $P(A \text{ OR } B)$ and makes the answer too small.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — a complement in disguise

Try It Now 3.2.4

Draw two cards with replacement. Find $P(\text{at least one black})$ . (Black suits: clubs, spades.)

"At least one black" is the complement of "both red." With replacement the draws are independent, so $P(\text{both red}) = \frac12 \cdot \frac12 = \frac14$ , and

P(\text{at least one black}) = 1 - \frac14 = \frac34

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.2 — five events on two coins

Example 3.2.3 — Flipping two coins: events and their relationships

Sample space $\{HH, HT, TH, TT\}$ , each outcome equally likely.

$A$ = at most one tail = $\{HH,HT,TH\}$ , $P(A)=\frac34$ . $B$ = all tails = $\{TT\}$ , $P(B)=\frac14$ — $B=A'$ .

$C$ = all heads = $\{HH\}$ : $C \cap B = \varnothing$ , so $P(C \text{ AND } B)=0$ — $C$ and $B$ are mutually exclusive.

$D$ = more than one tail = $\{TT\}=B$ . $E$ = heads first = $\{HH,HT\}$ . $P(\text{at least one tail}) = 1-\frac14=\frac34$ .

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — one white, one red, with replacement

Try It Now 3.2.5

A box has one white, one red ball; select one, replace, select again. $W$ = white twice, $D$ = different colors, $F$ = white first. Find $P(W), P(D), P(F)$ ; are $W,D$ mutually exclusive? Are $W,F$ independent?

Answer: $P(W)=\frac14$ , $P(D)=\frac12$ , $P(F)=\frac12$ . $W=\{WW\}$ and $D=\{WR,RW\}$ share no outcomes — mutually exclusive. But $P(F\mid W)=1 \neq \frac12=P(F)$ , so $W$ and $F$ are not independent.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.2 — five probability questions on two coins

Example 3.2.4 — Flipping two coins: five probability questions

$F$ = at most one tail = $\{HH,HT,TH\}$ , $P(F)=\frac34$ . $G$ = same face twice = $\{HH,TT\}$ , $P(G)=\frac12$ .

$F \cap G = \{HH\}$ , so $P(F \text{ AND } G)=\frac14 \neq 0$ — $F$ and $G$ are not mutually exclusive.

$J$ = all tails = $\{TT\}$ : $F \cap J = \varnothing$ , so $F$ and $J$ are mutually exclusive — you cannot flip "at most one tail" and "all tails" together.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Try it in rāSHio

Let the AND value decide it

Open rāSHio and choose Calculators → Probability to check the five answers above against $\{HH, HT, TH, TT\}$ . Enter the counts for $F$ , $G$ and $J$ , then read $P(F \text{ AND } G)$ and $P(F \text{ AND } J)$ off the panel.

Mutual exclusivity is nothing more than that AND value landing on zero — seeing $\frac14$ beside $0$ puts the whole distinction on one screen.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.2 — reading it off the panel

rāSHio: Calculators → Probability

Figure 3.2.1: reading $P(A \text{ AND } B)$ and the two conditional probabilities off rāSHio's Calculators → Probability panel; run the same steps on the two-coin events above.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — Spanish and German

Try It Now 3.2.6

$S$ = learning Spanish, $G$ = learning German. $P(S)=0.4$ , $P(G)=0.2$ , $P(S \text{ AND } G)=0.08$ . Are $S$ and $G$ independent?

P(S) \cdot P(G) = (0.4)(0.2) = 0.08 = P(S \text{ AND } G)

Yes — the product matches the AND, so $S$ and $G$ are independent.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.2 — a die, a complement, a conditional

Example 3.2.5 — Rolling a die: complements, intersections, a conditional

$A$ = odd = $\{1,3,5\}$ , $B$ = even = $\{2,4,6\}$ — complements, $P(A)=P(B)=\frac12$ . $C$ = odd $>2$ = $\{3,5\}$ , $D$ = even $<5$ = $\{2,4\}$ : $C \cap D = \varnothing$ — mutually exclusive.

$E$ = $<5$ = $\{1,2,3,4\}$ : $C \cap E = \{3\}$ , so $P(C \text{ AND } E) = \frac16 \neq 0$ — $C,E$ are not mutually exclusive.

Conditioning on $A$ shrinks the space to $\{1,3,5\}$ ; two of three are in $C$ , so $P(C \mid A) = \frac23$ .

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — a red-and-green marble bag

Try It Now 3.2.7

Six red marbles (1–6) and four green marbles (1–4), ten outcomes total. $R$ = red, $O$ = odd-numbered. Find $P(R \text{ AND } O)$ .

Red-and-odd marbles: $R1, R3, R5$ — three of ten outcomes.

P(R \text{ AND } O) = \frac{3}{10} = 0.3

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Context Pause — three tests, one answer

The conditional form, the multiplication form, and the reverse conditional all encode the same idea.

$P(M \mid H) = P(M)$ , $P(M) \cdot P(H) = P(M \text{ AND } H)$ , and the reverse conditional all say: knowing one event tells you nothing about the other. Use whichever lines up with the numbers you have.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.3 — confirming independence two ways

Example 3.2.6 — Showing independence two ways

$M$ = taking math, $H$ = taking science. $P(M)=0.6$ , $P(H)=0.5$ , $P(M \text{ AND } H)=0.3$ .

P(M \mid H) = \frac{0.3}{0.5} = 0.6 = P(M) \qquad P(M)\cdot P(H) = (0.6)(0.5) = 0.3 = P(M \text{ AND } H)

Both tests agree: $M$ and $H$ are independent — knowing a student takes science does not change the chance they take math.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — books and DVDs at the library

Try It Now 3.2.8

$B$ = checks out a book, $D$ = checks out a DVD. $P(B)=0.40$ , $P(D)=0.30$ , $P(B \text{ AND } D)=0.20$ . Find $P(B\mid D)$ , $P(D\mid B)$ ; are $B,D$ independent? mutually exclusive?

Answer: $P(B\mid D)=\frac{2}{3}\approx0.667$ , $P(D\mid B)=0.5$ . $P(B)\cdot P(D)=0.12 \neq 0.20$ — not independent. $P(B \text{ AND } D)=0.20\neq0$ — not mutually exclusive.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.3 — English and speech classes

Example 3.2.7 — Three questions about two class events

$C$ = English, $D$ = speech. $P(C)=0.75$ , $P(D)=0.3$ , $P(C\mid D)=0.75$ , $P(C \text{ AND } D)=0.225$ .

$P(C\mid D)=0.75=P(C)$ → independent. $P(C \text{ AND } D)=0.225\neq0$ → not mutually exclusive. $P(D\mid C) = \frac{0.225}{0.75} = 0.3$ .

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — fans, colors, and a recovered probability

Try It Now 3.2.9

70% root for the home team; 25% wear blue; 20% wear blue and root away; of away-team fans, 67% wear blue. $A$ = roots away, $L$ = wears blue. Independent? Mutually exclusive?

Answer: $P(A)=0.30$ . $P(A)\cdot P(L) = 0.075 \neq 0.20 = P(A \text{ AND } L)$ — not independent, and $0.20\neq0$ , so not mutually exclusive either.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.3 — a colored, numbered deck

Example 3.2.8 — Drawing one card from a colored, numbered deck

Three red (1–3), five blue (1–5) — eight outcomes. $R$ , $B$ , $E$ =even. $P(R)=\frac38$ , $P(B)=\frac58$ , $P(R \text{ AND } B)=0$ — mutually exclusive (a card can't be both colors).

$P(E)=\frac38$ . Let $G$ =greater than 3, $H$ =blue 1–4: $P(G)=\frac14=P(G\mid H)$ — $G,H$ are independent.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — Sam's commute

Try It Now 3.2.10

$P(I)=0.44$ , $P(F)=0.56$ , $P(I \text{ AND } F)=0$ — Sam takes only one route. Find $P(I \text{ OR } F)$ .

Zero AND means mutually exclusive, so the OR rule is simple addition:

P(I \text{ OR } F) = 0.44 + 0.56 = 1.00

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Context Pause — three tests for independence

Pick whichever test matches the numbers you were given.

$P(W \text{ AND } L) = P(W)\cdot P(L)$ , $P(L\mid W)=P(L)$ , and the reverse conditional $P(W\mid L)=P(W)$ all encode the same idea — use whichever is convenient.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.3 — women and long hair

Example 3.2.9 — A class on women and long hair

60% of students are women; 50% have long hair; 45% are women with long hair; of the women, 75% have long hair. $W$ = woman, $L$ = long hair.

P(W)\cdot P(L) = (0.60)(0.50) = 0.30 \neq 0.45 = P(W \text{ AND } L)

Both tests agree: not independent — knowing a student is a woman raises the chance of long hair from 0.50 to 0.75.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Try it in rāSHio

Run both independence tests on the same numbers

Open rāSHio and choose Calculators → Probability, then enter $P(W)=0.60$ , $P(L)=0.50$ and $P(W \text{ AND } L)=0.45$ so the panel shows the product beside the AND, and $P(L\mid W)$ beside $P(L)$ .

The tool will not tell you the events are dependent — it just sets $0.30$ next to $0.45$ and $0.75$ next to $0.50$ , which is exactly the comparison both tests are asking you to make.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.3 — Independence and Mutual Exclusivity Are Different

Two different questions

Mutually exclusive

Asks: can both happen? $P(A \text{ AND } B)=0$ . For events with positive probability, this rules out independence — the product $P(A)\cdot P(B)$ would have to be $0$ too, and it can't be.

Independent

Asks: does one tell you anything about the other? Independent events can have a non-empty intersection, so they are usually not mutually exclusive.

Mutually exclusive events with positive probability are never independent — the two labels describe different relationships, not two ends of the same scale.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Your turn — white and red, with replacement, three events

Try It Now 3.2.11

Box: one white, one red, sampled with replacement. $W$ =white twice, $F$ =white first, $S$ =white second. Find $P(W)$ , $P(S\mid F)$ ; are $F,S$ independent? mutually exclusive? are $W,F$ independent?

Answer: $P(W)=\frac14$ , $P(S\mid F)=\frac12=P(S)$ — $F,S$ independent, and both can happen so not mutually exclusive. $W$ implies $F$ , so $P(F\mid W)=1\neq\frac12$ — $W,F$ not independent.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2.3 — a coin and a die, built out

Example 3.2.10 — Coin flip and die roll: building a sample space

$2 \times 6 = 12$ outcomes: $\{T1,\dots,T6,H1,\dots,H6\}$ . $A$ =heads-then-even = $\{H2,H4,H6\}$ , $P(A)=\frac{3}{12}$ . $B$ =heads-then-3 = $\{H3\}$ , $P(B)=\frac{1}{12}$ .

$A \cap B = \varnothing$ — mutually exclusive. But $P(A)\cdot P(B) = \frac{1}{48} \neq 0$ — the AND and the product disagree, so $A,B$ are dependent.

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

Key Terminology — the vocabulary this section put to work

independent events — two events $A$ and $B$ such that knowing one occurred does not change the probability the other occurred; $P(A\mid B)=P(A)$ or $P(A \text{ AND } B)=P(A)\cdot P(B)$ .

dependent events — two events that are not independent; knowing one changes the probability of the other.

mutually exclusive events — two events $A$ and $B$ that cannot occur at the same time; $P(A \text{ AND } B)=0$ .

sampling with replacement — each picked member is returned before the next pick, so a member can appear more than once and the picks are independent.

sampling without replacement — each picked member is set aside, so no member can appear twice and the picks are dependent.

complement — for an event $A$ , the outcomes in the sample space not in $A$ , written $A'$ .

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

The headline result

The AND value is the whole story — for both tests

$P(A \text{ AND } B) = 0$ means mutually exclusive. $P(A \text{ AND } B) = P(A)\cdot P(B)$ means independent. The same joint probability answers two different questions.

Mutually exclusive events with positive probability are never independent — and independent events, having a non-empty intersection, are usually not mutually exclusive.

† Default to dependent and to not mutually exclusive when a problem is silent on the relationship — both defaults are the conservative read, and both are wrong far less often than assuming the convenient shortcut.

3.2

Independent and Mutually Exclusive Events · bookSHelf Intro Stats§3.2

§3.2 — conclusions

What §3.2 leaves you with

The core idea

Independence and mutual exclusivity are two different questions with one shared number, $P(A \text{ AND } B)$ . Read it against a product for independence, against zero for mutual exclusivity — never assume either without checking.

Watch for

With replacement keeps draws independent; without replacement makes them dependent. And a positive AND rules out mutual exclusivity even when independence also fails — the two tests can both say "no" at once.

Next: §3.3 — Two Basic Rules of Probability, where these same AND and OR values become the multiplication and addition rules.