3.2 Independent and Mutually Exclusive Events
SLO 3
Describe and apply probability concepts and distributions.
Two tests decide how a pair of events relate: independence (one event does not change the other's chance) and mutual exclusivity (they cannot both happen). You learn to tell them apart on coin, dice and card setups, and to see why drawing with or without replacement changes the answer.
Learning Objectives
By the end of this section, you will be able to:
- determine whether two events are independent by checking one of three equivalent conditions;
- decide whether two events are mutually exclusive by computing their intersection;
- distinguish between independence and mutual exclusivity, which are often confused;
- apply the multiplication rule for independent events to compute \(P(A \text{ AND } B)\);
- recognize when sampling with versus without replacement drives dependence between draws.
"Independent" and "mutually exclusive" are not the same thing. They are so often confused that the rest of this section is really about pulling the two ideas apart, learning the test for each, and seeing what changes when an event tells you something about another.
3.2.1 Independent Events
Two events \(A\) and \(B\) are independent if the knowledge that one occurred does not change the probability that the other occurs. Formally, \(A\) and \(B\) are independent when any one of the following holds (they are equivalent when \(P(A), P(B) > 0\)):
$$P(A \mid B) = P(A) \quad \text{or} \quad P(B \mid A) = P(B) \quad \text{or} \quad P(A \text{ AND } B) = P(A) \cdot P(B)$$If two events are NOT independent, they are called dependent.
Sampling may be done with replacement or without replacement, and the choice matters.
- With replacement: After you pick a member of the population, you put it back before the next pick. The same member can appear in the sample more than once, and the result of the first pick does not change the probabilities for the second pick. With-replacement draws are independent.
- Without replacement: Once you pick a member, you set it aside. The second pick happens from a smaller population, and its probabilities are affected by the result of the first pick. Without-replacement draws are dependent (not independent).
If the problem does not tell you 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 leads to probabilities bigger than 1.
Two events \(A\) and \(B\) are independent when knowing that one occurred does not change the chance the other occurs. Roll a fair die twice. Whatever the first roll shows, the second roll still has six equally likely faces. The two outcomes are independent.
The formal test is short. Two events are independent if any one of these three equations holds:
$$P(A \mid B) = P(A)$$ $$P(B \mid A) = P(B)$$ $$P(A \text{ AND } B) = P(A) \cdot P(B)$$The three are equivalent when \(P(A)\) and \(P(B)\) are both positive. Pick whichever is easiest given the numbers you have. If none of them holds, the events are dependent — knowing one changes the chance of the other.
The multiplication rule \(P(A \text{ AND } B) = P(A) \cdot P(B)\) is what independence actually buys you. When two events are independent, the joint probability is just the product of the individual probabilities. Sections 3.3 and 3.4 generalize that rule to events that are not independent, where the AND has to be reconstructed from a conditional probability instead.
You draw two cards from a fair, well-shuffled 52-card deck.
a. You draw the first card, note it, and put it back in the deck before drawing the second card. Are the two draws independent?
b. You draw the first card and set it aside before drawing the second card. Are the two draws independent?
Solution
Part a — with replacement is independent. Putting the first card back restores the deck to its full 52 cards before the second draw, so the result of the first draw has no effect on the second. The two events are independent.
Part b — without replacement is dependent. After the first card is set aside, only 51 cards remain, and the probabilities for the second draw depend on which card you pulled first. The two events are dependent.
Answer: (a) independent, (b) dependent.
You have a fair, well-shuffled deck of 52 cards. It consists of four suits (clubs, diamonds, hearts, spades), with 13 cards in each suit — the numbers 1 through 10, plus the jack, queen, and king of that suit.
a. Sampling with replacement. Pick three cards with replacement. The first card is the jack of spades. You put it back, shuffle, and draw again — the ten of clubs. You put it back, shuffle, and draw a third time — the jack of spades again. Your three picks are {jack of spades, ten of clubs, jack of spades}. You picked the jack of spades twice. Each draw happens from a full 52-card deck, so the three events are independent.
b. Sampling without replacement. Pick three cards without replacement. The first card is the seven of hearts. You set it aside, so 51 cards remain. The second card is the three of diamonds. You set it aside, so 50 cards remain. The third card is the jack of spades. Your three picks are {seven of hearts, three of diamonds, jack of spades}. Because you did not put the first card back, you cannot pick the same card twice, and the probabilities for each new draw depend on what was already removed. The three events are dependent.
Solution
The two scenarios differ in only one detail: what happens to the first card. Returning the card to the deck keeps the sample space the same size for every draw, which keeps the events independent. Setting the card aside shrinks the sample space by one, which couples the second and third draws to the first.
Answer: (a) the three draws are independent; (b) the three draws are dependent.
You pick three cards at random from a fair, well-shuffled 52-card deck.
a. Suppose you know the three picked cards are {jack of spades, seven of hearts, jack of spades}. Was the sampling with or without replacement?
b. Suppose you know the three picked cards are {jack of spades, seven of hearts, five of diamonds}. Was the sampling with or without replacement?
Solution
Part a — look for repeats. The jack of spades appears twice, and the deck has only one jack of spades. You can only see the same physical card twice if you put it back between draws.
Part b — every card is distinct. The three cards are three different cards, which is consistent with either sampling plan. The information given is not enough to decide.
Answer: (a) with replacement; (b) cannot be determined from the cards alone.
A fair, well-shuffled deck of 52 cards. Spades, hearts, diamonds, and clubs. The values are 1 through 10 plus jack, queen, and king. Use \(J\) for jack, \(Q\) for queen, \(K\) for king, and \(A\) for ace when needed.
a. Suppose you pick four cards and do not put them back. Your cards are \(\{J\spadesuit,\; 5\heartsuit,\; 5\diamondsuit,\; K\clubsuit\}\).
b. Suppose you pick four cards and put each card back before the next pick. Your cards are \(\{J\spadesuit,\; 7\heartsuit,\; 6\heartsuit,\; K\spadesuit\}\).
Which is with replacement and which is without replacement?
Solution
Part a. The two 5s are different suits (hearts and diamonds), so you can draw two 5s without replacement. The hand has no card that appears twice, which is consistent with without-replacement sampling.
Part b. The hand has a 7 and a 6, both hearts — that is fine without replacement too, but the only way to be sure is the surrounding context. In this problem, the text says you put each card back before the next pick, so it is with replacement.
Answer: (a) without replacement; (b) with replacement.
Suppose you sample four cards without replacement from a fair, well-shuffled 52-card deck. Which of the following outcomes are possible? Answer the same question for sampling with replacement.
a. \(\{J\spadesuit,\; 5\heartsuit,\; 5\diamondsuit,\; K\clubsuit\}\)
b. \(\{Q\spadesuit,\; 7\heartsuit,\; 6\heartsuit,\; K\spadesuit\}\)
c. \(\{J\spadesuit,\; 7\heartsuit,\; 6\heartsuit,\; K\spadesuit\}\)
Solution
Part a. Each card is distinct and the deck has one of each, so this is possible under either sampling plan.
Part b. Each card is distinct, so this is possible under either sampling plan. (Two cards are spades, but the deck has four jacks, four queens, and four kings, so there is no repeat.)
Part c. The jack and king are both spades, which is fine. The 7 and 6 are both hearts — also fine without replacement. Every card is distinct, so it is possible under either plan.
Answer: All three hands are possible under both sampling plans. The hands in (a), (b), and (c) contain no repeated cards, so they give no information to distinguish with-replacement from without-replacement sampling.
3.2.2 Mutually Exclusive Events
Two events \(A\) and \(B\) are mutually exclusive if they cannot occur at the same time. Equivalently, \(P(A \text{ AND } B) = 0\).
Two events \(A\) and \(B\) are mutually exclusive if they cannot occur at the same time. In set language, \(A\) and \(B\) share no outcomes, so
$$P(A \text{ AND } B) = 0$$When two events are mutually exclusive, \(A \cap B = \varnothing\), the empty set. The OR rule for them collapses to the addition rule because there is no overlap to subtract.
For example, take the sample space \(S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\). Let \(A = \{1, 2, 3, 4, 5\}\), \(B = \{4, 5, 6, 7, 8\}\), and \(C = \{7, 9\}\). Then \(A \cap B = \{4, 5\}\), so
$$P(A \text{ AND } B) = \frac{2}{10} \neq 0$$and \(A\) and \(B\) are not mutually exclusive. But \(A \cap C = \varnothing\) and \(B \cap C = \varnothing\), so neither pair shares any outcomes and both pairs are mutually exclusive.
When in doubt, assume two events are not mutually exclusive. The test is the same as for independent events — one line of arithmetic — and assuming mutual exclusivity when the events actually overlap silently under-counts \(P(A \text{ OR } B)\) and makes the answer too small.
Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card. (The black suits are clubs and spades.)
Solution
Step 1 — compute the complement. "At least one black" is the complement of "no black cards", which is the same as "both cards are red". With replacement, the two draws are independent.
Step 2 — find the probability of one red draw. Half the deck is red, so \(P(\text{red}) = \frac{26}{52} = \frac{1}{2}\).
Step 3 — apply the complement rule. Because the two draws are independent,
$$P(\text{both red}) = \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) = \frac{1}{4}$$Step 4 — subtract from 1.
$$P(\text{at least one black}) = 1 - P(\text{both red}) = 1 - \frac{1}{4} = \frac{3}{4}$$Answer: \(\frac{3}{4}\).
Flip two fair coins. The sample space is \(\{HH,\; HT,\; TH,\; TT\}\), where the first letter is the first coin and the second letter is the second coin. \(H\) is heads, \(T\) is tails. So \(HT\) means heads-then-tails and \(TH\) means tails-then-heads; the two are different outcomes.
Let \(A\) be the event of getting at most one tail — zero tails or one tail. Then
$$A = \{HH,\; HT,\; TH\}$$The outcome \(HH\) shows zero tails, and \(HT\) and \(TH\) each show one tail.
Let \(B\) be the event of getting all tails. Then \(B = \{TT\}\). The events \(A\) and \(B\) are complements of each other, so \(B = A'\) and
$$P(A) + P(B) = P(A) + P(A') = 1$$The four outcomes are equally likely, so
$$P(A) = \frac{3}{4} \quad \text{and} \quad P(B) = \frac{1}{4}$$Let \(C\) be the event of getting all heads. Then \(C = \{HH\}\). Since \(C \cap B = \{HH\} \cap \{TT\} = \varnothing\),
$$P(C \text{ AND } B) = 0$$and \(C\) and \(B\) are mutually exclusive. You cannot flip both heads and tails on every coin at the same time.
Let \(D\) be the event of getting more than one tail. Then \(D = \{TT\}\), so \(D = B\), and
$$P(D) = \frac{1}{4}$$Let \(E\) be the event of getting a head on the first flip. This implies the second flip can be either heads or tails, so
$$E = \{HH,\; HT\}$$The probability of getting at least one tail in two flips is \(1 - P(\text{no tails}) = 1 - P(\text{all heads}) = 1 - P(\{HH\}) = 1 - \frac{1}{4} = \frac{3}{4}\).
Solution
Every probability here is just a count out of four. The four outcomes \(\{HH, HT, TH, TT\}\) are equally likely because the two flips are independent, so each has probability \(\frac{1}{4}\).
- \(P(A) = \frac{3}{4}\) because \(A\) contains three of the four outcomes.
- \(P(B) = \frac{1}{4}\) because \(B = \{TT\}\) is one outcome.
- \(P(C \text{ AND } B) = 0\) because \(C = \{HH\}\) and \(B = \{TT\}\) share no outcomes; \(C\) and \(B\) are mutually exclusive.
- \(P(D) = \frac{1}{4}\) because \(D = \{TT\}\) is one outcome.
- \(P(\text{at least one tail}) = \frac{3}{4}\).
Answer: \(P(A) = \frac{3}{4},\; P(B) = \frac{1}{4},\; P(C \text{ AND } B) = 0,\; P(D) = \frac{1}{4},\; P(\text{at least one tail}) = \frac{3}{4}\).
A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
a. Let \(W\) be the event of getting the white ball twice.
b. Let \(D\) be the event of getting two balls of different colors.
c. Let \(F\) be the event of getting white on the first pick.
d. Are \(W\) and \(D\) mutually exclusive?
e. Are \(W\) and \(F\) independent?
Solution
Part a — both draws are white. The two draws are independent, and \(P(\text{white}) = \frac{1}{2}\) on each draw, so
$$P(W) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$Part b — different colors. Either white then red or red then white, each with probability \(\frac{1}{4}\), so
$$P(D) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$Part c — white on the first draw.
$$P(F) = \frac{1}{2}$$Part d — are W and D mutually exclusive? \(W = \{WW\}\) and \(D = \{WR,\; RW\}\). They share no outcomes, so \(W \cap D = \varnothing\) and \(W\) and \(D\) are mutually exclusive.
Part e — are W and F independent? If \(W\) happens, the first pick is white, so \(F\) definitely happened; \(P(F \mid W) = 1\), but \(P(F) = \frac{1}{2}\). The two are not equal, so \(W\) and \(F\) are not independent.
Answer: (a) \(\frac{1}{4}\), (b) \(\frac{1}{2}\), (c) \(\frac{1}{2}\), (d) yes — mutually exclusive, (e) no — not independent.
Flip two fair coins. The sample space is \(\{HH,\; HT,\; TH,\; TT\}\).
a. Let \(F\) be the event of getting at most one tail (zero tails or one tail).
b. Let \(G\) be the event of getting two faces that are the same.
c. Let \(H\) be the event of getting a head on the first flip followed by a head or a tail on the second flip.
d. Are \(F\) and \(G\) mutually exclusive?
e. Let \(J\) be the event of getting all tails. Are \(F\) and \(J\) mutually exclusive?
Solution
Part a — count zero-tail and one-tail outcomes. Zero tails gives \(\{HH\}\); one tail gives \(\{HT,\; TH\}\). Together,
$$F = \{HH,\; HT,\; TH\}, \qquad P(F) = \frac{3}{4}$$Part b — same-face outcomes. Two faces the same means both heads or both tails:
$$G = \{HH,\; TT\}, \qquad P(G) = \frac{2}{4} = \frac{1}{2}$$Part c — heads on the first flip. The first flip is heads, the second can be either face:
$$H = \{HH,\; HT\}, \qquad P(H) = \frac{2}{4} = \frac{1}{2}$$Part d — are F and G mutually exclusive? Their intersection is \(\{HH\}\), which is not empty, so
$$P(F \text{ AND } G) = \frac{1}{4} \neq 0$$and \(F\) and \(G\) are not mutually exclusive.
Part e — are F and J mutually exclusive? \(J = \{TT\}\) and \(F = \{HH,\; HT,\; TH\}\). The two sets share no outcomes, so \(F \cap J = \varnothing\) and
$$P(F \text{ AND } J) = 0$$which means \(F\) and \(J\) are mutually exclusive. You cannot flip "at most one tail" and "all tails" on the same two flips.
Answer: \(P(F) = \frac{3}{4},\; P(G) = \frac{1}{2},\; P(H) = \frac{1}{2};\; F\) and \(G\) are not mutually exclusive; \(F\) and \(J\) are mutually exclusive.
Let the AND value decide it
Open rāSHio and choose Calculators → Probability to check the five answers above against the four-outcome sample space \(\{HH,\; HT,\; TH,\; TT\}\). Enter the counts you found 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, and seeing \(\tfrac{1}{4}\) beside \(0\) puts the whole distinction on one screen.
Figure 3.2.1 — Reading P(A AND B) and the two conditional probabilities off rāSHio's Calculators → Probability panel; run the same steps on the two-coin events above.
Let \(S\) be the event that a randomly chosen student is learning Spanish. Let \(G\) be the event that the student is learning German. Then \(S \text{ AND } G\) is the event that the student is learning both languages. Suppose \(P(S) = 0.4\), \(P(G) = 0.2\), and \(P(S \text{ AND } G) = 0.08\). Are \(S\) and \(G\) independent?
Solution
Step 1 — pick the easiest test. The numbers give us the AND and the two individual probabilities, so the multiplication form is the natural test:
$$P(S) \cdot P(G) = (0.4)(0.2) = 0.08$$Step 2 — compare. \(P(S \text{ AND } G) = 0.08\) and \(P(S) \cdot P(G) = 0.08\). The two are equal, so the events are independent.
Step 3 — sanity-check with the other forms. \(P(S \mid G) = \frac{0.08}{0.2} = 0.4 = P(S)\). The conditional equals the marginal, which is what the independence test says should happen.
Answer: Yes — \(S\) and \(G\) are independent.
Roll one fair, six-sided die. The sample space is \(\{1, 2, 3, 4, 5, 6\}\).
Let \(A\) be the event of rolling an odd face. Then \(A = \{1, 3, 5\}\).
Let \(B\) be the event of rolling an even face. Then \(B = \{2, 4, 6\}\).
Complement of \(A\). The complement of \(A\) is \(A' = B\), because \(A\) and \(B\) together make up the sample space. By the complement rule,
$$P(A) + P(B) = 1$$and since each outcome is equally likely, \(P(A) = P(B) = \frac{3}{6} = \frac{1}{2}\).
Let \(C\) be the event of rolling an odd face larger than two. Then \(C = \{3, 5\}\).
Let \(D\) be the event of rolling an even face smaller than five. Then \(D = \{2, 4\}\).
Are \(C\) and \(D\) mutually exclusive? A face cannot be both odd and even at the same time, so \(C \cap D = \varnothing\),
$$P(C \text{ AND } D) = 0$$and \(C\) and \(D\) are mutually exclusive.
Let \(E\) be the event of rolling a face less than five. Then \(E = \{1, 2, 3, 4\}\).
Problem. Are \(C\) and \(E\) mutually exclusive? Why or why not?
Solution. No. \(C = \{3, 5\}\) and \(E = \{1, 2, 3, 4\}\). The two events share the outcome 3, so
$$P(C \text{ AND } E) = \frac{1}{6} \neq 0$$To be mutually exclusive, the AND has to be zero.
Find \(P(C \mid A)\). This is a conditional probability. \(C = \{3, 5\}\) and \(A = \{1, 3, 5\}\). To find \(P(C \mid A)\), find the probability of \(C\) inside the reduced sample space \(A\). The original sample space \(\{1, 2, 3, 4, 5, 6\}\) shrinks to \(A = \{1, 3, 5\}\), so
$$P(C \mid A) = \frac{2}{3}$$Solution
Three things to read off the die.
- \(A\) and \(B\) are complements and each has probability \(\frac{1}{2}\).
- \(C\) and \(D\) are mutually exclusive because an outcome cannot be both odd and even.
- \(C\) and \(E\) are not mutually exclusive: outcome 3 belongs to both, so their AND has probability \(\frac{1}{6}\).
- Conditioning on \(A\) shrinks the sample space from \(\{1, 2, 3, 4, 5, 6\}\) to \(\{1, 3, 5\}\), and two of the three are in \(C\), so \(P(C \mid A) = \frac{2}{3}\).
Answer: \(C\) and \(E\) are not mutually exclusive; \(P(C \mid A) = \frac{2}{3}\).
In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4.
- Let \(R\) be the event of drawing a red marble.
- Let \(G\) be the event of drawing a green marble.
- Let \(O\) be the event of drawing an odd-numbered marble.
The sample space is \(S = \{R1, R2, R3, R4, R5, R6, G1, G2, G3, G4\}\) — ten outcomes.
What is \(P(R \text{ AND } O)\)?
Solution
Step 1 — identify the favorable outcomes. A red AND odd marble is one of the six red marbles whose number is odd: \(R1, R3, R5\).
Step 2 — count and divide. Three favorable outcomes out of ten total, so
$$P(R \text{ AND } O) = \frac{3}{10} = 0.3$$Answer: \(\frac{3}{10} = 0.3\).
The conditional form \(P(M \mid H) = P(M)\), the multiplication form \(P(M) \cdot P(H) = P(M \text{ AND } H)\), and the reverse conditional all encode the same idea — the events are independent when knowing one tells you nothing about the other. Use whichever one lines up with the information you have.
Let \(M\) be the event that a student is taking a math class. Let \(H\) be the event that the student is taking a science class. Then \(M \text{ AND } H\) is the event that the student is taking both. Suppose \(P(M) = 0.6\), \(P(H) = 0.5\), and \(P(M \text{ AND } H) = 0.3\).
a. Show that \(P(M \mid H) = P(M)\). By the definition of conditional probability,
$$P(M \mid H) = \frac{P(M \text{ AND } H)}{P(H)} = \frac{0.3}{0.5} = 0.6 = P(M)$$b. Show that \(P(M \text{ AND } H) = P(M) \cdot P(H)\).
$$P(M) \cdot P(H) = (0.6)(0.5) = 0.3 = P(M \text{ AND } H)$$Since \(M\) and \(H\) are independent, knowing a student is taking a science class does not change the chance they are taking a math class. If the two events had not been independent, then knowing one would change the probability of the other.
Solution
Both tests pass:
- \(P(M \mid H) = 0.6 = P(M)\).
- \(P(M) \cdot P(H) = 0.3 = P(M \text{ AND } H)\).
Either one is enough to confirm independence. Doing both is good practice and confirms the same answer two different ways.
Answer: \(M\) and \(H\) are independent.
A student goes to the library. Let \(B\) be the event that the student checks out a book and \(D\) be the event that the student checks out a DVD. Suppose \(P(B) = 0.40\), \(P(D) = 0.30\), and \(P(B \text{ AND } D) = 0.20\).
a. Find \(P(B \mid D)\).
b. Find \(P(D \mid B)\).
c. Are \(B\) and \(D\) independent?
d. Are \(B\) and \(D\) mutually exclusive?
Solution
Part a — \(P(B \mid D)\). Use the definition of conditional probability:
$$P(B \mid D) = \frac{P(B \text{ AND } D)}{P(D)} = \frac{0.20}{0.30} = \frac{2}{3} \approx 0.667$$Part b — \(P(D \mid B)\).
$$P(D \mid B) = \frac{P(B \text{ AND } D)}{P(B)} = \frac{0.20}{0.40} = 0.5$$Part c — independence. Compare the product to the AND:
$$P(B) \cdot P(D) = (0.40)(0.30) = 0.12 \neq 0.20 = P(B \text{ AND } D)$$The product and the AND disagree, so \(B\) and \(D\) are not independent. The same answer falls out of the conditional: \(P(B \mid D) = \frac{2}{3} \neq 0.4 = P(B)\).
Part d — mutual exclusivity. The AND is \(0.20\), not zero, so \(B\) and \(D\) are not mutually exclusive.
Answer: (a) \(\frac{2}{3}\), (b) 0.5, (c) not independent, (d) not mutually exclusive.
Let \(C\) be the event that a student is taking an English class. Let \(D\) be the event that a student is taking a speech class. Suppose \(P(C) = 0.75\), \(P(D) = 0.3\), \(P(C \mid D) = 0.75\), and \(P(C \text{ AND } D) = 0.225\). Justify your answers numerically.
a. Are \(C\) and \(D\) independent?
b. Are \(C\) and \(D\) mutually exclusive?
c. What is \(P(D \mid C)\)?
Solution
Part a — independence. The conditional is given and equals the marginal:
$$P(C \mid D) = 0.75 = P(C)$$So \(C\) and \(D\) are independent.
Part b — mutual exclusivity. The AND is positive, so they share outcomes:
$$P(C \text{ AND } D) = 0.225 \neq 0$$So \(C\) and \(D\) are not mutually exclusive.
Part c — find \(P(D \mid C)\). Use the definition of conditional probability:
$$P(D \mid C) = \frac{P(C \text{ AND } D)}{P(C)} = \frac{0.225}{0.75} = 0.3$$Answer: (a) independent, (b) not mutually exclusive, (c) \(P(D \mid C) = 0.3\).
In a basketball arena:
- 70% of the fans are rooting for the home team.
- 25% of the fans are wearing blue.
- 20% of the fans are wearing blue and are rooting for the away team.
- Of the fans rooting for the away team, 67% are wearing blue.
Let \(A\) be the event that a fan is rooting for the away team. Let \(L\) be the event that a fan is wearing blue. Are the events \(A\) and \(L\) independent? Are they mutually exclusive?
Solution
Step 1 — recover the missing probability. 70% root for the home team, so 30% root for the away team: \(P(A) = 0.30\). \(P(L) = 0.25\) and \(P(L \text{ AND } A) = 0.20\).
Step 2 — check independence.
$$P(A) \cdot P(L) = (0.30)(0.25) = 0.075$$ $$P(A \text{ AND } L) = 0.20$$The product is 0.075, the AND is 0.20 — they disagree, so \(A\) and \(L\) are not independent. (The given fact that 67% of away-team fans wear blue is the same conclusion from the conditional side: \(P(L \mid A) = 0.67 \neq 0.25 = P(L)\).)
Step 3 — check mutual exclusivity. \(P(A \text{ AND } L) = 0.20 \neq 0\), so \(A\) and \(L\) are not mutually exclusive.
Answer: Not independent and not mutually exclusive.
In a box there are three red cards and five blue cards. The red cards are marked 1, 2, 3 and the blue cards are marked 1, 2, 3, 4, 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.
Let \(R\) be the event of drawing a red card, \(B\) the event of drawing a blue card, and \(E\) the event of drawing an even-numbered card.
The sample space is \(S = \{R1, R2, R3, B1, B2, B3, B4, B5\}\) — eight outcomes.
Marginal probabilities.
$$P(R) = \frac{3}{8}, \qquad P(B) = \frac{5}{8}, \qquad P(R \text{ AND } B) = 0$$The last probability is zero because a card cannot be both red and blue.
The even-numbered cards are \(R2, B2, B4\) — three of them, so \(P(E) = \frac{3}{8}\).
Conditioning on color.
$$P(E \mid B) = \frac{2}{5}$$(There are five blue cards: 1, 2, 3, 4, 5. Of those, two are even: 2 and 4.)
$$P(B \mid E) = \frac{2}{3}$$(There are three even-numbered cards: \(R2, B2, B4\). Of those, two are blue: \(B2\) and \(B4\).)
Mutual exclusivity. \(R\) and \(B\) are mutually exclusive because \(P(R \text{ AND } B) = 0\).
Independence. Let \(G\) be the event of drawing a card with a number greater than 3. Then \(G = \{B4, B5\}\), and \(P(G) = \frac{2}{8} = \frac{1}{4}\). Let \(H\) be the event of drawing a blue card numbered between 1 and 4 inclusive. Then \(H = \{B1, B2, B3, B4\}\), and \(P(G \mid H) = \frac{1}{4}\). Since
$$P(G) = \frac{1}{4} = P(G \mid H)$$\(G\) and \(H\) are independent.
Solution
- \(P(R) = \frac{3}{8}\), \(P(B) = \frac{5}{8}\), \(P(E) = \frac{3}{8}\).
- \(P(E \mid B) = \frac{2}{5}\), \(P(B \mid E) = \frac{2}{3}\).
- \(R\) and \(B\) are mutually exclusive.
- \(G\) and \(H\) are independent because \(P(G \mid H) = P(G) = \frac{1}{4}\).
Answer: The marginals are \(\frac{3}{8}, \frac{5}{8}, \frac{3}{8}\); the conditionals are \(\frac{2}{5}\) and \(\frac{2}{3}\); \(R\) and \(B\) are mutually exclusive; \(G\) and \(H\) are independent.
Sam is deciding which route to take to work. Their choices are \(I\) = the Interstate and \(F\) = Fifth Street.
- \(P(I) = 0.44\) and \(P(F) = 0.56\).
- \(P(I \text{ AND } F) = 0\) because Sam will take only one route to work.
What is the probability of \(I \text{ OR } F\)?
Solution
Step 1 — recognize the structure. Because \(P(I \text{ AND } F) = 0\), the two events are mutually exclusive, and the OR rule for mutually exclusive events is the simple addition rule.
Step 2 — add.
$$P(I \text{ OR } F) = P(I) + P(F) = 0.44 + 0.56 = 1.00$$This is the right answer — Sam must take one of the two routes.
Answer: \(1.00\).
The multiplication form \(P(W \text{ AND } L) = P(W) \cdot P(L)\), the conditional form \(P(L \mid W) = P(L)\), and the reverse conditional \(P(W \mid L) = P(W)\) all encode the same idea. Use whichever one is convenient given what the problem gives you.
In a particular college class, 60% of the students are women. Fifty percent of all students in the class have long hair. Forty-five percent of the students are women and have long hair. Of the women students, 75% have long hair.
Let \(W\) be the event that a student is a woman and \(L\) be the event that a student has long hair. One student is picked randomly. Are the events of being a woman and having long hair independent?
The given probabilities are:
$$P(W) = 0.60, \qquad P(L) = 0.50, \qquad P(W \text{ AND } L) = 0.45, \qquad P(L \mid W) = 0.75$$Solution
Test 1 — multiplication form. We are given \(P(W \text{ AND } L) = 0.45\), and
$$P(W) \cdot P(L) = (0.60)(0.50) = 0.30$$The AND and the product are not equal, so the events are not independent.
Test 2 — conditional form. We are given \(P(L \mid W) = 0.75\), and \(P(L) = 0.50\). They are not equal, so the events are not independent.
Both tests give the same answer: the events of being a woman and having long hair are not independent, because knowing a student is a woman changes the chance the student has long hair (from 0.50 to 0.75).
Answer: Not independent. Knowing a student is a woman raises the probability of long hair from 0.50 to 0.75.
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 \(P(W) \cdot P(L)\) beside the AND you were given, 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 in this example are asking you to make.
3.2.3 Independence and Mutual Exclusivity Are Different
Pick whichever of the two equivalent statements is easiest to check, given the numbers in front of you. The first uses a conditional: the chance of \(A\) given \(B\) should match the chance of \(A\) on its own. The second uses the AND: the chance of both together should equal the product of their individual chances. Either one is enough. If the equation you chose is true, \(A\) and \(B\) are independent. If it is false, they are dependent.
$$P(A \mid B) = P(A) \qquad P(A \text{ AND } B) = P(A) \cdot P(B)$$Independence and mutual exclusivity are different relationships. Two events are mutually exclusive if they cannot happen at the same time — their AND is zero. Two events are independent if knowing one tells you nothing about the other — the conditional equals the marginal, or equivalently the AND equals the product.
For two events with positive probability:
- Mutually exclusive implies \(P(A \text{ AND } B) = 0\), while independence would require \(P(A \text{ AND } B) = P(A) \cdot P(B) > 0\). So mutually exclusive events with positive probability are never independent.
- Independent events can have a non-empty intersection, so they are usually not mutually exclusive.
The two ideas are about different questions: mutual exclusivity asks "can both happen?" and independence asks "does one tell you anything about the other?".
A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let \(W\) be the event of getting the white ball twice, \(F\) the event of getting the white ball on the first draw, and \(S\) the event of getting the white ball on the second draw.
a. Compute \(P(W)\).
b. Compute \(P(S \mid F)\).
c. Are \(F\) and \(S\) independent?
d. Are \(F\) and \(S\) mutually exclusive?
e. Are \(W\) and \(F\) independent?
Solution
Part a — both draws are white.
$$P(W) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$Part b — white on the second draw given white on the first. The first ball is replaced, so the second draw is from a full box:
$$P(S \mid F) = \frac{1}{2}$$Part c — independence of F and S. With replacement, the two draws are independent:
$$P(S \mid F) = \frac{1}{2} = P(S)$$So \(F\) and \(S\) are independent.
Part d — mutual exclusivity of F and S. They can both happen (white-white), so \(P(F \text{ AND } S) = \frac{1}{4} \neq 0\), and \(F\) and \(S\) are not mutually exclusive.
Part e — independence of W and F. \(W\) implies \(F\), so \(P(F \mid W) = 1 \neq \frac{1}{2} = P(F)\), and \(W\) and \(F\) are not independent.
Answer: (a) \(\frac{1}{4}\), (b) \(\frac{1}{2}\), (c) independent, (d) not mutually exclusive, (e) not independent.
a. Toss one fair coin (the coin has two sides, \(H\) and \(T\)). The outcomes are \(H\) and \(T\). Count the outcomes: there are 2 outcomes.
b. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5, or 6 dots on a side). The outcomes are 1, 2, 3, 4, 5, 6. Count the outcomes: there are 6 outcomes.
c. Multiply the two numbers of outcomes. The answer is \(2 \times 6 = 12\).
d. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in part c is the number of outcomes (size of the sample space). The outcomes are:
$$T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6$$e. Event \(A\) = heads (\(H\)) on the coin followed by an even number (2, 4, 6) on the die. Then
$$A = \{H2, H4, H6\}, \qquad P(A) = \frac{3}{12}$$f. Event \(B\) = heads on the coin followed by a three on the die. Then
$$B = \{H3\}, \qquad P(B) = \frac{1}{12}$$g. Are \(A\) and \(B\) mutually exclusive? Hint: what is \(P(A \text{ AND } B)\)? If \(P(A \text{ AND } B) = 0\), then \(A\) and \(B\) are mutually exclusive.
h. Are \(A\) and \(B\) independent? Hint: is \(P(A \text{ AND } B) = P(A) \cdot P(B)\)? If yes, the events are independent; if not, they are dependent.
Solution
Part a. Two outcomes, \(H\) and \(T\).
Part b. Six outcomes, 1 through 6.
Part c. \(2 \times 6 = 12\) outcomes.
Part d. The twelve outcomes are \(\{T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6\}\).
Part e. \(A = \{H2, H4, H6\}\), so \(P(A) = \frac{3}{12} = \frac{1}{4}\).
Part f. \(B = \{H3\}\), so \(P(B) = \frac{1}{12}\).
Part g. \(A \cap B = \varnothing\) (no outcome is both an even-coin-die result and the singleton \(H3\)). So \(P(A \text{ AND } B) = 0\), and \(A\) and \(B\) are mutually exclusive.
Part h. Compare the AND to the product:
$$P(A) \cdot P(B) = \frac{3}{12} \cdot \frac{1}{12} = \frac{3}{144} = \frac{1}{48}$$The AND is 0 and the product is \(\frac{1}{48}\); they are not equal, so \(A\) and \(B\) are dependent.
Answer: (a) 2, (b) 6, (c) 12, (d) 12 outcomes listed, (e) \(P(A) = \frac{1}{4}\), (f) \(P(B) = \frac{1}{12}\), (g) mutually exclusive, (h) dependent.
Problem Set 3.2
Problem 1. A fair coin is flipped twice. Let \(A\) be the event of getting heads on the first flip and \(B\) the event of getting heads on the second flip. Are \(A\) and \(B\) independent? Are they mutually exclusive?
Problem 1 Solution
Step 1 — list the sample space. Two coin flips give four equally likely outcomes: \(\{HH,\; HT,\; TH,\; TT\}\).
Step 2 — read off the events. \(A\) = heads on the first flip = \(\{HH,\; HT\}\). \(B\) = heads on the second flip = \(\{HH,\; TH\}\). Both have probability \(\frac{1}{2}\).
Step 3 — test independence. The two flips are independent coin flips, so \(P(B \mid A) = P(B) = \frac{1}{2}\). Equivalently, \(P(A \text{ AND } B) = P(\{HH\}) = \frac{1}{4} = (\frac{1}{2})(\frac{1}{2}) = P(A) \cdot P(B)\). Either test says the events are independent.
Step 4 — test mutual exclusivity. \(A \cap B = \{HH\}\), so \(P(A \text{ AND } B) = \frac{1}{4} \neq 0\). The events are not mutually exclusive.
Answer: Independent, not mutually exclusive.
Problem 2. A card is drawn at random from a standard 52-card deck. Let \(H\) be the event that the card belongs to the heart suit and \(F\) the event that the card is a face card (jack, queen, or king). Are \(H\) and \(F\) independent?
Problem 2 Solution
Step 1 — find the marginal probabilities. Of 52 cards, 13 are hearts and 12 are face cards (4 jacks, 4 queens, 4 kings). So \(P(A) = \frac{13}{52} = \frac{1}{4}\) and \(P(B) = \frac{12}{52} = \frac{3}{13}\).
Step 2 — find the joint probability. A card that is both a heart and a face card is one of the three face cards in the heart suit (jack, queen, king of hearts). So \(P(A \text{ AND } B) = \frac{3}{52}\).
Step 3 — test independence. \(P(A) \cdot P(B) = \frac{1}{4} \cdot \frac{3}{13} = \frac{3}{52} = P(A \text{ AND } B)\). The product matches the AND, so \(A\) and \(B\) are independent.
Answer: Yes, \(A\) and \(B\) are independent.
Problem 3. A box contains 5 red balls and 7 blue balls. Two balls are drawn without replacement. Let \(A\) be the event that the first ball is red and \(B\) the event that the second ball is red. Are \(A\) and \(B\) independent?
Problem 3 Solution
Step 1 — set up the marginals. \(P(\text{first red}) = \frac{5}{12}\). After drawing one red ball, 11 balls remain (4 red, 7 blue), so \(P(\text{second red} \mid \text{first red}) = \frac{4}{11}\). Since \(\frac{4}{11} \neq \frac{5}{12}\), the two draws are not independent.
Step 2 — verify with the multiplication form. \(P(A) \cdot P(B) = \frac{5}{12} \cdot \frac{5}{12} = \frac{25}{144}\). \(P(A \text{ AND } B) = P(\text{both red}) = \frac{5}{12} \cdot \frac{4}{11} = \frac{20}{132} = \frac{5}{33} = \frac{60}{396}\), while \(\frac{25}{144} = \frac{68.75}{396}\). The two are not equal.
Answer: No, \(A\) and \(B\) are not independent (the sampling is without replacement, so the events are dependent).
Problem 4. A single fair die is rolled. Let \(A\) be the event of getting an even number and \(B\) the event of getting a number greater than 3. Are \(A\) and \(B\) independent?
Problem 4 Solution
Step 1 — find the marginals and joint. Even numbers: \(\{2, 4, 6\}\), so \(P(A) = \frac{3}{6} = \frac{1}{2}\). Numbers greater than 3: \(\{4, 5, 6\}\), so \(P(B) = \frac{3}{6} = \frac{1}{2}\). Numbers both even and greater than 3: \(\{4, 6\}\), so \(P(A \text{ AND } B) = \frac{2}{6} = \frac{1}{3}\).
Step 2 — test independence. \(P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \neq \frac{1}{3} = P(A \text{ AND } B)\). The product and the AND disagree.
Answer: No, \(A\) and \(B\) are not independent.
Problem 5. A family has two children. Assume each child is independently equally likely to be a boy or a girl. Let \(A\) be the event that both children are girls and \(B\) the event that the older child is a girl. Are \(A\) and \(B\) independent?
Problem 5 Solution
Step 1 — list the sample space. Two children, each independently a boy (B) or a girl (G), give four equally likely outcomes: \(\{BB,\; BG,\; GB,\; GG\}\).
Step 2 — read the events. \(A = \{GG\}\) (both girls), so \(P(A) = \frac{1}{4}\). \(B = \{GG,\; GB\}\) (older is a girl), so \(P(B) = \frac{2}{4} = \frac{1}{2}\).
Step 3 — test independence. \(P(A) \cdot P(B) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}\). \(P(A \text{ AND } B) = P(\{GG\}) = \frac{1}{4}\). The product \(\frac{1}{8}\) and the AND \(\frac{1}{4}\) disagree.
Answer: No, \(A\) and \(B\) are not independent. (Knowing both are girls automatically tells you the older is a girl, so the conditional has to be 1, not \(\frac{1}{2}\).)
Problem 6. \(P(A) = 0.3\), \(P(B) = 0.4\), and \(P(A \text{ AND } B) = 0.12\). Are \(A\) and \(B\) independent?
Problem 6 Solution
Test — multiplication form. \(P(A) \cdot P(B) = (0.3)(0.4) = 0.12\) and \(P(A \text{ AND } B) = 0.12\). The two are equal, so \(A\) and \(B\) are independent.
Answer: Yes, \(A\) and \(B\) are independent.
Problem 7. \(P(A) = 0.5\), \(P(B) = 0.7\), and \(P(A \text{ AND } B) = 0\). Are \(A\) and \(B\) mutually exclusive? Are they independent?
Problem 7 Solution
Mutual exclusivity. \(P(A \text{ AND } B) = 0\), so \(A\) and \(B\) are mutually exclusive.
Independence. If \(A\) and \(B\) were independent, we would need \(P(A) \cdot P(B) = P(A \text{ AND } B)\). Here \(P(A) \cdot P(B) = (0.5)(0.7) = 0.35\), which is not zero. So the events are not independent.
Answer: Mutually exclusive, not independent.
Problem 8. \(P(A) = 0.6\), \(P(B) = 0.5\), and \(P(A \text{ OR } B) = 0.8\). Find \(P(A \text{ AND } B)\).
Problem 8 Solution
Step 1 — apply the addition rule. \(P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B)\), so \(P(A \text{ AND } B) = P(A) + P(B) - P(A \text{ OR } B) = 0.6 + 0.5 - 0.8 = 0.3\).
Answer: \(P(A \text{ AND } B) = 0.3\).
Problem 9. \(P(A) = 0.4\), \(P(A \text{ AND } B) = 0.1\), and \(P(A \text{ OR } B) = 0.7\). Find \(P(B)\).
Problem 9 Solution
Step 1 — apply the addition rule. \(P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B)\), so \(0.7 = 0.4 + P(B) - 0.1\), giving \(P(B) = 0.4\).
Answer: \(P(B) = 0.4\).
Problem 10. \(P(A) = 0.2\) and \(P(B) = 0.3\). If \(A\) and \(B\) are mutually exclusive, what is \(P(A \text{ OR } B)\)?
Problem 10 Solution
Step 1 — apply the addition rule for mutually exclusive events. When \(P(A \text{ AND } B) = 0\), the rule simplifies to \(P(A \text{ OR } B) = P(A) + P(B)\).
Step 2 — add. \(P(A \text{ OR } B) = 0.2 + 0.3 = 0.5\).
Answer: \(P(A \text{ OR } B) = 0.5\).
Problem 11. A coin lands heads with probability 0.5. The coin is flipped 3 times. Let \(A\) be the event of getting at least one head and \(B\) the event of getting at least one tail. Are \(A\) and \(B\) mutually exclusive?
Problem 11 Solution
Step 1 — find the complement of each event. \(A'\) = no heads = all three flips are tails, so \(P(A') = (\frac{1}{2})^3 = \frac{1}{8}\). \(B'\) = no tails = all heads, so \(P(B') = \frac{1}{8}\).
Step 2 — test mutual exclusivity. \(A\) and \(B\) both happen when there is at least one head and at least one tail, so the only outcome excluded is \(\{HHH, TTT\}\), but the outcome \(HHT\) (and others) belongs to both \(A\) and \(B\). So \(A \cap B \neq \varnothing\) and \(A\) and \(B\) are not mutually exclusive.
Answer: Not mutually exclusive.
Problem 12. A bag contains 4 white and 6 black marbles. Two marbles are drawn with replacement. What is the probability that both are white?
Problem 12 Solution
Step 1 — find the probability of one white marble. \(P(\text{white}) = \frac{4}{10} = \frac{2}{5}\).
Step 2 — apply independence (with replacement). \(P(\text{both white}) = \frac{2}{5} \cdot \frac{2}{5} = \frac{4}{25}\).
Answer: \(\frac{4}{25} = 0.16\).
Problem 13. A bag contains 4 white and 6 black marbles. Two marbles are drawn without replacement. What is the probability that both are white?
Problem 13 Solution
Step 1 — find the probability of the first draw. \(P(\text{first white}) = \frac{4}{10} = \frac{2}{5}\).
Step 2 — find the conditional probability for the second draw. After one white marble is removed, 3 white and 6 black remain out of 9, so \(P(\text{second white} \mid \text{first white}) = \frac{3}{9} = \frac{1}{3}\).
Step 3 — multiply. \(P(\text{both white}) = \frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15}\).
Answer: \(\frac{2}{15} \approx 0.133\).
Problem 14. Two fair dice are rolled. Let \(A\) be the event that the first die shows a 3 and \(B\) the event that the second die shows a 3. Are \(A\) and \(B\) independent?
Problem 14 Solution
Step 1 — find the marginals. Each die roll is independent, so \(P(A) = \frac{1}{6}\) and \(P(B) = \frac{1}{6}\).
Step 2 — test independence. \(P(A) \cdot P(B) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}\). \(P(A \text{ AND } B) = P(\{3, 3\}) = \frac{1}{36}\). The two are equal.
Answer: Yes, \(A\) and \(B\) are independent.
Problem 15. A student takes two courses. The probability of passing the first course is 0.7 and the probability of passing the second course is 0.8. If passing one course has no effect on passing the other, what is the probability of passing both?
Problem 15 Solution
Step 1 — identify the structure. "Passing one course has no effect on passing the other" means the events are independent.
Step 2 — multiply. \(P(\text{both}) = 0.7 \cdot 0.8 = 0.56\).
Answer: \(0.56\).
Problem 16. A test for a disease is 99% accurate: it gives a positive result 99% of the time when the disease is present and 1% of the time when it is not. Suppose 1% of the population has the disease. Let \(D\) be the event that a randomly chosen person has the disease and \(+\) the event that the test is positive. Are \(D\) and \(+\) independent?
Problem 16 Solution
Step 1 — test independence via the multiplication form. \(P(D) = 0.01\) and \(P(+) = P(+ \mid D) \cdot P(D) + P(+ \mid D') \cdot P(D') = 0.99 \cdot 0.01 + 0.01 \cdot 0.99 = 0.0198\).
Step 2 — compute the product. \(P(D) \cdot P(+) = 0.01 \cdot 0.0198 = 0.000198\).
Step 3 — compare with the AND. \(P(D \text{ AND } +) = P(+ \mid D) \cdot P(D) = 0.99 \cdot 0.01 = 0.0099\). The product \(0.000198\) and the AND \(0.0099\) disagree by a factor of 50, so \(D\) and \(+\) are not independent. (This is the standard result behind Bayesian screening: a positive test is far more likely when the disease is present than when it is not.)
Answer: No, \(D\) and \(+\) are not independent.
Problem 17. A coin lands heads with probability 0.5. The coin is flipped 3 times. What is the probability of getting exactly two heads?
Problem 17 Solution
Step 1 — count favorable outcomes. Exactly two heads in three flips means one of the patterns \(HHT, HTH, THH\), three outcomes.
Step 2 — divide by the total. Eight equally likely outcomes total, so
$$P(\text{exactly two heads}) = \frac{3}{8} = 0.375$$Answer: \(\frac{3}{8} = 0.375\).
Problem 18. A die is rolled and a coin is flipped. Let \(A\) be the event of rolling a 5 or 6 and \(B\) the event of flipping heads. Are \(A\) and \(B\) independent?
Problem 18 Solution
Step 1 — find the marginals. Rolling a 5 or 6 has probability \(\frac{2}{6} = \frac{1}{3}\). Flipping heads has probability \(\frac{1}{2}\). The die roll and the coin flip are independent.
Step 2 — test independence. \(P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}\). \(A\) and \(B\) can both happen, and the joint probability is exactly the product. So the events are independent.
Answer: Yes, \(A\) and \(B\) are independent.
Problem 19. A standard 52-card deck is shuffled and one card is dealt. Let \(A\) be the event that the card is a spade and \(B\) the event that the card is a 7. Are \(A\) and \(B\) independent? Are they mutually exclusive?
Problem 19 Solution
Step 1 — find the marginals. \(P(A) = \frac{13}{52} = \frac{1}{4}\) (13 spades). \(P(B) = \frac{4}{52} = \frac{1}{13}\) (four 7s, one per suit).
Step 2 — find the joint. The 7 of spades is the only card that is both a spade and a 7, so \(P(A \text{ AND } B) = \frac{1}{52}\).
Step 3 — test independence. \(P(A) \cdot P(B) = \frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52} = P(A \text{ AND } B)\). The product matches the AND, so the events are independent.
Step 4 — test mutual exclusivity. \(P(A \text{ AND } B) = \frac{1}{52} \neq 0\), so the events are not mutually exclusive.
Answer: Independent, not mutually exclusive.
Problem 20. A bag contains 3 red, 4 white, and 5 blue marbles. One marble is drawn, its color noted, and it is returned to the bag. A second marble is drawn. Are the color of the first marble and the color of the second marble independent events?
Problem 20 Solution
Step 1 — recognize the structure. The first marble is returned to the bag, so the second draw happens from the same population of 12 marbles. The first draw has no effect on the second.
Step 2 — verify with a conditional. \(P(\text{red on second} \mid \text{anything on first}) = \frac{3}{12} = \frac{1}{4}\), regardless of what the first draw was. The conditional equals the marginal, so the events are independent.
Answer: Yes, the two draws are independent.
Problem 21. The probability that it rains on Saturday is 0.3. The probability that it rains on Sunday is 0.4. If the weather on the two days is independent, what is the probability that it rains on at least one of the two days?
Problem 21 Solution
Step 1 — find the complement. "At least one day of rain" is the complement of "no rain on either day." The two days are independent, so
$$P(\text{no rain on either day}) = (1 - 0.3)(1 - 0.4) = (0.7)(0.6) = 0.42$$Step 2 — subtract from 1.
$$P(\text{at least one rainy day}) = 1 - 0.42 = 0.58$$Answer: \(0.58\).
Problem 22. A class has 30 students: 18 women and 12 men. 15 students are sophomores. 10 students are women sophomores. Let \(W\) be the event that a randomly chosen student is a woman and \(S\) the event that they are a sophomore. Are \(W\) and \(S\) independent?
Problem 22 Solution
Step 1 — find the marginals. \(P(W) = \frac{18}{30} = 0.6\). \(P(S) = \frac{15}{30} = 0.5\).
Step 2 — find the joint. \(P(W \text{ AND } S) = \frac{10}{30} = \frac{1}{3} \approx 0.333\).
Step 3 — test independence. \(P(W) \cdot P(S) = (0.6)(0.5) = 0.30\). The product \(0.30\) and the AND \(0.333\) disagree (slightly), so \(W\) and \(S\) are not independent.
Answer: Not independent.
Problem 23. A die is rolled twice. What is the probability that the sum of the two rolls is 7?
Problem 23 Solution
Step 1 — count favorable outcomes. Sum equals 7 when the two dice show (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — six outcomes.
Step 2 — divide by the total. 36 equally likely outcomes total, so
$$P(\text{sum is 7}) = \frac{6}{36} = \frac{1}{6}$$Answer: \(\frac{1}{6} \approx 0.167\).
Problem 24. Two events \(A\) and \(B\) are independent. \(P(A) = 0.4\) and \(P(B) = 0.5\). What is \(P(A \text{ AND } B)\)? What is \(P(A \text{ OR } B)\)?
Problem 24 Solution
Step 1 — apply independence to find the AND. Independent events multiply:
$$P(A \text{ AND } B) = P(A) \cdot P(B) = (0.4)(0.5) = 0.20$$Step 2 — apply the addition rule.
$$P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) = 0.4 + 0.5 - 0.2 = 0.7$$Answer: \(P(A \text{ AND } B) = 0.20\) and \(P(A \text{ OR } B) = 0.70\).
Problem 25. A pair of fair dice is rolled. Let \(A\) be the event that the sum is 7 and \(B\) the event that at least one die is a 6. Are \(A\) and \(B\) mutually exclusive?
Problem 25 Solution
Step 1 — find the events. \(A\) = sum is 7: \(\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}\), six outcomes. \(B\) = at least one die is 6: 11 outcomes (one 6, 11; two 6s, 1; total 11).
Step 2 — find the intersection. Both happen means the sum is 7 AND at least one die is 6: \(\{(1,6), (6,1)\}\), two outcomes.
Step 3 — test mutual exclusivity. \(P(A \text{ AND } B) = \frac{2}{36} = \frac{1}{18} \neq 0\), so \(A\) and \(B\) are not mutually exclusive.
Answer: Not mutually exclusive.
Problem 26. A coin is flipped and a card is drawn from a standard deck. Let \(A\) be the event of getting heads and \(B\) the event of drawing a heart. Are \(A\) and \(B\) independent? Are they mutually exclusive?
Problem 26 Solution
Step 1 — find the marginals. \(P(A) = \frac{1}{2}\) (heads on a fair coin). \(P(B) = \frac{13}{52} = \frac{1}{4}\) (heart drawn from a full deck).
Step 2 — test independence. The coin flip and the card draw are independent actions, so \(P(A \text{ AND } B) = P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}\). The product matches the AND, so the events are independent.
Step 3 — test mutual exclusivity. \(P(A \text{ AND } B) = \frac{1}{8} \neq 0\), so the events are not mutually exclusive (a coin flip and a card draw can both happen).
Answer: Independent, not mutually exclusive.
Problem 27. The probability of event \(A\) is 0.6 and the probability of event \(B\) is 0.7. What is the smallest possible value of \(P(A \text{ AND } B)\)? What is the largest possible value?
Problem 27 Solution
Step 1 — find the smallest possible value of the AND. The AND is at least \(P(A) + P(B) - 1 = 0.6 + 0.7 - 1 = 0.3\) by the addition rule (this is the Bonferroni lower bound).
Step 2 — find the largest possible value of the AND. The AND cannot exceed either marginal, so it is at most \(\min(P(A), P(B)) = 0.6\).
Answer: Smallest: \(0.3\). Largest: \(0.6\).
Problem 28. A medical test has a 5% false positive rate (5% of healthy people test positive) and a 2% false negative rate (2% of sick people test negative). In a certain population, 10% have the condition. Let \(S\) be the event that a person has the condition and \(+\) the event that the test is positive. Are \(S\) and \(+\) independent?
Problem 28 Solution
Step 1 — find the marginals. \(P(S) = 0.10\) (10% of the population has the condition).
Step 2 — find \(P(+)\). The test can be positive in two ways. True positive: \(P(+ \mid S) = 1 - 0.02 = 0.98\). False positive: \(P(+ \mid S') = 0.05\).
$$P(+) = 0.98 \cdot 0.10 + 0.05 \cdot 0.90 = 0.098 + 0.045 = 0.143$$Step 3 — find the product. \(P(S) \cdot P(+) = 0.10 \cdot 0.143 = 0.0143\).
Step 4 — find the AND. \(P(S \text{ AND } +) = P(+ \mid S) \cdot P(S) = 0.98 \cdot 0.10 = 0.098\).
Step 5 — compare. \(0.0143 \neq 0.098\), so the events are not independent.
Answer: Not independent.
Problem 29. A survey of 200 customers finds that 120 bought product A and 80 bought product B. 40 customers bought both. Let \(A\) be the event of buying product A and \(B\) the event of buying product B. Are \(A\) and \(B\) independent?
Problem 29 Solution
Step 1 — find the marginals. \(P(A) = \frac{120}{200} = 0.6\). \(P(B) = \frac{80}{200} = 0.4\).
Step 2 — find the joint. \(P(A \text{ AND } B) = \frac{40}{200} = 0.2\).
Step 3 — test independence. \(P(A) \cdot P(B) = 0.6 \cdot 0.4 = 0.24\). The product \(0.24\) and the AND \(0.20\) disagree.
Answer: Not independent.
Problem 30. A family has three children. Assume each child is independently equally likely to be a boy or a girl. What is the probability that they have exactly two girls?
Problem 30 Solution
Step 1 — count favorable outcomes. Two girls in three children means exactly one of the three children is a boy. The patterns are \(GGB,\; GBG,\; BGG\) — three outcomes.
Step 2 — divide by the total. \(2^3 = 8\) equally likely outcomes, so
$$P(\text{exactly two girls}) = \frac{3}{8} = 0.375$$Answer: \(\frac{3}{8} = 0.375\).
Key Terms
independent events — two events \(A\) and \(B\) such that knowing one occurred does not change the probability that the other occurred; equivalently \(P(A \mid B) = P(A)\) 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 — a sampling scheme in which each picked member is returned to the population before the next pick, so the same member can appear more than once and the events are independent.
sampling without replacement — a sampling scheme in which each picked member is set aside, so no member can appear more than once and the events are dependent.
complement — for an event \(A\), the set of all outcomes in the sample space that are not in \(A\), written \(A'\).