Independent events practice problems with full solutions

Name the events and list what you are given. Let $G$ be "glaze crack" and $W$ be "warped", so $P(G) = 0.15$, $P(W) = 0.20$, and $P(G \cap W) = 0.03$.

(a) The product test compares the joint probability against the product of the two separate probabilities: $P(G) \cdot P(W) = 0.15 \times 0.20 = 0.03$.

(a) That product equals the given joint probability, since $0.03 = 0.03$, so $G$ and $W$ are independent. A cracked glaze tells you nothing about whether the piece is warped.

(b) The conditional test asks whether $P(W \mid G) = P(W)$. Compute the conditional probability: $P(W \mid G) = \frac{P(G \cap W)}{P(G)} = \frac{0.03}{0.15} = 0.20$.

(b) Compare the two numbers: $P(W \mid G) = 0.20$ and $P(W) = 0.20$. They match, so the conditional test reaches the same verdict as the product test.

(c) "Neither" is the complement of "at least one", so find the union first: $P(G \cup W) = P(G) + P(W) - P(G \cap W) = 0.15 + 0.20 - 0.03 = 0.32$. Then $P(\text{neither}) = 1 - 0.32 = 0.68$.

(c) Check it by the product route, which independence makes available: $P(\text{no crack}) \cdot P(\text{not warped}) = 0.85 \times 0.80 = 0.68$. The two routes agree.

Name the events. Let $A$ be "adopted within two weeks" and $U$ be "puppy", so $P(A) = 0.40$, $P(U) = 0.30$, and $P(A \cap U) = 0.18$.

(a) Product test: $P(A) \cdot P(U) = 0.40 \times 0.30 = 0.12$.

(a) Compare that against the joint probability the shelter reported, $P(A \cap U) = 0.18$. Since $0.18 \ne 0.12$, the two events are not independent.

(b) Use the definition of conditional probability: $P(A \mid U) = \frac{P(A \cap U)}{P(U)} = \frac{0.18}{0.30} = 0.60$.

(b) Interpret it. Among puppies, 60 percent are adopted within two weeks, against 40 percent of dogs overall. Learning that a dog is a puppy raises its two-week adoption chance by 20 percentage points, and that shift is exactly what dependence means.

(c) General addition rule: $P(A \cup U) = P(A) + P(U) - P(A \cap U) = 0.40 + 0.30 - 0.18 = 0.52$.

Notice that the two tests had to agree here. Rearranging $P(A \mid U) = \frac{P(A \cap U)}{P(U)}$ gives $P(A \cap U) = P(A \mid U) \cdot P(U) = 0.60 \times 0.30 = 0.18$, so a conditional above $P(A)$ forces a joint probability above $P(A) \cdot P(U)$.

Read the three probabilities you need straight off the table. Let $S$ be "used self-checkout" and $M$ be "member". Then $P(S) = \frac{140}{400} = 0.35$, $P(M) = \frac{240}{400} = 0.60$, and the top left cell gives $P(S \cap M) = \frac{84}{400} = 0.21$.

(a) Product test: $P(S) \cdot P(M) = 0.35 \times 0.60 = 0.21$, which equals $P(S \cap M) = 0.21$, so the events are independent.

(b) Conditional test: $P(S \mid M) = \frac{84}{240} = 0.35$, and $P(S) = 0.35$. They match, so the conditional test also says independent.

(b) The other conditional confirms it: $P(S \mid \text{not a member}) = \frac{56}{160} = 0.35$ as well. Membership moves the self-checkout rate nowhere at all.

(c) Start from the definition of conditional probability, $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$, which is available whenever $P(B) > 0$.

(c) Suppose the conditional test passes, so $P(A \mid B) = P(A)$. Substituting the definition gives $\frac{P(A \cap B)}{P(B)} = P(A)$, and multiplying both sides by $P(B)$ gives $P(A \cap B) = P(A) \cdot P(B)$, which is the product test.

(c) Every step reverses. Dividing $P(A \cap B) = P(A) \cdot P(B)$ through by $P(B)$ returns $P(A \mid B) = P(A)$. So one test passes exactly when the other does, and you only ever need to run whichever is easier from the numbers in front of you.

Check the model first. Every project gets exactly one award level, and $0.15 + 0.25 + 0.30 + 0.30 = 1.00$, so the four levels are mutually exclusive and cover every project.

(a) Mutually exclusive: yes. One project cannot hold two award levels at once, so $P(\text{gold} \cap \text{silver}) = 0$.

(a) Independent: no. The product test gives $P(\text{gold}) \cdot P(\text{silver}) = 0.15 \times 0.25 = 0.0375$, and $0 \ne 0.0375$. The conditional test says it more bluntly: $P(\text{gold} \mid \text{silver}) = 0$, far below $P(\text{gold}) = 0.15$, because learning a project took silver rules gold out entirely.

(b) Mutually exclusive: no. Independence is given, so $P(\text{gold} \cap \text{first-time}) = 0.15 \times 0.40 = 0.06$, and $0.06 \ne 0$, so a first-time entrant can certainly win gold.

(b) Independent: yes, by the fact the problem states, and the joint probability $0.06$ is what that independence produces.

(c) Gold and silver are mutually exclusive, so the overlap term in the addition rule is $0$: $P(\text{gold} \cup \text{silver}) = 0.15 + 0.25 - 0 = 0.40$.

(d) Gold and first-time are not mutually exclusive, so the overlap has to come back out: $P(\text{gold} \cup \text{first-time}) = 0.15 + 0.40 - 0.06 = 0.49$.

Compare the two pairs. Mutual exclusivity pushed the joint probability down to $0$ and destroyed independence, while independence pushed the joint probability up to the positive product $0.06$ and ruled out mutual exclusivity. The two conditions pull in opposite directions on the same quantity, which is why a pair of events almost never satisfies both.

(a) The van takes exactly one route, so it cannot be on the coastal route and the valley route on the same morning. The events are mutually exclusive, and $P(C \cap V) = 0$.

(b) Multiply the two separate probabilities: $P(C) \cdot P(V) = 0.36 \times 0.45$. Work the whole numbers first, $36 \times 45 = 1620$, then place the decimal point four places in, giving $0.36 \times 0.45 = 0.162$.

(c) The product test compares $P(C \cap V)$ against $P(C) \cdot P(V)$, which here is $0$ against $0.162$. Since $0 \ne 0.162$, the events $C$ and $V$ are not independent.

(d) $P(C \mid V) = \frac{P(C \cap V)}{P(V)} = \frac{0}{0.45} = 0$, while $P(C) = 0.36$. Knowing the van took the valley route drops the coastal probability from $0.36$ all the way to $0$, which is the largest drop possible, since no probability can fall below $0$. Learning that one of two mutually exclusive events happened always rules the other out completely.

(e) Set up the proof. Suppose $A$ and $B$ are mutually exclusive with $P(A) > 0$ and $P(B) > 0$, and suppose for contradiction that they are also independent.

(e) Mutual exclusivity gives $P(A \cap B) = 0$. Independence gives $P(A \cap B) = P(A) \cdot P(B)$. Setting the two expressions for the same quantity equal forces $P(A) \cdot P(B) = 0$.

(e) A product of two real numbers is zero only when at least one factor is zero, so $P(A) = 0$ or $P(B) = 0$. Both were assumed positive, so this is a contradiction, and the two events cannot be independent. The only loophole is the trivial one: if $P(A) = 0$, then $P(A \cap B) = 0 = P(A) \cdot P(B)$ holds automatically, so an impossible event is technically independent of everything.

Start with the single-draw probability: $\frac{10}{25} = 0.40$ of the packets are the wildflower mix.

(a) With replacement the crate is identical for the second draw, so the two draws are independent and the product applies directly: $P(\text{both wildflower}) = 0.40 \times 0.40 = 0.16$.

(b) Without replacement the second draw happens in a crate of 24 packets holding 9 wildflower packets, so use the general multiplication rule: $P(\text{both}) = \frac{10}{25} \times \frac{9}{24}$.

(b) Evaluate each factor: $\frac{10}{25} = 0.40$ and $\frac{9}{24} = 0.375$. Then $0.40 \times 0.375 = 0.15$, which is lower than the $0.16$ from part (a).

(c) The conditional test settles it. $P(\text{second wildflower} \mid \text{first wildflower}) = \frac{9}{24} = 0.375$, while $P(\text{first wildflower}) = \frac{10}{25} = 0.40$. Since $0.375 \ne 0.40$, the draws are not independent without replacement. With replacement the same conditional is $\frac{10}{25} = 0.40$, which matches, so only sampling with replacement keeps the draws independent.

(d) Split the second draw by what happened on the first. If the first packet was wildflower, with probability $\frac{10}{25}$, the second is wildflower with probability $\frac{9}{24}$. If the first was herb, with probability $\frac{15}{25}$, the second is wildflower with probability $\frac{10}{24}$.

(d) Add the two routes. The first gives $\frac{10}{25} \times \frac{9}{24} = \frac{90}{600} = 0.15$, the second gives $\frac{15}{25} \times \frac{10}{24} = \frac{150}{600} = 0.25$, and $0.15 + 0.25 = 0.40$.

(d) Notice what did and did not move. The unconditional probability for the second draw is still $0.40$, identical to the first draw. Dependence does not have to shift a marginal probability; it shows up in the conditional, which slid from $0.40$ to $0.375$ once you knew what the first draw was.

(a) Find the single-draw probability first: $\frac{24000}{40000} = 0.60$. Because the staff member never selects the same cardholder twice, each selection changes the pool available to the next one, so the 5 selections are dependent by construction.

(b) If the first cardholder used the app, the pool for the second is 39999 cardholders of whom 23999 used the app, so $P(\text{second} \mid \text{first}) = \frac{23999}{39999}$.

(b) Divide. Since $39999 \times 0.6 = 23999.4$, the fraction sits just below $0.6$, and carrying the division out gives $\frac{23999}{39999} = 0.59999$ to five decimal places, against $P(\text{first}) = 0.60000$. The dependence is real, and it moves the probability by about $0.00001$.

(c) Treat the 5 selections as independent draws each with probability $0.60$, so the answer is $0.60^5$. Build it up: $0.6^2 = 0.36$, $0.6^3 = 0.216$, $0.6^4 = 0.1296$, and $0.6^5 = 0.07776$, about $0.078$.

(c) The exact without-replacement value also rounds to $0.078$ at three decimal places, so nothing of consequence is lost by using the independence shortcut on a population this large.

(d) At the branch the pool shrinks fast, so write out all five factors: $P(\text{all 5 used the app}) = \frac{12}{20} \times \frac{11}{19} \times \frac{10}{18} \times \frac{9}{17} \times \frac{8}{16}$.

(d) Multiply the numerators: $12 \times 11 = 132$, then $132 \times 10 = 1320$, then $1320 \times 9 = 11880$, then $11880 \times 8 = 95040$. Multiply the denominators: $20 \times 19 = 380$, then $380 \times 18 = 6840$, then $6840 \times 17 = 116280$, then $116280 \times 16 = 1860480$.

(d) Simplify and divide: $\frac{95040}{1860480} = \frac{33}{646} = 0.0511$ to four decimal places. The independence estimate is still $0.60^5 = 0.07776$, so it overshoots by $0.07776 - 0.0511 = 0.0267$, which is $\frac{0.0267}{0.0511} = 0.52$, about 52 percent too high.

(e) The AP course calls it the 10% condition: when you sample without replacement, the population must be at least 10 times the sample size, that is $n \le \frac{N}{10}$. In the city, $n = 5$ and $\frac{N}{10} = \frac{40000}{10} = 4000$, so the condition passes with enormous room and the draws behave like independent trials. At the branch, $n = 5$ and $\frac{N}{10} = \frac{20}{10} = 2$, so the condition fails badly and the independence shortcut is not usable.

Pull the three probabilities off the table. Let $B$ be "booked in advance" and $L$ be "used the laser cutter". Then $P(B) = \frac{200}{500} = 0.40$, $P(L) = \frac{200}{500} = 0.40$, and the top left cell gives $P(B \cap L) = \frac{120}{500} = 0.24$.

(a) Product test: $P(B) \cdot P(L) = 0.40 \times 0.40 = 0.16$. The table's joint probability is $0.24$, and $0.24 \ne 0.16$, so the two events are not independent.

(b) Conditional test: $P(L \mid B) = \frac{120}{200} = 0.60$, against $P(L) = 0.40$. Visitors who booked ahead used the laser cutter 60 percent of the time, against 40 percent of visitors overall, a gap of 20 percentage points, which is what you would see if people book ahead precisely because they want that machine.

(b) The walk-in side confirms it: $P(L \mid \text{walked in}) = \frac{80}{300} = 0.2667$ to four decimal places, well below $P(L) = 0.40$. Both conditionals move away from the marginal, in opposite directions, which is the signature of dependence.

(c) Independence requires the joint probability to equal the product of the marginals, which is $0.40 \times 0.40 = 0.16$. Convert that back to a count out of 500: $0.16 \times 500 = 80$.

(c) Fill in the rest and check the totals still hold. Booked without the laser cutter becomes $200 - 80 = 120$, walk-ins with the laser cutter becomes $200 - 80 = 120$, and walk-ins without becomes $300 - 120 = 180$. Row totals are $80 + 120 = 200$ and $120 + 180 = 300$; column totals are $80 + 120 = 200$ and $120 + 180 = 300$. The conditional now matches as well, since $\frac{80}{200} = 0.40 = P(L)$.

(d) "At least one" is fastest through the complement. A single visit avoids the laser cutter with probability $1 - 0.40 = 0.60$, and independence across visits is what lets you multiply: $P(\text{none of the 4}) = 0.60^4$.

(d) Evaluate: $0.6^2 = 0.36$ and $0.36^2 = 0.1296$, so $P(\text{none}) = 0.1296$ and $P(\text{at least one}) = 1 - 0.1296 = 0.8704$.

(e) Say what each claim is about. Part (a) asks whether two attributes of the same visit, booking and laser cutter use, carry information about each other, and the table says they do. Part (d) asks whether separate visits carry information about each other, which is what licenses multiplying $0.60$ four times. A table can refute the first kind of independence and say nothing at all about the second, so part (d) rests on taking the makerspace at its word, and a class or club arriving together would break it.