Probability rules practice problems

Name the events and list what you are given. Let $D$ be 'on the debate team' and $C$ be 'in the chess club'. Then $P(D) = 0.30$, $P(C) = 0.20$, and $P(D \cap C) = 0.08$.

(a) The two events overlap, so use the general addition rule and subtract the overlap once: $P(D \cup C) = P(D) + P(C) - P(D \cap C)$.

(a) Substitute the numbers: $P(D \cup C) = 0.30 + 0.20 - 0.08 = 0.42$. Skipping the subtraction would count the 8% who do both twice, giving $0.50$.

(b) 'Neither' is the complement of 'at least one of the two', so $P(\text{neither}) = 1 - P(D \cup C) = 1 - 0.42 = 0.58$.

(c) 'Exactly one' takes the union and removes the overlap: $P(\text{exactly one}) = P(D \cup C) - P(D \cap C) = 0.42 - 0.08 = 0.34$.

(c) Check it the long way. Debate only is $0.30 - 0.08 = 0.22$, chess only is $0.20 - 0.08 = 0.12$, and $0.22 + 0.12 = 0.34$, which matches.

Check the setup first. The three outcomes are mutually exclusive and cover every possibility, and $0.62 + 0.24 + 0.14 = 1.00$, so this is a legitimate probability model.

(a) 'Late at all' means either 1 to 5 minutes late or more than 5 minutes late. Those two cannot both describe the same day, so add them directly: $P(\text{late}) = 0.24 + 0.14 = 0.38$.

(b) 'Not more than 5 minutes late' is the complement of 'more than 5 minutes late', so $P(\text{not more than 5 late}) = 1 - 0.14 = 0.86$.

(b) Confirm by adding the two outcomes it contains: $0.62 + 0.24 = 0.86$, which matches the complement calculation.

(c) The general addition rule is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Here the two lateness categories are mutually exclusive, so $P(A \cap B) = 0$ and the subtraction term drops out. Plain addition is the general rule with a zero overlap, not a separate rule.

(a) Identify the complement. 'At least one failure' is everything except 'zero failures', so $P(\text{at least one}) = 1 - P(\text{no failures})$.

(a) Find the probability a single bag passes: $1 - 0.03 = 0.97$.

(a) The 12 bags are independent, so the probability all 12 pass is the product of 12 copies of $0.97$: $P(\text{no failures}) = 0.97^{12}$.

(a) Evaluate: $0.97^{12} = 0.6938$ to four decimal places.

(a) Subtract from 1: $P(\text{at least one failure}) = 1 - 0.6938 = 0.3062$, about $0.306$.

(b) Adding $12 \times 0.03$ applies the addition rule to 12 events as though only one bag could fail, but two or more bags can fail in the same batch. Those overlapping outcomes get counted repeatedly, which is why $0.36$ overshoots the correct $0.306$. Push the same reasoning to 40 bags and it would return $1.2$, an impossible probability.

Name the events. Let $C$ be 'buys coffee' and $R$ be 'buys a pastry', so $P(C) = 0.5$, $P(R) = 0.3$, and $P(C \cap R) = 0.15$.

(a) Mutually exclusive events cannot both happen, which would force $P(C \cap R) = 0$. Here $P(C \cap R) = 0.15$, which is not $0$, so the events are not mutually exclusive.

(b) Independence is a different condition: it requires $P(C \cap R) = P(C) \times P(R)$. Compute the product: $0.5 \times 0.3 = 0.15$.

(b) The product equals the given joint probability, since $0.15 = 0.15$, so $C$ and $R$ are independent. The conditional check agrees: $P(R \mid C) = \frac{P(C \cap R)}{P(C)} = \frac{0.15}{0.5} = 0.3 = P(R)$, so learning that a customer bought coffee does not change the chance of a pastry.

(c) Use the general addition rule: $P(C \cup R) = P(C) + P(R) - P(C \cap R) = 0.5 + 0.3 - 0.15 = 0.65$.

(d) 'Neither' is the complement of the union: $P(\text{neither}) = 1 - 0.65 = 0.35$.

Note the pattern in parts (a) and (b). These events are independent and not mutually exclusive, and that pairing is forced: two events with nonzero probabilities can never be both, because mutual exclusivity puts the joint probability at $0$ while independence puts it at the positive product $P(C) \times P(R)$.

Read the pieces you need from the table. The marginals are $P(\text{bus}) = \frac{80}{200} = 0.4$ and $P(\text{sport}) = \frac{120}{200} = 0.6$, and the top left cell gives the joint probability $P(\text{bus} \cap \text{sport}) = \frac{48}{200} = 0.24$.

(a) Apply the general addition rule: $P(\text{bus} \cup \text{sport}) = 0.4 + 0.6 - 0.24 = 0.76$.

(a) Check with raw counts: $80 + 120 - 48 = 152$ students, and $\frac{152}{200} = 0.76$. The no bus, no sport cell confirms it too, since the 48 students counted there are the only ones the union leaves out, and $200 - 48 = 152$.

(b) Product check: $P(\text{bus}) \times P(\text{sport}) = 0.4 \times 0.6 = 0.24$, which equals the joint probability $0.24$, so the events are independent.

(b) Conditional check for confirmation: $P(\text{sport} \mid \text{bus}) = \frac{48}{80} = 0.6$, and $P(\text{sport}) = 0.6$. Knowing that a student rides the bus does not change the chance they play a sport, so both checks agree.

(c) Mutually exclusive would mean no student does both, but 48 students ride the bus and play a sport, so the events are not mutually exclusive. This table is the standard reminder that independent events do overlap; it is the size of that overlap, exactly $P(A) \times P(B)$, that makes them independent.

(a) Take the complement of each light being red: the probabilities of not red are $1 - 0.40 = 0.60$, $1 - 0.25 = 0.75$, and $1 - 0.50 = 0.50$.

(a) The three lights are independent, so multiply those three probabilities: $P(\text{none red}) = 0.60 \times 0.75 \times 0.50$.

(a) Work left to right: $0.60 \times 0.75 = 0.45$, then $0.45 \times 0.50 = 0.225$.

(b) 'At least one red' is the complement of 'none red', so $P(\text{at least one red}) = 1 - 0.225 = 0.775$.

(c) Adding the three probabilities uses the addition rule as though the three red events were mutually exclusive, but the driver can hit two or three red lights on the same trip, and those overlaps get counted more than once. The giveaway is the answer itself: $1.15$ is greater than $1$, and no probability can exceed $1$.

Read the pieces you need. $P(\text{pool}) = \frac{100}{250} = 0.40$, $P(\text{trainer}) = \frac{80}{250} = 0.32$, and $P(\text{pool} \cap \text{trainer}) = \frac{40}{250} = 0.16$.

(a) Apply the general addition rule: $P(\text{pool} \cup \text{trainer}) = 0.40 + 0.32 - 0.16 = 0.56$.

(a) Check with counts: $100 + 80 - 40 = 140$ members, and $\frac{140}{250} = 0.56$.

(b) Product check: $P(\text{pool}) \times P(\text{trainer}) = 0.40 \times 0.32 = 0.128$, but the joint probability from the table is $0.16$. Since $0.16 \ne 0.128$, the events are not independent.

(b) Conditional check for confirmation: $P(\text{trainer} \mid \text{pool}) = \frac{40}{100} = 0.40$, while $P(\text{trainer}) = 0.32$. Pool users are more likely than members overall to have a trainer, so both comparisons reach the same verdict.

(c) Independence requires the joint probability to equal the product of the marginals, which is $0.40 \times 0.32 = 0.128$. Convert that back to a count out of 250: $0.128 \times 250 = 32$.

(c) Verify that 32 is consistent with the fixed totals. The rest of the table follows: 68 pool users without a trainer, 48 non-pool members with a trainer, and 102 members with neither, which still gives row totals of 100 and 150 and column totals of 80 and 170. The conditional now matches: $\frac{32}{100} = 0.32 = P(\text{trainer})$.

Name the events. Let $F$ be 'fog delay' and $M$ be 'mechanical delay', with $P(F) = 0.12$, $P(M) = 0.05$, and $F$ and $M$ independent.

(a) Independence is what licenses the multiplication: $P(F \cap M) = P(F) \times P(M) = 0.12 \times 0.05 = 0.006$.

(b) General addition rule, using the joint probability from part (a): $P(F \cup M) = 0.12 + 0.05 - 0.006 = 0.164$.

(b) Complement confirmation: a day has neither delay with probability $(1 - 0.12)(1 - 0.05) = 0.88 \times 0.95 = 0.836$, so $P(\text{at least one}) = 1 - 0.836 = 0.164$, matching the addition rule.

(c) Not mutually exclusive. Mutual exclusivity would require $P(F \cap M) = 0$, but part (a) gives $0.006$, so both kinds of delay can strike the same day. More generally, two events with nonzero probabilities that are independent can never be mutually exclusive, since independence fixes the joint probability at the positive product $P(F) \times P(M)$.

(d) Turn the multi-day question into a complement. A single day has no fog delay with probability $1 - 0.12 = 0.88$, so 5 independent days all avoiding fog has probability $0.88^5$.

(d) Evaluate: $0.88^5 = 0.5277$ to four decimal places, so $P(\text{at least one foggy day}) = 1 - 0.5277 = 0.4723$, about $0.472$.

(d) State the assumption: fog delays occur independently from day to day, each with the same probability $0.12$. That assumption is worth doubting here, because weather arrives in multi-day systems, so fog delays tend to cluster and $0.472$ should be read as a rough estimate rather than an exact figure.