Two events A and B satisfy P(A) = 0.4, P(B) = 0.3, and P(A ∩ B) = 0.1. What is P(A ∪ B)?
A0.7 — simply add the probabilities since both events can occur
B0.6 — subtract the intersection to correct for double-counting
C0.12 — multiply the probabilities since they are not independent
D0.5 — take the average of the two probabilities
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.4 + 0.3 − 0.1 = 0.6. Option A is the classic error: adding without subtracting the intersection counts the overlap twice. The intersection region appears in both P(A) and P(B), so it must be subtracted once to give each outcome exactly one count.
Question 2 Multiple Choice
You roll a fair 6-sided die 4 times. What is the most efficient approach to find P(at least one 6)?
AAdd the probabilities of getting exactly 1 six, exactly 2 sixes, exactly 3 sixes, and exactly 4 sixes
BUse the complement: 1 − P(no sixes) = 1 − (5/6)⁴
CApply the addition rule to four separate roll events
DMultiply the probability of a 6 on one roll by 4
The complement rule turns 'at least one' into 'none': P(no 6 in 4 rolls) = (5/6)⁴, so P(at least one 6) = 1 − (5/6)⁴ ≈ 0.518. This is one calculation. The direct approach (option A) requires four separate binomial terms. Whenever 'at least one' appears, ask yourself whether the complement — 'none' — is easier to compute.
Question 3 True / False
If two events are mutually exclusive, they are also complementary.
TTrue
FFalse
Answer: False
Mutually exclusive means P(A ∩ B) = 0 — the events can't both occur simultaneously. Complementary is stricter: A and Aᶜ must also together cover the entire sample space, so P(A) + P(Aᶜ) = 1. Every complementary pair is mutually exclusive, but most mutually exclusive pairs are not complementary. Rolling a 1 and rolling a 2 on a die are mutually exclusive but not complementary — the other four outcomes belong to neither.
Question 4 True / False
Using the complement rule is often the most efficient way to compute the probability of 'at least one' occurrence.
TTrue
FFalse
Answer: True
The complement of 'at least one' is 'none at all' — typically a single multiplicative calculation. The direct approach requires summing probabilities over all possible counts (1, 2, 3, ..., n occurrences), which grows rapidly. The complement shortcut is almost always simpler, especially when outcomes are independent.
Question 5 Short Answer
Explain why P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and not just P(A) + P(B).
Think about your answer, then reveal below.
Model answer: When you add P(A) and P(B), any outcome in the intersection A ∩ B is counted twice — once as part of A and once as part of B. Subtracting P(A ∩ B) removes the extra count, giving each outcome exactly one contribution to the total. The simpler formula P(A ∪ B) = P(A) + P(B) is correct only when A and B are mutually exclusive, because then the intersection is empty and there is nothing to remove.
This is the inclusion-exclusion principle for two sets. Visualize a Venn diagram: the left circle is A, the right is B, and the middle lens is A ∩ B. Adding the full left circle and the full right circle counts the lens region twice. Subtracting the lens once gives the correct total area of the union. The same logic extends to three or more events with more alternating addition and subtraction terms.