The axioms of probability and their corollaries
Answers are for parents and teachers.
1
P(A) = 0.62, P(B) = 0.35 and P(A ∩ B) = 0.2. Find P(A′), P(A − B), P(B − A), P(A ∪ B) and the probability that neither event occurs.
P(A′) = 0.38; P(A − B) = 0.62 − 0.2 = 0.42; P(B − A) = 0.35 − 0.2 = 0.15; P(A ∪ B) = 0.62 + 0.35 − 0.2 = 0.77; neither: 1 − 0.77 = 0.23.
2
At a school café a customer orders tea with probability 0.5, juice with probability 0.35, and both with probability 0.15. Find the probability of at least one of the two, neither, tea only, exactly one and at most one.
At least one: 0.5 + 0.35 − 0.15 = 0.70. Neither: 0.30. Tea only: 0.35. Exactly one: 0.70 − 0.15 = 0.55. At most one: 1 − 0.15 = 0.85.
3
A and B are exclusive, P(B) = 2P(A) and P(A ∪ B) = 0.54. Find P(A) and P(B). Then find them again if A ⊂ B instead.
Exclusive: 3P(A) = 0.54, so P(A) = 0.18 and P(B) = 0.36. With A ⊂ B: P(B) = 0.54 and P(A) = 0.27.
4
A four-sector spinner has chances in the ratio 1 : 1 : 2 : 4 for the outcomes a, b, c, d. Find the four probabilities and P(c or d).
8x = 1, so x = 1/8: P(a) = 1/8, P(b) = 1/8, P(c) = 1/4, P(d) = 1/2, and P(c or d) = 3/4.
5
Is it possible that P(A) = 0.9, P(B) = 0.6 and P(A ∩ B) = 0.4? Give the allowed range of P(A ∩ B).
No. The overlap is at least 0.9 + 0.6 − 1 = 0.5 and at most 0.6, so 0.4 is outside the allowed range 0.5 to 0.6.