Probability problems with unknowns and in context, and checking whether stated probabilities are possible
Statistics Sec 3 Term 1, Unit 3, Lesson 1
GoalSolve probability problems with unknowns and stories, and check whether stated probabilities can be true.
New words
unknown · a letter such as x for a probability we do not yet knowratio of probabilities · probabilities in the proportion 1 : 2 : 3, so x, 2x, 3xpossible range · the values P(A ∩ B) is allowed to take
Explanation
Many problems give a relation instead of a number. Call the unknown probability x, write the other probabilities in terms of x, and use an axiom to form an equation: exclusive events give a sum, and the outcomes of S always add up to 1. A second skill is checking whether the stated numbers can be true at all. For any two events P(A ∩ B) cannot be bigger than the smaller of P(A) and P(B), and because P(A ∪ B) cannot exceed 1, we need P(A) + P(B) − P(A ∩ B) ≤ 1, that is, P(A ∩ B) ≥ P(A) + P(B) − 1. So the overlap lies between max(0, P(A) + P(B) − 1) and min(P(A), P(B)). For P(A) = 0.6 and P(B) = 0.7 the allowed values run from 0.3 to 0.6, as the picture shows. A coach who claims P(win both) = 0.2 for these chances has made a mistake, however confident she sounds.
Worked examples
P(B) = 3P(A), P(A ∪ B) = 0.6. If A and B are exclusive, P(A) + 3P(A) = 0.6, so P(A) = 0.15 and P(B) = 0.45. If instead A ⊂ B, then A ∪ B = B, so P(B) = 0.6 and P(A) = 0.2. The same words give different answers for different relations.
A spinner has three colour zones whose chances are in the ratio 1 : 2 : 3. Write P(a) = x, P(b) = 2x and P(c) = 3x. The outcomes add up to 1, so 6x = 1 and x = 1/6. The probabilities are 1/6, 1/3 and 1/2.
A coach says: P(win on Saturday) = 0.6, P(win on Sunday) = 0.7 and P(win both) = 0.2. The overlap must be at least 0.6 + 0.7 − 1 = 0.3 and at most 0.6, so 0.2 is impossible. The statement is false, and it would stay false whatever the coach believes.
Class activity
Invent a two-event story about your class, for instance pupils who have a bicycle and pupils who bring a packed lunch. Choose P(A), P(B) and P(A ∩ B) yourself. Use the possible range to say which values of P(A ∩ B) you are allowed to choose, then fill the four regions.
Practice
1
A and B are exclusive, P(B) = 2P(A) and P(A ∪ B) = 0.45. Find P(A) and P(B).
3P(A) = 0.45, so P(A) = 0.15 and P(B) = 0.30.
2
Same relation P(B) = 2P(A) and P(A ∪ B) = 0.45, but now A ⊂ B. Find P(A) and P(B).
A ∪ B = B, so P(B) = 0.45 and P(A) = 0.225.
3
S = {a, b, c, d} with P(a) = x, P(b) = 2x, P(c) = 0.3 and P(d) = 0.1. Find x, P(a) and P(b).
3x + 0.4 = 1, so x = 0.2: P(a) = 0.2 and P(b) = 0.4.
4
Can P(A) = 0.5, P(B) = 0.8 and P(A ∩ B) = 0.2 all be true? Give the allowed range for P(A ∩ B).
No. The overlap must be at least 0.5 + 0.8 − 1 = 0.3 and at most 0.5, so the allowed range is 0.3 to 0.5, and 0.2 is outside it.
5
Of 50 pupils, 30 like football, 25 like volleyball and 12 like both. One pupil is chosen at random. Find P(likes at least one) and P(likes neither).
At least one: (30 + 25 − 12)/50 = 43/50 = 0.86. Neither: 1 − 0.86 = 0.14.
