F71SM STATISTICAL METHODS
Tutorial on Section 2 PROBABILITY: SOLUTIONS
1. (a)
P(A∪B∪C) =
= P(A)+(P(B)+P(C)−P(B∩C))
P(A)+P(B∪C)−P(A∩(B∪C))(additionruleforeventsAandB∪C)
− P((A ∩ B) ∪ (A ∩ C)) (event identity A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
= P(A)+P(B)+P(C)−P(B∩C)
−(P(A∩B)−P(A∩C)+P((A∩B)∩(A∩C)
= P(A)+P(B)+P(C)−P(B∩C)−P(A∩B)−P(A∩C)
+ P(A ∩ B ∩ C) (event identity (A ∩ B) ∩ (A ∩ C) = A ∩ B ∩ C
= P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)
(b) P(A∩B∩C) =
= P (A)P (B|A)P (C|A ∩ B)
P(A∩B)P(C|A∩B)(multiplicationrulefor2eventsA∩BandC)
(i)P(A∪(B∩C))=P(A)+P(B∩C)−P(A∩B∩C)=0.3+(0.6×0.5)−0=0.6 (ii) P((A ∪ B) ∩ C) = P((A ∩ C) ∪ (B ∩ C)) = (0.3 × 0.5) + (0.6 × 0.5) − 0 = 0.45
2. (a)
(b) P(A ∩ B) = 0 while P(A)P(B) ̸= 0, so A and B are not independent. It follows that
3.
4. 5.
6.
A, B, C are not independent.
P(A) = 30/150 = 1/5, P(B) = 100/150 = 2/3, P(A ∩ B) = 20/150 = 2/15 = P(A)P(B), so
A and B are independent.
P (A ∩ B) ̸= 0, so A and B are not mutually exclusive.
P (1st selected is male) = P (3rd selected is male) = P (5th selected is male) = 12/20 = 0.6 The promoter has divided the 9 balls into three groups: group W , consisting of 4 winning balls,
group B, consisting of the single bonus ball, and group L, consisting of 4 losing balls. (i)4×3×2×1= 1 =0.0079
(ii) WWWB,WWBW,WBWW,BWWW so4×4 ×3 ×2 ×1 = 4 =0.0317 9 8 7 6 126
Let A be event ‘at least 2 on male lives’ and B be event ‘at least one on each sex’. Choosing 4 from 15, consisting of 10 on male and 5 on female lives:
9 8 7 6 126
10 2 5 1 44
P(All male) = 15 = 13, P(All female) = 15 = 273 44
⇒ P (B) = 1 − 2/13 − 1/273 = 230/273
A ∩ B is the event ‘2 males and 2 females’ ∪ ‘3 males and 1 female’
⇒ P(A|B) = 10/13 = 21 = 0.9130 230/273 23
105 105 22 31
P(A∩B)= 15 + 15 =30/91+40/91=70/91=10/13 44
1
7.
P (Blue|Con) = =
P (Blue)P (Con|Blue) P (Con)
0.3 × 0.46 = 0.3 × 0.46 + 0.7 × 0.36
0.138 = 0.39
0.3538
8. First stage: 3 branches, probs 0.3, 0.6, 0.1
Second stage: each branch has 3 sub-branches; #1: probs 0.01, 0.1, 0.89; #2: probs 0.1, 0.3, 0.6; #3: probs 0.4, 0.6, 0.
(a) (0.3×0.89)+(0.6×0.6)+(0.1×0)=0.267+0.36=0.627
(b) P(cat 3|loss of life) = P(cat 3)P(loss of life|cat 3)/P(loss of life) = (0.1 × 0.4)/((0.3 ×
0.01) + (0.6 × 0.1) + (0.1 × 0.4)) = 0.04/0.103 = 40/103 = 0.3883
9. P(Athrows5)=P{(1,4)∪(4,1)∪(2,3)∪(3,2)}=4/36=1/9
Writing {1, 2, 4} to mean values 1, 2, 4 are thrown in any order, then
P (B throws 7) = P {{1, 1, 5} ∪ {1, 2, 4} ∪ {1, 3, 3} ∪ {2, 2, 3}} = (3 + 6 + 3 + 3)/216 = 5/72
⇒P(Awins)=1+8×671+8×6721+···=11−67−1=9 =0.6429 997299729 98114
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