Answer: The Answer Is Not Letter B or Letter C
A. P′ (0, −1)
Step-by-step explanation:
Math 370, Actuarial Problemsolving Problems on General Probability Rules
Problems on general probability rules, independence,
conditional probability
1. Assuming A, B, C are mutually independent, with P(A) = P(B) = P(C) = 0.1,
compute:
(a) P(A ∪ B) Solution: P(A) + P(B) − P(A)P(B) = 0.19
(b) P(A ∪ B ∪ C)
Solution: By formula the formula for P(A∪B∪C) and indep., P(A∪B∪C) =
3 · 0.1 − 3 · 0.1
2 + 0.1
3 = 0.271
(c) P(A \ (B ∪ C))
Solution: P(A) − P(A ∩ B) − P(A ∩ C) + P(A ∩ B ∩ C) = 0.081
2. Given that P(A) = 0.3, P(A|B) = 0.4, and P(B) = 0.5, compute:
(a) P(A ∩ B) Solution: P(A|B)P(B) = 0.4 · 0.5 = 0.2
(b) P(B|A) Solution: P(B ∩ A)/P(A) = 0.2/0.3 = 0.666
(c) P(A0
|B) Solution: P(A0 ∩ B)/P(B) = ((P(B) − P(A ∩ B))/P(B) = 0.6
(d) P(A|B0
) Solution: P(A∩B0
)/P(B0
) = (P(A)−P(A∩B))/(1−P(B)) = 0.2
3. Assume A and B are independent events with P(A) = 0.2 and P(B) = 0.3. Let C be
the event that at least one of A or B occurs, and let D be the event that exactly
one of A or B occurs.
(a) Find P(C).
Solution: The event C is just the union of A and B, so P(C) = P(A ∪ B) =
P(A) + P(B) − P(A)P(B) = 0.44
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Math 370, Actuarial Problemsolving Problems on General Probability Rules
(b) Find P(D).
Solution: Drawing a Venn diagram, we see that D consists of the union of
A and B minus the overlap. Thus, P(D) = P(A ∪ B \ A ∩ B) = P(A ∪ B) −
P(A)P(B) = 0.38
(c) Find P(A|D) and P(D|A).
Solution: P(A|D) = P(A ∩ D)/P(D) = P(A \ A ∩ B)/P(D) = (0.2 − 0.2 ·
0.3)/0.38 = 7/19 . P(D|A) = P(A\A∩B)/P(A) = (0.2−0.2 · 0.5)/0.2 = 0.7 .
(d) Determine whether A and D are independent.
Solution: A and D are not independent since by the previous part, P(A|D) 6=
P(A).
Alternative solution: From above, P(A∩D) = 0.14, P(A)P(D) = 0.2·0.38 =
0.076, so P(A ∩ D) 6= P(A)P(D), and therefore A and D are not independent.
4. Given that P(A ∪ B) = 0.7 and P(A ∪ B0
) = 0.9, find P(A).
Solution: By De Morgan’s law, P(A0 ∩ B0
) = P((A ∪ B)
0
) = 1 − P(A ∪ B) =
1 − 0.7 = 0.3 and similarly P(A0 ∩ B) = 1 − P(A ∪ B0
) = 1 − 0.9 = 0.1. Thus,
P(A0
) = P(A0 ∩ B0
) + P(A0 ∩ B) = 0.3 + 0.1 = 0.4, so P(A) = 1 − 0.4 = 0.6 .
5. Given that A and B are independent with P(A) = 2P(B) and P(A ∩ B) = 0.15, find
P(A0 ∩ B0
).
Solution: By independence and the given data, 0.15 = P(A ∩ B) = P(A)P(B) =
2P(B)
2
, so P(B) = √
0.075 = 0.273, and P(A) = 2P(B) = 0.546. Hence P(A0∩B0
) =
P(A0
)P(B0
) = (1 − 0.546)(1 − 0.273) = 0.33 . (Note the use of the “independence of
complements” property here.)
6. Given that A and B are independent with P(A ∪ B) = 0.8 and P(B0
) = 0.3, find
P(A).
Solution: We have P(B) = 1 − 0.3 = 0.7 and 0.8 = P(A) + P(B) − P(A ∩
B) = P(A) + P(B) − P(A)P(B) = P(A)(1 − 0.7) + 0.7. Solving for P(A) gives
P(A) = (0.8 − 0.7)/0.3 = 0.33 .
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Math 370, Actuarial Problemsolving Problems on General Probability Rules
7. Given that P(A) = 0.2, P(B) = 0.7, and P(A|B) = 0.15, find P(A0 ∩ B0
).
Solution: By De Morgan’s Law, P(A0 ∩ B0
) = P((A ∪ B)
0
) = 1 − P(A ∪ B) =
1−P(A)−P(B)+P(A∩B). Using the given values of P(A) and P(B) and P(A∩B) =
P(A|B)P(B) = 0.15 · 0.7 = 0.105 (the multiplication formula), we get P(A0 ∩ B0
) =
1 − 0.2 − 0.7 + 0.105 = 0.205 .
8. Given P(A) = 0.6, P(B) = 0.7, P(C) = 0.8, P(A ∩ B) = 0.3, P(A ∩ C) = 0.4,
P(B ∩ C) = 0.5, P(A ∩ B ∩ C) = 0.2, find P(A ∩ B0 ∩ C
0
).
Solution: If A, B0 and C
0 were independent, we could apply the product formula,
and the answer would be immediate, but we don’t know this (in fact, they are not).
However, from a Venn diagram we see that P(A∩B0∩C
0
) is equal to to P(A)−P(A∩
B)−P(A∩C)+P(A∩B∩C). Inserting the given values, we get 0.6−0.3−0.4+0.2 =
0.1 as answer.
