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ESI 6321 – Spring 2007
Exam 1 solutions
Problem 1
a) Denote by A, B, C, D, and E respectively, the following events:
A : heart failure is due to induced substances B : heart failure is due to outside factors C : heart failure is due to natural occurrences D : heart failure is due to disease E : heart failure is due to infection
P (A) = P (A | B) · P (B) + P (A | C) · P (C) = 0. 73 · 0 .13 + 0 · 0. 87 = 0. 0949
b) P (D ∪ E) = P (D) + P (E). Now,
P (D) = P (D | B) · P (B) + P (D | C) · P (C) = 0 · 0 .13 + 0. 27 · 0. 87 = 0. 2349 , and
P (E) = P (E | B) · P (B) + P (E | C) · P (C)
Thus P (D ∪ E) = 0.2349 + 0.1479 = 0. 3828.
Problem 2
a)
P (B | A) =
P (B) =
P (B | A) 6 = P (B) ⇒ events A and B are not independent
b)
P (B | A) = 2230 = 1130 = 0. 733
Problem 3
Let A, B, C be the following events:
A = user is legitimate B = user is fraudulent C = call is originated from two or more metropolitan areas
From the problem statement we have:
P (C | B) = 0. 3 P (C | A) = 0. 01 P (B) = 0. 0001
Through Bayes’ theorem, we can now compute the required probability:
P (B | C) =
P (C | B) · P (B)
P (C | B) · P (B) + P (C | A) · P (A)
P (C | B) · P (B)
P (C | B) · P (B) + P (C | A) · (1 − P (B))
Problem 4
Define PA, PB , PC to be the probabilities that respectively A, B, and C function properly.
(a) PA = P (all 20 ICs of A work) = (1 − 0 .01)^20 = 0.8179.
(b) PB = 1 − P (none of the 20 ICs of B works) = 1 − 0. 0120 = 1 − 10 −^40.
(c) PC = P (20 ICs of C work) + P (19 ICs of C work) = (1 − 0 .01)^20 + C 1920 · (1 − 0 .01)^19 · 0. 011 =
- 98
(d) Let PD = P(product functions properly). Then PD = PA · PB · PC =. 80.
(e) Apply Bayes rule.
P (set A fails|product fails) = P (product fails|set A fails) × P (set A fails) P (product fails)
1 × (1 − PA)
1 − PD
