University Solutions to First Hour Exam ECE 313
of Illinois Spring 2000
1.(a) P(A ∪ B) = P(A) + P(B) -–P(A ∩ B) = P(A) + P(B) – P(A|B)P(B).
Hence, P(B) = [P(A ∪ B) – P(A)]/[1 – P(A|B)] = 2/5.
(b) P(C|A ∪ B) = P(C ∩ (A ∪ B))
P(A ∪ B) = P((C ∩ A) ∪ (C ∩ B))
P(A ∪ B) = P(C ∩ A) + P (C ∩ B)
P(A) + P(B) since A and
B are disjoint. Now, P(C ∩ A) = P(C|A)P(A), P(C ∩ B) = P(C|B)P(B), and P(A) = 2P(B) giving that
P(C|A ∪ B) = P(C|A)P(A) + P(C|B)P(B)
P(A) + P(B) = P(C|A)2P(B) + P(C|B)P(B)
2P(B) + P(B) = 2P(C|A) + P(C|B)
3 = 8
21
2. The probability that the alarm will ring is 9
10×2
3 = 3
5, and thus P(no ring) = 2
5. It follows that
P{wake on time} = P{wake|no ring}P{no ring} + P{wake|ring}P{ring} = 1
4×2
5 + 5
6×3
5 = 3
5.
3.(a) The likelihood matrix and maximum-likelihood decision rule is as shown below.
H0: student is from Section A 2
63
61
6
H1: student is from Section D 3
10 4
10 3
10
Excellent Good Average
Sanity check: Each row adds up to 1, just as it should for a likelihood matrix. The desired conditional
probabilities can be read off from the likelihood matrix as shaded.
P{student from Section A is assumed to be in Section D} = 1
6
P{student from Section D is assumed to be in Section A} = 7
10.
Since the probability that the student is in Section A is 3
8, we get P{incorrect decision} = 1
6×3
8 + 7
10×5
8 = 1
2.
More simply, 5 students from Section A and 35 students from Section D, that is, a total of 40 out of the
80 students, are mis-classified.
(c) The joint probability matrix and minimum-error-probability decision rule are as shown.
H0: student is from Section A 2
6×3
8 = 2
16 3
6×3
8 = 3
16 1
6×3
8=1
16
H1: student is from Section D 3
10×5
8 = 3
16 4
10×5
8= 4
16 3
10×5
8 = 3
16
Excellent Good Average
Sanity check: The sum of all the entries is 1. Note that we always decide the student is from Section D!
(d) P{incorrect decision} = P(chosen student is from Section A} = 3
8 < 1
2, the answer in part (b). It better be!
(e) P{excellent student from A and good student from B} = 1
3×2
5 = 2
15
P{excellent student from D and good student from A} = 3
10×1
2 = 3
20 > 2
15. Hence, the maximum-likelihood
decision is that the excellent student is from Section D and the good student from Section A, i.e. H1.
4.(a) A binomial random variable with parameters (n,p) has mean np. Hence, E[X] = 8×1
2 = 4.
(b) Y = 1, 2, 3 passengers are left behind according as X = 6, 7, or 8. Since X takes on values 6, 7, 8 with
with probabilities 28
256, 8
256, 1
256 respectively, we readily find that E[Y] = 1×28 + 2×8 + 3×1
256 = 47
256.