Page 119 - Fundamentals of Communications Systems

Review of Probability and Random Variables 3.33

3.5 Example Solutions
Problem 3.26. Using 24 hour time keeping to produce a mapping from time to a
real number results in a uniform dialing model for the placement of calls having
a PDF given as
1
f T (t) = 0 ≤ t ≤ 24
24
= 0 elsewhere (3.43)

1 24

(a) P (A) = P (0 ≤ t ≤ 1 ∪ 19 ≤ t ≤ 24) = f T (t)dt + f T (t)dt = 1/24 +
0 19
5/24 = 1/4.
(b) Here we use total probability over events A and B.
P (C) = P (C|A)P (A) + P (C|B)P (B) = 0.1 × 0.25 + 0.75 × 0.75 = 0.5875.
(3.44)
Note here use was made of P (B) = 1 − P (A) = 0.75.
(c) Using the definition of conditional probability gives P (A∩ C) = P (A, C) =
P (C|A)

P (A) = 0.25 × 0.1 = 0.025.
(d) Here we use the definition of conditional probability.
P (A, C) 0.025
P (A|C) = = = 0.04255 (3.45)
P (C) 0.5875
Note, if we had not completed parts (b) and (c), then we could have used
Bayes rule
P (C|A)P (A) P (C|A)P (A)
P (A|C) = = (3.46)
P (C) P (C|A)P (A) + P (C|B)P (B)
Problem 3.27.

φ
(φ) = P ( p ≤ φ) = (x)dx. Given that
(a) F p f p
−∞
1
(x) = − π ≤ x ≤ π
f p
2π
= 0 elsewhere (3.47)
gives
(φ) = 0 φ ≤−π
F p
φ + π
= − π ≤ φ ≤ π
2π
= 1 φ ≥ π (3.48)