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Math127121 assignment 9solutions
Course: Discrete Mathematics (Mathematics 1271)
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University: Lakehead University
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Math 1271/3071 Assignment 9
Solutions
1. Given P(A∪B) = 0.9, P (A∩B) = 0.3,and P(A) = 0.5,find P(A|B) and
P(B|A).
Solution: We have P(B|A) = P(A∩B)/P (A) = 0.3/0.5 = 3/5.Since P(A∪B) =
P(A) + P(B)−P(A∩B),we have 0.9 = 0.5 + P(B)−0.3,so P(B) = 0.7.Thus
P(A|B) = P(A∩B)/P (B) = 0.3/0.7 = 3/7.
2. When a fair die is rolled five times, what is the probability that it comes up six
exactly three times, given that the first roll was a two?
Solution: Let Adenote the event that the die comes up six exactly three times, and
let Bdenote the event that the first roll was a two. Our sample space Shas |S|= 65.
the probability we want is P(A|B) = P(A∩B)/P (B) = |A∩B|/|S|
|B|/|S|=|A∩B|
|B|.Given
that the first roll was a two, there are 64possible outcomes for the next four rolls.
We have 4
3ways to choose which rolls come up six, and for each of these 5 ways
to choose a different outcome for the other roll, so we have |A∩B|= 54
3and our
probability is 54
3/64.Alternatively, we could view this as a case of four independent
Bernoulli trials with getting a six being a success. (The first one having already
been a failure.) We then get the probability of getting exactly three successes being
4
3(1/6)3(5/6)1= 54
3/64.
3. A fair coin is tossed four times. Let Adenote the event “exactly one of the first
two tosses is heads” and let Bdenote the event “exactly two of the four tosses are
heads”. Are Aand Bindependent events?
Solution: To check whether the events are independent, we need to check if P(A∩
B) = P(A)P(B).We have that the successive coin tosses are independent Bernoulli
trials, so P(A) = 2
1(1/2)1(1/2)1= 2/4 = 1/2 and P(B) = 4
2(1/2)2(1/2)2=
6/16 = 3/8.We have A∩B={HT T H, HT HT, T HHT, T HT H},so |A∩B|=
4.Since |S|= 24= 16,we have P(A∩B) = 4/16 = 1/4.Since P(A)P(B) =
(1/2)(3/8) = 3/16 6= 1/4 = P(A∩B),we see that the events are not independent.
4. A pair of fair dice are rolled. Given the following events: A the sum is six, B
the sum is even, C a least one die shows a 2, find P(C|A), P (C|B), P (B|C),and
P(A|C).
Solution: We have |S|= 62= 36.Also, A={(5,1),(1,5),(2,4),(4,2),(3,3)},so
|A|= 5 and P(A) = 5/36; B={(6,6),(6,4),(6,2),(4,6),(2,6),(4,4),(4,2),(2,4),
(2,2),(5,5),(5,3),(5,1),(3,5),(1,5),(3,3),(3,1),(1,3),(1,1)},so |B|= 18 and
P(B) = 1/2; and C={(2,6),(6,2),(2,5),(5,2),(2,4),(4,2),(3,2),(2,3),(2,2),
(2,1),(1,2)},so |C|= 11 and P(C) = 11/36.We have A∩C={(2,4),(4,2)},so |A∩
C|= 2 and P(A∩C) = 2/36 = 1/18; and B∩C={(2,6),(6,2),(2,4),(4,2),(2,2)},
so |B∩C|= 5 and P(B∩C) = 5/36.Thus P(C|A) = P(C∩A)/P (A) =
(1/18)/(5/36) = 2/5, P (C|B) = P(C∩B)/P (B) = (5/36)/(1/2) = 5/18, P (B|C) =
P(C∩B)/P (C) = (5/36)/(11/36) = 5/11,and P(A|C) = P(C∩A)/P (C) =
(1/18)/(11/36) = 2/11.
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