Math 302, Assignment 2 Solutions
1. In a small town, there are three bakeries. Each of the bakeries bakes twelve cakes per day. Bakery 1
has two different types of cake, bakery 2 three different types, and bakery 3 four different types. Every
bakery bakes equal amounts of cakes of each type.
You randomly walk into one of the bakeries, and then randomly buy two cakes.
(a) What is the probability that you will buy two cakes of the same type?
(b) Suppose you have bought two different types of cake. Given this, what is the probability that you
went to bakery 2?
Solution: (a) Define the events Fi = {choose bakery i}, and E = {buy different cakes}. Then P(Fi ) = 13 ,
and we compute the conditional probabilities
6 2
6
P(E|F1 ) = 112 =
2
11
2
3 · 41 8
P(E|F2 ) = 12 =
2
11
3 2
6· 9
P(E|F3 ) = 121 =
2
11
By the law of total probability,
1 6 8 9 23
P(E) = + + = .
3 11 11 11 33
10
Therefore P(buy same type) = 33 .
(b) By Bayes,
P(F2 ) 8
P(F2 |E) = P(E|F2 ) =
P(E) 23
2. An assembly line produces a large number of products, of which 1% are faulty in average. A quality
control test correctly identifies 98% of the faulty products, and 95% of the flawless products. For every
product that is identified as faulty, the test is run a second time, independently.
(a) Suppose that a product was identified as faulty in both tests. What is the probability that it is,
indeed, faulty?
(b) What if the quality control test is only performed once?
Solution: Let
F = the event that a product is faulty,
E1 = the event that the first test result is “faulty”
E2 = the event that the second test result is “faulty”
Then, P(F ) = 1% and, since the second test is independent of the first,
P(E1 ∩ E2 |F ) = P(E1 |F )P(E2 |F ) = 0.982
P(E1 ∩ E2 |F c ) = P(E1 |F c )P(E2 |F c ) = 0.052
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, By Bayes’ theorem,
P(E1 ∩ E2 |F )P(F )
P(F |E1 ∩ E2 ) =
P(E1 ∩ E2 |F )P(F ) + P(E1 ∩ E2 |F c )P(F c )
0.982 · 0.01
= ≈ 80%.
0.982 · 0.01 + 0.052 · 0.99
If that test had only be run once, we would get
0.98 · 0.01
P(F |E1 ) = ≈ 17%,
0.98 · 0.01 + 0.05 · 0.99
so even a very reliable test cannot identify a rare fault with satisfactory accuracy in a single try.
3. Let m be an integer chosen uniformly from {1, . . . , 100}. Decide whether the following events are inde-
pendent:
(a) E = {m is even} and F = {m is divisible by 5}
(b) E = {m is prime} and F = {at least one of the digits of m is a 2}
(c) Can you replace the number 100 by a different number, in such a way that your answer to (a) changes?
(E.g., if your answer was “dependent”, try to change the number 100 in such a way your answer becomes
“independent”).
Solution: (a) P(E)P(F ) = (1/2)(1/5) = 1/10 = P(E ∩ F ), so E and F are independent.
(b) P(E) = 25/100, P(F ) = 19/100, and P(E ∩ F ) = 3/100. The events E and F are not independent
since P(E ∩ F ) 6= P(E)P(F ).
50 20 10
(c) Already replacing it by 101 makes the events dependent: 101 · 101 6= 101 . The fact that E and F
of (a) were dependent was a pure coincidence, and had nothing to do with number theoretic properties
of divisibility by 2 and 5. Whether one could change the 100 in part (b) to make the events mentioned
there independent would be a much harder question!
4. Let X be a discrete random variable with values in N = {1, 2, . . .}. Prove that X is geometric with
parameter p = P(X = 1) if and only if the memoryless property
P(X = n + m | X > n) = P(X = m)
holds.
Hint: To show that the memoryless property implies that X is geometric, you need to prove that the
p.m.f. of X has to be P(X = k) = p(1 − p)k−1 . For this, use P(X = k) = P(X = k + 1|X > 1) repeatedly.
Solution: We first show that a geometric RV has the memoryless property: We learned that P(X =
m) = p(1 − p)m−1 and that P(X > m) = (1 − p)m , therefore by the definition of conditional probability
we obtain
P(X = n + m) p(1 − p)n+m−1
P(X = n + m | X > n) = =
P(X > n) (1 − p)n
= p(1 − p)m−1 = P(X = m)
2
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1. In a small town, there are three bakeries. Each of the bakeries bakes twelve cakes per day. Bakery 1
has two different types of cake, bakery 2 three different types, and bakery 3 four different types. Every
bakery bakes equal amounts of cakes of each type.
You randomly walk into one of the bakeries, and then randomly buy two cakes.
(a) What is the probability that you will buy two cakes of the same type?
(b) Suppose you have bought two different types of cake. Given this, what is the probability that you
went to bakery 2?
Solution: (a) Define the events Fi = {choose bakery i}, and E = {buy different cakes}. Then P(Fi ) = 13 ,
and we compute the conditional probabilities
6 2
6
P(E|F1 ) = 112 =
2
11
2
3 · 41 8
P(E|F2 ) = 12 =
2
11
3 2
6· 9
P(E|F3 ) = 121 =
2
11
By the law of total probability,
1 6 8 9 23
P(E) = + + = .
3 11 11 11 33
10
Therefore P(buy same type) = 33 .
(b) By Bayes,
P(F2 ) 8
P(F2 |E) = P(E|F2 ) =
P(E) 23
2. An assembly line produces a large number of products, of which 1% are faulty in average. A quality
control test correctly identifies 98% of the faulty products, and 95% of the flawless products. For every
product that is identified as faulty, the test is run a second time, independently.
(a) Suppose that a product was identified as faulty in both tests. What is the probability that it is,
indeed, faulty?
(b) What if the quality control test is only performed once?
Solution: Let
F = the event that a product is faulty,
E1 = the event that the first test result is “faulty”
E2 = the event that the second test result is “faulty”
Then, P(F ) = 1% and, since the second test is independent of the first,
P(E1 ∩ E2 |F ) = P(E1 |F )P(E2 |F ) = 0.982
P(E1 ∩ E2 |F c ) = P(E1 |F c )P(E2 |F c ) = 0.052
This study source was downloaded by 100000858936669 from CourseHero.com on 02-02-2023 23:44:43 GMT -06:00
https://www.coursehero.com/file/46056110/HW-2-Solutionpdf/
, By Bayes’ theorem,
P(E1 ∩ E2 |F )P(F )
P(F |E1 ∩ E2 ) =
P(E1 ∩ E2 |F )P(F ) + P(E1 ∩ E2 |F c )P(F c )
0.982 · 0.01
= ≈ 80%.
0.982 · 0.01 + 0.052 · 0.99
If that test had only be run once, we would get
0.98 · 0.01
P(F |E1 ) = ≈ 17%,
0.98 · 0.01 + 0.05 · 0.99
so even a very reliable test cannot identify a rare fault with satisfactory accuracy in a single try.
3. Let m be an integer chosen uniformly from {1, . . . , 100}. Decide whether the following events are inde-
pendent:
(a) E = {m is even} and F = {m is divisible by 5}
(b) E = {m is prime} and F = {at least one of the digits of m is a 2}
(c) Can you replace the number 100 by a different number, in such a way that your answer to (a) changes?
(E.g., if your answer was “dependent”, try to change the number 100 in such a way your answer becomes
“independent”).
Solution: (a) P(E)P(F ) = (1/2)(1/5) = 1/10 = P(E ∩ F ), so E and F are independent.
(b) P(E) = 25/100, P(F ) = 19/100, and P(E ∩ F ) = 3/100. The events E and F are not independent
since P(E ∩ F ) 6= P(E)P(F ).
50 20 10
(c) Already replacing it by 101 makes the events dependent: 101 · 101 6= 101 . The fact that E and F
of (a) were dependent was a pure coincidence, and had nothing to do with number theoretic properties
of divisibility by 2 and 5. Whether one could change the 100 in part (b) to make the events mentioned
there independent would be a much harder question!
4. Let X be a discrete random variable with values in N = {1, 2, . . .}. Prove that X is geometric with
parameter p = P(X = 1) if and only if the memoryless property
P(X = n + m | X > n) = P(X = m)
holds.
Hint: To show that the memoryless property implies that X is geometric, you need to prove that the
p.m.f. of X has to be P(X = k) = p(1 − p)k−1 . For this, use P(X = k) = P(X = k + 1|X > 1) repeatedly.
Solution: We first show that a geometric RV has the memoryless property: We learned that P(X =
m) = p(1 − p)m−1 and that P(X > m) = (1 − p)m , therefore by the definition of conditional probability
we obtain
P(X = n + m) p(1 − p)n+m−1
P(X = n + m | X > n) = =
P(X > n) (1 − p)n
= p(1 − p)m−1 = P(X = m)
2
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