Solutions for Stat-571-Homework #4 Problem 1 Problem 2

STAT 571 Lilun Du dulilun@stat.wisc.edu
Problem 1
Solutions for Stat-571-Homework #4
Denote B i be the event that the ith Bucket i is select, i = 1, 2, 3, then B 1 , B 2 , B 3 is a partition of Ω and
they are mutually exclusive.
(a) Let A 1 be the event that the ball we selected is black, then by the Law of Total Probability, we
can obtain
P (A 1 ) =
3∑
P (B i )P (A 1 |B i )
i=1
= 0.1 ∗
= 0.61
4
4 + 5 + 3 + 0.2 ∗ 1
1 + 4 + 7 + 0.7 ∗ 80
80 + 20
(b) Similarly,let A 2 denote the event that the ball we slected is red, then P (A 2 ) = 0.1 ∗ 5
12 + 0.2 ∗ 4
0.7 ∗ 0 = 0.108.
(c) Similarly,let A 3 denote the event that the ball we slected is white,then P (A 3 ) = 0.1 ∗ 3 12 + 0.2 ∗ 7 12 +
0.7 ∗
20
80+20 = 0.232.
(d) From the Bayes’ theorem, we can get the inverse probability as
P (B i |A 1 ) = P (A 1|B i )P (B i )
P (A 1 )
Thus, from the previous parts, we got immediately that
1/3 ∗ 0.1
P (B 1 |A 1 ) = = 0.055
0.61
1/12 ∗ 0.2
P (B 2 |A 1 ) = = 0.027
0.61
8/10 ∗ 0.7
P (B 3 |A 1 ) = = 0.918
0.61
12 +
Problem 2
For this problem, we denote C i the event that the ith bucket is selected.
(a) When n = 1, using the Law of The Total Probability, we can get the followng derivation,
P (B = 1) = P (B = 1|C 1 )P (C 1 ) + P (B = 1|C 2 )P (C 2 )
17
=
17 + 6 + 8 ∗ 0.5 + 40
40 + 30 + 25 ∗ 0.5
= 0.485
Similarly, we get P (R = 1) = 0.255 and P (W = 1) = 0.261
(b) Let n=10, what is the distribution of B? For each trial, B=1 can be viwed as a successs with
probability 0.485. Thus B is binomail(10, 0.485). Similarly, R is also followed as Binomial(10, 0.255)
and Biomial(10, 0.261)
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STAT 571 Lilun Du dulilun@stat.wisc.edu
Problem 3
We denote A, B, AB, O be the events that a person has the blood type A, B, AB, O respectively.
Similarly, typeA, typeB, typeAB, typeO for the bloodtype that are typed. The we can get immediately
that P (typeA|O) = 4%, which is the probabilty of a soldier who was typed as having blood type A
and his actual blood type is O. Likewise, P (typeA|A) = 88%, P (typeA|B) = 4%, P (typeA|AB) = 10%
By the Law of The Total Probability, we have
P (typeA) = P (typeA|A)P (A) + P (typeA|B)P (B) + P (typeA|AB)P (AB) + P (typeA|O)P (O)
= 0.88 ∗ 0.43 + 0.04 ∗ 0.07 + 0.1 ∗ 0.04 + 0.04 ∗ 0.46
= 0.4036
From the question we need to find the P(A|typeA). Using Bayes Theorem, we can easily derive the
results as
Problem 4
P (typeA|A)P (A)
P (A|typeA) =
P (typeA)
0.88 ∗ 0.43
=
0.4036
= 0.9376
(a) Since the number of tall plants with green pods are 900, ”tall“ and ”green“ are not mutually
exclusive.
(b) Denoting event A be the tree which is tall and B be the tree with green pods, then P (A) =
900+300
1600
= 0.75, P (B) = 900+300
1600
= 0.75 and P (A ∩ B) = 900
1600
= 0.5625. By cheching the equation
that P (A ∩ B) = P (A)P (B), we know that ”tall“ and ”green“ are independent for this collection
of plants.
Problem 5
(a) Denote P (A1) be the probality of choosing a nucleotide A from region I(it is the conditional probability),
and P(A1)=0.2. Similarly, we can define P (G2)...
P (two nucletides are the match)
= P (A1, A2) + P (T 1, T 2) + P (G1, G2) + P (C1, C2)
= P (A2)P (A1) + P (T 2)P (T 1) + P (G2)P (G1) + P (C2)P (C1)
= 0.2 ∗ 0.25 + ∗0.2 ∗ 0.25 + 0.3 ∗ 0.25 + 0.3 ∗ 0.25
= 0.25
(b) First, we know that the triplet can be replicated, for example we might have AAA. Denote GG be
all possible triplets choosing from A, T, G, C. In fact, we have 64 elements in GG. For region I,
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STAT 571 Lilun Du dulilun@stat.wisc.edu
we similarly define GG1, GG2. Thus,
∑
P (all two triplets matched) =
=
=
triplet(i)∈GG1
∑
triplet(i)∈GG1
∑
triplet(i)∈GG1
= 1
64
P (all two triplets mathced|triplet(i) ∈ GG1)P (treplet(i) ∈ GG1)
1
× P (treplet(i) ∈ GG1)
64
P (treplet(i) ∈ GG1) × 1
64
Where the second equation holds because P (all two triplets mathced|triplet(i) ∈ GG1) = 0.25 ∗
0.25 ∗ 0.25 = 1
64 .
Problem 6
(a) Please see the probobiity tree in the last part of this solution.
(b) Denote A 1 be the event that the first coin flipping is Yes and A 2 be complementary event. Then,
P (A 1 ) = P (A 2 ) = 0.5. Furthermore, we can get P (yes|A 1 ) = 0.5 and P (yes|A 2 ) = 0.2. Then,
by Law of The Total Probability , we have P (yes) = P (yes|A 1 )P (A 1 ) + P (yes|A 2 )P (A 2 ) =
0.5 ∗ 0.5 + 0.2 ∗ 0.5 = 0.35.
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