Download MATH 461/661 Homework 4 Solutions

MATH 461/661 Homework 4
Solutions
1. 2.7.11 Assume nobody wants to take an elevator ride just to get off on the 1st floor. Then
there are 7 floors as possible destinations. Assume all floors are equally likely, and consider the
process as randomly assigning floors to individuals. There are 77 possible floor assignments.
The probability of 7 individuals all getting different floors is
P (all different floors) =
7!
77
Similarly, the probability they all get off on the same floor is
P (all on the same floor) =
7
.
77
Various factors call into question the assumption that all floors are equally likely (e.g., number
of residents per floor, number who use stairs vs. elevator, number of people who often come
and go with family or friends, whether this elevator services all floors or not, etc.)
2. 2.7.15 Assuming all birthdays are equally likely, there are k2 ways to pick the pair, 365
possibilities for the shared date, and 364 · · · (364 − (k − 2) + 1) ways of picking the remaining
birthdays (without replacement) for the other individuals, therefore
365 k2 · 364 Pk−2
.
P (1 pair has same birthday) =
365k
3. 3.2.4 Let p ≡ P (audit) = 0.153 for each of the n = 6 corporations. Then P (≥ 2 audited)
should follow the binomial distribution with parameters n and p. That implies
P (≥ 2 audited) = 1 − P (0 or 1 audited) = 1 − (1 − p)6 − 6p(1 − p)5 ≈ 0.23
4. Question: Consider the experiment of rolling six six-sided dice. The sample space (S) is all
length-six sequences made up of integers 1 to 6, with replacement. Find (a) the probability of
all dice yielding the same number, and (b) the probability that all the numbers are distinct.
(a) There are 66 equally likely possible outcomes, hence,
P (all same) = P (all 1, all 2, all 3, ..., all 6) =
1
6
= 5
6
6
6
(b) Likewise, counting all rearrangements of 6 distinct numbers yields
P (all distinct) =
6!
5!
= 5
6
6
6
5. Question: Let Yi each be random variables given by the following functions of the outcomes
in the experiment described above (in problem 4). For each of these new random variables Yi
given below, describe (1) the new sample space associated with Yi (i.e., SY = Y (S)) and (2)
the Probability function P (Yi = k) for appropriate values of k.
1
(a) Y1 is the number of even integers in the sequence.
(1) S1 = {0, 1, 2, 3, 4, 5, 6}
(2) P (Y1 = k) follows a binomial distribution with n = 6 and p = 0.5 (half of 1-6 are
even, the other half odd).
(b) (461 only) Y2 is the number of integers greater than 3 in the sequence.
(1) S2 = {0, 1, 2, 3, 4, 5, 6}
(2) P (Y2 = k) is binomial distribution with n = 6 and p = 0.5 (half of 1-6 are greater
than 3).
(c) Y3 is the number of 2s in the sequence.
(1) S3 = {0, 1, 2, 3, 4, 5, 6}
(2) P (Y2 = k) is binomial with n = 6 and p = 1/6 (since P (rolling a 2) = 1/6).
(d) (661 only) Y4 is the indicator function that tests if the sequence sum is less than 8.
(1) As with all indicator functions, the possible outcomes are S4 = {0, 1}.
(2) P (Y4 = k) follows a Bernoulli distribution with
7
p = P ({111111}, {211111}, ..., {111112}) = 6
|
{z
}
6
6 permutations
1
6. 3.2.11
The number of girls is binomial with
n = 4, p =
. This gives that P (2, 2) =
2
4 2
4
4
3
2
3
1
p (1 − p) = 8 and P ({1, 3}, {3, 1}) = 3 p (1 − p) + 1 p (1 − p)3 = 21 , therefore having
2
one of one sex and three of the other is more likely.
7. 3.2.22 Let N = 4050, u = 514, v = 4050−514 = 3536 and n = 65. Then P (k not vaccinated)
is hypergometric with N total and u “successes” possible with n trials, i.e.,
3536 v 514
u
k
P (k not vaccinated) =
n−k
N
n
=
k
65−k
4050
65
8. 3.2.26 This is hypergeometric with N = 80, r = 20, n = 10. Therefore
20 60
P (6) =
6
80
10
4
≈ 0.0115
9. 3.2.34 (461 only) First, following the hint,
P (one or more groups diseased) = 1 − P (all groups diseased).
With four infected individuals and three groups, the only possible configuration of individuals
that gives ”all groups diseased” is 2 infecteds in one group, and 1 in each of the others. So
what is P (all groups diseased)?
Answer #1: If you recognized that the formula given in the text for the answer to 3.2.33
applies to this case, you see there are 4 infected individuals (the “red balls” in 3.2.33) and
2
we are dividing the 21 individuals into 3 groups of 7. Noting that there are 3 ways to assign
which group has 2 infecteds, we have
7 7 7
7 7 7
7 7 7
7 7 7
P (all diseased) =
2
1 1
21
4
+
1
2 1
21
4
1
+
1 2
21
4
=3
2
1 1
21
4
Answer #2: Another approach is to apply the formula used in 2.3.36 (the answer to 2.3.35
given in the back of the text) as follows. Think of dividing the population into three groups
of seven, but blind to who is infected. Then considering group as a ”kind”. Now count the
ways of picking which positions are held by infected individuals, again accounting for which
group gets 2 infecteds. As above, this gives
7 7 7
P (all diseased) = 3
2
1 1
21
4
In any case, the answer is
P (at least 1 group free of disease) = 1 − 3
7
2
7 7
1 1
21
4
≈ 0.484
3.2.35 (661 only)
Answer #1: Begin by focusing on one type. That is, noting that there are ni individuals of
type i, and N − ni of the other types, then by the hypergeometeric distribution
ni N −ni
ki
P (ki ) =
n−ki
N
n
.
By the law of conditional probability,
n1
k1
P (k1 , k2 ) = P (k2 |k1 )P (k1 ) = P (k2 |k1 )
N −n1
n−k1
N
n
Recognizing that P (k2 |k1 ) is equivalent to the hypergeometric distribution but with N − n1
total options, n2 individuals of type 2, and a sample size of n − k1 , we have that
n2 N −n1 −n2
k2
P (k2 |k1 ) =
n−k1 −k2
N −n1
n−k1
Thus,
P (k1 , k2 ) =
n2
k2
N −n1 −n2
n−k1 −k2
N −n1
n−k1
n1
k1
N −n1
n−k1
N
n
=
n1
k1
n2
k2
P
N− 3
N −n1 −n2
n−k1 −k2
N
n
n
(nk11 )(nk22 )(nk33 )( n−P3i=1 kii )
i=1
Continuing this argument yields P (k1 , k2 , k3 ) =
and so on, which yields
(Nn )
the desired form of the generalized hypergeometric distribution.
3
Answer #2: The argument above can also be used to construct a proof by induction.
Proof. Consider
some i ∈ {2, ..., t − 1}. Let the total number of types i and higher
P
P be denoted
by mi = ts=i ns , and likewise denote the sum of ki , ki+1 and higher as ji = ts=i ks . By the
hypergeometric distribution, our “base case” for induction is
n1 m2
P (k1 ) = P (k1 , j2 ) =
k1
N
n
j2 .
Next, assume that for some i ∈ {2, .., t − 1} the following holds:
n1 n2
ni−1 mi
·
·
·
k
k
k
j
P (k1 , ..., ki−1 , ji ) = 1 2 N i−1 i .
n
Noting that, as explained in Answer #1 above,
ni
ki
P (ki |k1 , ..., ki−1 ) =
P
N − i−1
s=1 ns −ni
Pi−1
n− s=1 ks −ki
P
N − i−1
s=1 ns
Pi−1
n− s=1 ks
=
ni
ki
mi+1
ji+1
mi
ji
it therefore follows that
P (k1 , ..., ki ) = P (ki |k1 , ..., ki−1 )P (k1 , ..., ki−1 ) = P (ki |k1 , ..., ki−1 )
=
ni
ki
mi+1
ji+1
mi
ji
n1
k1
n2
k2
ni−1
ki−1
···
N
n
mi
ji
=
n1
k1
n2
k2
ni
ki
···
N
n
n1
k1
n2
k2
N
n
mi+1
ji+1
ni−1
ki−1
···
Since this holds for i = 2, by induction it holds for i = t − 1, and thus it holds that
n1 n2
· · · nktt
k1
k2
P (n1 = k1 , ..., nt = kt ) =
N
n
4
mi
ji