MATH 411 – Mathematical Probability (Summer 2015)
Section 100
Homework Set 2
due June 15, Monday
June 8, Monday
1. (a) Two fair dice are thrown n times in succession. Compute the probability that
“double 6” appears at least once. How large need n be to make this probability at
least 21 ?
(b) What is the probability that at least one of the pair of fair dice lands on 6 given
that
1. the sum of the dice is 9?
2. the largest of the two is 6?
2. Let S = {1, 2, . . . , n} and suppose that A and B are, independently, equally likely
to be any of the 2n subsets (including the null set and S itself) of S. Compute the
probability of the event: “A is contained in B”.
3. A die is not fair, there is a probability 0.2 that it will land on 6. We roll the die
4 times and let X be the number of times that 6 appears. Compute E[X].
4. Let X be a random variable with E[X] = 0 and Var(X) = 1. Find E[|X − 2|2 ] and
Var(2X − 7).
5. A total of 2n people, consisting of n married couples, are randomly seated (all
possible ordering being equally likely) at a round table. Let Ci denote the event that
the members of couple i are seated next to each other, i = 1, 2, . . . , n.
(a) Find P(Ci ).
(b) For j , i, find P(C j |Ci ).
(c) Approximate the probability that there are no married couples who are seated
next to each other.
Hints – Solutions
1. Hint. (a) If Ei : “doubles six in i-th throwing”, then P(Ei ) = 1/36. Thus we may write:
P(∪ni=1 Ei ) = 1 − P(E1c . . . Enc ) = 1 −
n
Y
P(Eic ) = 1 −
i=1
This probability is at least 1/2 as long as n ≥
!n
n
Y
35
.
(1 − P(Ei )) = 1 −
36
i=1
ln 2
ln(36/35)
≈ 24.6.
(b) Let E : “at least one of the pair of fair dice lands on 6” and F1 :“the sum of the dice is 9” and
F2 :“the largest of the two is 6”. Then, P(E|F1 ) = P(EF1 )/P(F1 ). Note that EF1 = {(3, 6), (6, 3)}
and F1 = {(3, 6), (4, 5), (5, 4), (6, 3)}. Thus, we find P(E|F1 ) = 1/2. Note that F2 ⊂ E , so
P(E|F2 ) = 1.
2. Hint. Since, the sets are chosen uniformly and independently, we have that each pair (A, B)
of subsets of S occurs with probability 1/4n . Moreover, the event {A ⊂ B} occurs if we choose
any subset B of S and then any subset A of B. There exist
!
n
X
n k
2 = 3n
k
k=0
possible such pairs, hence the desired probability is P({A ⊂ B}) =
write:
P({A ⊂ B}) =
n
X
n
3
4
. Alternatively, we may
P({A ⊂ B} | B ∈ Fk )P(B ∈ Fk ),
k=0
where Fk is the set of all subsets of S with exactly k elements. Therefore, we obtain:
P({A ⊂ B}) =
n
X
2k
k=0
n
k
2n 2n
=
!
n
1 X n k
3n
2 = n,
4n k=0 k
4
by the binomial expansion formula.
3. Hint. Since the rollings of the die are indpependent, if we identify “success=lands on 6” and
“probability of success p=0.2” then, the random variable “X =number of times that 6 appears“,
is the numbers of successes in four independent trials. This is binomial random variable with
parameters n = 4, p = 0.2. Therefore,
E(X) = np = 0.8.
4. Hint. Since EX = 0 and VarX = 1, it follows that EX 2 = 1. Therefore, we get:
E(X − 2)2 = EX 2 − 4EX + 4 = 5,
and
Var(2X − 7) = 4Var(X) = 4.
2
2
and P(C j | Ci ) = 2n−
for j , i.
5. Hint. A standard counting argument yields: P(Ci ) = 2n−
1
2
Note that the events Ci , C j are not independent, but they are weakly dependent in the sense:
P(C j | Ci )
P(Ci C j )
2n − 1 n→∞
=
=
−→ 1.
P(Ci )P(C j )
P(C j )
2n − 2
Hence, if n is sufficiently large the probability:
P(no pairings) = P(no successes),
where the success in the i-th trial is considered the members of the i-th couple to be seated
2
next to each other and the probability of success in the i-th trial is pi = P(Ci ) = 2n−
. Therefore,
1
when n is large we may approximate by independent trials, i.e. a binomial distribution with
2
parameters n and p = 2n−
. Again, when n is large the binomial can be approximated further
1
2n
by a Poisson with parameter λ = np = 2n−
, thus we obtain:
1
P(no pairings) = P(no successes in n trials)
≈ P (no successes in binomial with (n, p))
2n
≈ P(X = 0) = e−λ = e− 2n−1 ≈ e−1 ,
where X is Poisson random variable with parameter λ.