MATH 431: HOMEWORK 6
3.4.2 Solution.
a) P (D = k) =
1 k−1
,
2
because after the first ball this is just a geometric
random variable with parameter 1/2. (The success is getting a ball with a different
color than the first one.) In particular D − 1 is a geometric(1/2).
b) From the definition:
∞
X
k−1
1
E(D) =
k·
2
k=2
0
k !
∞
X
1
1
=
(k + 1) ·
−1·
2
2
k=0
= Σ1 − 1 =
1
−1=3
(1 − 12 )2
using the formula for Σ1 from the text with q = 12 .
Alternate proof : D − 1 is a geometric(1/2) so E(D − 1) =
1
1/2
= 2 which gives ED = 3.
c) From the definitions:
∞
X
k−1
1
k ·
E(D ) =
2
k=2
k !
0
∞
X
1
1
2
2
=
(k + 1) ·
−1 ·
2
2
2
2
k=0
= Σ2 − 1 =
1 + 12
− 1 = 11
(1 − 12 )3
Var(D) = E(D2 ) − (E(D))2 = 11 − 32 = 2
p
√
SD(D) = Var(D) = 2
using the formula for Σ2 from the text with q = 12 .
Alternate proof : the variance of D − 1 is the same as the variance of D. So V arD =
V ar(D − 1) =
q
p2
= 2 (since p = q = 1/2).
3.4.4 Solution.
a) P (one is odd one out) =
6
8
=
3
4,
given there are eight possibilities, two of
which involve all getting the same.
r−1 3
b) P (T = r) = 14
· 4 , by geometric distribution for the number of times T played,
with p = 43 .
c) E(T ) =
1
p
=
4
3
1
2
MATH 431: HOMEWORK 6
3.4.10 Solution. This is similar to problem 3.4.2, but now with arbitrary p in place of p = 12 , and
separately for q = 1 − p.
a) P (X = k) = P (first k − 1 success then a failure) + P (first k − 1 failure then a success)
= pk−1 q + q k−1 p. Note that k can be any integer which is at least 2.
b)
E(X) =
∞
X
k · (pk−1 q + q k−1 p) =
k=2
∞
X
!
(k + 1) · (pk q + q k p)
− (p + q)
k=0
= qΣ1,p + pΣ1,q − 1
=
p
1 1
q
+
− 1 = + − 1,
2
2
(1 − p)
(1 − q)
q p
using the formula for Σ1 from the text with both p and q.
Alternate solution: One can also get the solution by noting that if the sum went from
k = 1 then we would get the sum of expectations of two geometric random variables with
parameters p and q. This would give p1 + 1q and the contribution of the k = 1 term is p+q = 1.
c)
E(X 2 ) =
∞
X
k 2 · (pk−1 q + q k−1 p) =
k=2
∞
X
!
(k + 1)2 · (pk q + q k p)
− (p + q)
k=0
= qΣ2,p + pΣ2,q − 1
=
q(1 + p) p(1 + q)
1+p 1+q
+
−1=
+ 2 −1
3
3
(1 − p)
(1 − q)
q2
p
Var(X) = E(X 2 ) − (E(X))2
1+p 1+q
=
+ 2 −1−
q2
p
1 1
+ −1
q p
2
=
p3 + q 3 − 2p2 q 2
,
p2 q 2
using the formula for Σ2 from the text with both p and q.
Alternate solution: Again, if we include the k = 1 term in the sum for EX 2 then we
would get EY 2 + EZ 2 for geometric random variables Y, Z with parameters p and q. This
1+q
+ 1+p
and the contribution
p2
q2
1+q
1+p
+ q2 − 1. From this V arX can
p2
would give
of the k = 1 term is again p + q = 1. Thus
EX 2
be computed the same way.
=
3.4.12 Solution. In all of the parts we will use the same idea: we break up the events in question
as the union of events {W1 = a, W2 = b}, use independence and then apply the geometric
sum formula. We can also use that P (Wi > k) = qik and P (Wi < k) = 1 − P (Wi > k − 1) =
1 − qik−1 .
MATH 431: HOMEWORK 6
3
a)
P (W1 = W2 ) =
=
∞
X
k=1
∞
X
P (W1 = k, W2 = k) =
∞
X
P (W1 = k)P (W2 = k)
k=1
q1k−1 p1 · q2k−1 p2 =
k=1
=
p1 p2
1 − q1 q2
b)
P (W1 < W2 ) =
=
∞
X
∞
X
k=1
l=k+1
∞
X
!
P (W1 = k, W2 = l)
P (W1 = k)
k=1
=
∞
X
∞
X
=
∞
X
∞
X
k=1
l=k+1
!
P (W2 = l)
l=k+1
q1k−1 p1 q2k =
∞
X
=
∞
X
!
P (W1 = k)P (W2 = l)
q1k−1 p1 P (W2 > k)
k=1
q1k p1 q2k+1
k=0
k=1
p 1 q2
=
1 − q1 q2
c) Symmetrically to part b), P (W1 > W2 ) =
p 2 q1
1 − q1 q2
d)
P (min(W1 , W2 ) = k) = P (W1 = k)P (W2 = k)
+ P (W1 = k)P (W2 > k) + P (W2 = k)P (W1 > k)
= p1 p2 q1k−1 q2k−1 + p1 q1k−1 q2k + p2 q2k−1 q1k
= (1 − q1 q2 )q1k−1 q2k−1
e)
P (max(W1 , W2 ) = k) = P (W1 = k)P (W2 = k) + P (W1 = k)P (W2 < k) + P (W2 = k)P (W1 < k)
= p1 p2 (q1 q2 )k−1 + p1 q1k−1 (1 − q2k−1 ) + p2 q2k−1 (1 − q1k−1 )
3.5.2 Solution. The number of raisins R in a cookie can be approximated by a Poisson random
variable with parameter µ, so we need to find µ such that
.99 ≥ P (R > 0) = 1 − P (R = 0) = 1 − e−µ ,
giving e−µ ≤ .01 and so −µ ≤ ln(.01) ≈ −4.605, i.e., µ ≥ 4.605.
3.5.3 Solution.
1
8
a) 32 raisins per pound of dough results in an average of 4 raisins per 2 ounces (=
pound). Thus the probability of a single cookie to contain no raisin is approximately
p = e−4 ≈ .0183. Since we have 12 cookies in a bag the probability of getting a cookie
without a raisin is 1 − (1 − e−4 )12 ≈ 0.199. This is because the number of no-raisin
4
MATH 431: HOMEWORK 6
cookies in a bag is just a binomial with parameters 12 and e−4 .
(Note that the numerical answer in the back of the book is off.)
b) Note that the average (µ) per pound translates to the average per cookie as µ/8. Now
we want
P (A bag has a cookie with no raisins) ≤ .05 ⇒ 1 − (1 − e−µ/8 )12 ≤ .05.
Completing the algebra of this gives that µ ≥ 43.65, so we must add 44 raisins to attain
the desired probability of non-complaint.
3.5.4 Solution. The probability of a single page to have at least 5 misprints is
p = 1 − e−1 (1 +
11 12 13 14
65
+
+
+ ) = 1 − e−1 ·
≈ 0.00366.
1!
2!
3!
4!
24
The probability of a 300 page book to contain a page with at least 5 misprints is thus
approximately 1 − (1 − 0.00366)300 ≈ 0.667. Note that this could also be computed with
Poisson approximation again: the number of pages with at least 5 errors is approximately
Poisson with parameter 0.00366 × 300 = 1.098 so the probability of at least one such page
is approximately 1 − e−1.098 ≈ 0.666.
3.5.9 (a) Since X and Y are independent we have
P(X = 1 and Y = 2) = P(X = 1)P(Y = 2) = e−1
11 −2 22
·e
= 2e−3 .
1!
2!
(b) First of all, this is equivalent to P(X + Y ≥ 2). Second, X and Y are independent, and
both Poisson distributed. The sum of independent Poisson random variables is also
Poisson, with the new rate being the sum of the individual rates (see p. 226 of Pitman
for reference). So X + Y has a Poisson distribution with rate 1 + 2 = 3. Thus we have
P(X + Y ≥ 2) = 1 − P(X + Y < 2) = 1 − P(X + Y = 0) − P(X + Y = 1)
= 1 − e−3 − 3e−3 ≈ .801.
(c) Using what we know from the previous part, we can approach this directly.
P(X = 1|X + Y = 4) =
=
P((X = 1) ∩ (X + Y = 4)
P((X = 1) ∩ (Y = 3))
=
P(X + Y = 4)
P(X + Y = 4)
e−1 e−2 8/6
32
=
≈ .395.
−3
4
e 3 /24
81
3.6.2 Solution. We argue by symmetry:
a) P (tenth card is a queen) = P (first card is a queen) =
1
13
b) P (twentieth card is a spade) = P (first card is a spade) =
1
4
c) P (last five cards are spades) = P (first five cards are spades) =
.05%
(13
5)
=
52
(5)
13·12·11·10·9
52·51·50·49·48
≈
MATH 431: HOMEWORK 6
5
d) P (last king appears on the 48th card) = P (3 kings in the first 47, king on 48th)
(4)(48)
= 3 5244 · 51 ≈ 0.0599.
(47)
Alternate solution: Using symmetry and reversing the order:
P (last king appears on the 48th card) = P (first king appears on the 5th card)
= P (no kings in the first 4 and a king for the fifth)
48
4
4
= 52 ·
48
4
which gives the same numerical answer.
3.6.5 Solution. Let Ii be the indicator variable for the ith box being empty, then by independence,
we get
1 n
P (Ii = 1) = 1 −
b
Since X =
Pb
i=1 Ii ,
this yields
E(X) =
b
X
i=1
1 n
E(Ii ) = b 1 −
b
In order to compute Var(X) we need E(X 2 ):

!
!
b
b
X
X
X
E(X 2 ) = E
Ii2 = E
Ii2 + 2E 
i=1
i=1

Ii Ij 
1≤i<j≤b
Since Ii2 = Ii , the first term is just E(X). For the second term we can use symmetry:


X
2E 
Ii Ij  = b(b − 1)E(I1 I2 ) = b(b − 1)P (first two boxes are empty)
1≤i<j≤b
2 n
= b(b − 1) 1 −
b
This gives
2 n
1 2n
1 n
2
Var(X) = b 1 −
+ b(b − 1) 1 −
−b 1−
b
b
b