Math 180A, Winter 2015
Homework 7 Solutions
3.3.14. (a) If X is the income of a family chosen at random, then by Markov’s inequality
P (X > 50, 000) ≤
E(X)
10, 000
=
= .2.
50, 000
50, 000
(b) With X as for part (a) but now using Chebyshev’s inequality (to take advantage of the
additional information σ 2 = Var(X) = (8000)2 )
P (X > 50, 000) = P (X − 10, 000 > 40, 000) ≤ P (|X − 10, 000| > 40, 000)
1
= P (|X − µ| > 5σ) ≤ 2 = .04.
5
3.3.19. Let X1 , X2 , . . . , X30 be the weights of the various customers on the elevator. These random
variables are independent and identically distributed, with E(Xk ) = 150 and Var(Xk ) = (55)2 .
Let S = X1 + · · · + X30 be the total weight of the customers on the elevator, so that E(S) = 4500
√
and Var(S) = 30 ∗ (55)2 , whence the standard deviation of S is 55 30 = 301.23. Using the normal
approximation (a.k.a. the Central Limit Theorem) we compute the chance of overloading the
elevator as
5000 − 4500
S − 4500
>
= 1 − Φ(1.66) = 1 − .9515 = .0485,
P (S > 5000) = P
301.23
301.23
a little less than a 5% chance.
3.3.20. (a) Your profit will be $8000 or more if and only if the profit in your single stock is $200
or $100 per share—in the first case your total profit will be $20,000 while in the second case it
will be $10,000. Since each of these possibilities has probability 1/4, the probability of a profit of
$8000 or more is 1/2.
(b) The expected profit in any one of the 100 stocks is 200 · 14 + 100 · 41 + 0 · 14 − 100 · 14 = 50
dollars, and the variance of that profit is
1
1
1
1
2002 · +1002 · + 02 · + (−100)2 · − 502
4
4
4
4
= (40, 000 + 10, 000 + 0 + 10, 000)/4 − 2500
= 12, 500
dollars. For your total profit, call it S, we therefore have E(S) = 100 · 50 = 5000, Var(S) =
√
100 · 12, 500 = 1, 250, 000, and SD(S) = 1, 250, 000 = 1118.03. Using the normal approximation
we find that
S − 5000
8000 − 5000
= 1 − Φ(2.68) = 1 − .9963 = .0037.
P (S ≥ 8000) = P
>
1118.03
1118.03
1
3.4.4. (a) The chance that no one is “odd one out” is (1/2)3 + (1/2)3 = 1/4, so the chance that
someone is “odd one out” is 3/4.
(b) The game lasts for exactly r tosses if and only if no one is “odd one out” for the first r − 1
tosses, and then someone is “odd one out” on the rth toss; this has probability (1/4)r−1 (3/4).
(c) The duration of the game is a random variable with the geometric distribution (on
{1, 2, . . .}) with success probability p = 3/4. The expected value of such a random variable is
1/p = 34 .
3.4.6. (a) The probability that the number of failures before the first success (in independent
Bernoulli (p) trials) is k is q k p (k failures followed by a success), where k = 0, 1, 2, . . ..
(b) Using the formula for the sum of a geometric series:
P (W > k) =
∞
X
qj p =
j=k+1
q k+1 p
= q k+1 .
1−q
(c) Let T be the number of trials until the first success. From class discussion we know that
E(T ) = 1/p. (See also example 3, starting on page 212.) Notice that W = T − 1. Therefore,
E(W ) = E(T − 1) = E(T ) − 1 = (1/p) − 1 = q/p.
(d) From class discussion or page 213 of the text, Var(T ) = q/p2 . Therefore E(T 2 ) = q/p2 +
1/p2 = (1 + q)/p2 . Consequently,
E(W 2 ) = E(T 2 − 2T + 1) =
(1 + q) 2
2 − 3p + p2
−
+
1
=
p2
p
p2
and so
2
Var(W ) = E(W 2 ) − [E(W )] =
2 − 3p + p2
−
p2
2
q
q
= 2.
p
p
[Alternatively, W and T have the same variance because they differ by a constant.]
3.4.11. (a) A wins on toss k if and only if A tosses k − 1 tails and then a head and B tosses k
k−1
k
. Therefore,
tails. This has probability qA
pA qB
P (A wins) =
∞
X
P (A wins on toss k) =
k=1
∞
X
k−1
k
qA
pA qB
=
k=1
pA qB
.
1 − qA qB
(b) Reversing the roles of A and B we find that
P (B wins) =
pB qA
.
1 − qA qB
(c) In view of (a) and (b),
P (draw) = 1 −
pA qB
pB qA
1 − qA − qB + qA qB
−
=
.
1 − qA qB
1 − qA qB
1 − qA qB
(d) A and B toss until one or the other (or both) get a head. The distribution of the time
until they stop is thus the distribution of the number of tosses until the first success, in which the
2
success probability is p = pA + pB − pA pB (the probability that at least one of A and B get a head
on any one toss). The distribution of the number N of times A and B must toss is therefore
P (N = k) = p · q k−1 ,
k = 1, 2, 3, . . . ,
where p is as above and q = 1 − p.
3.4.15. (a) Starting with the left side of the asserted identity (and using the results of 3.4.6),
P (F = k + m, F ≥ k)
P (F ≥ k)
P (F = k + m)
q k+m p
=
=
= q m p = P (F = m),
P (F ≥ k)
qk
P (F − k = m|F ≥ k) =
as required.
(b) Conversely, suppose F is a random variable with possible values the set {0, 1, 2, . . .} of
non-negative integers, such that
P (F = k + m|F ≥ k) = P (F = m),
m, k = 0, 1, 2, . . . .
The above conditional probability is
P (F = k + m, F ≥ k)
P (F = k + m)
=
,
P (F ≥ k)
P (F ≥ k)
so the stated identity holds if and only if
(†)
P (F = k + m) = P (F = m)P (F ≥ k),
m, k = 0, 1, 2, . . . .
In particular, if we take k = 1 in (†) and define q = P (F ≥ 1) and β = P (F = 0), then
P (F = m + 1) = q · P (F = m),
m = 0, 1, 2, . . . .
Therefore
P (F = 1) = q · P (F = 0) = q · β,
P (F = 2) = q · P (F = 1) = q 2 · β,
and then by induction
P (F = n) = q n · β,
But
1=
∞
X
P (F = n) =
n=0
n = 0, 1, 2, . . . .
∞
X
n=0
qn · β =
β
,
1−q
so β must equal 1 − q (which we usually call p, and then q = 1 − p), and therefore
P (F = n) = q n · p,
3
n = 0, 1, 2, . . . .
the geometric distribution on {0, 1, 2, . . .} of the random variable W in exercise 3.4.6. .
(c) The corresponding characterization of the geometric (p) distribution on {1, 2, . . .} is
P (G = k + m|G > k) = P (G = m),
k = 0, 1, 2, . . . , m = 1, 2, . . . .
The verification of this assertion is similar to the discussion above, and so is omitted.
3.5.3. Observe that there are, on average, 4 raisins per cookie.
(a) We seek the probability that a bag of one dozen cookies contains at least one cookie with
no raisins. Such a bag will generate a complaint. The number of raisins in a given cookie has
the Poisson distribution with µ = 4. For such a cookie, the chance that there are no raisins is
e−4 40 /0! = e−4 = 0.0183. Thus the chance that a given cookie has some raisins is (1 − e−4 ) =
0.9817. By independence, the chance that all 12 cookies in a given bag contain at least one raisin
is (1 − e−4 )12 = 0.8011. This is the probability that the bag does not generate a complaint.
By complementation, the chance that a bag of a dozen cookies does generate a complaint is
1 − (1 − e−4 )12 = 0.1989.
(b) Using the reasoning of part (a), we want to find µ such that 1 − (1 − e−µ )12 = .05. That
is, (1 − e−µ )12 = .95, or (1 − e−µ ) = .951/12 = 0.9957. Solving this for µ we obtain
µ = − log(1 − .9957) = − log(0.004265) = 5.46.
So an average of just under 5.5 raisins per cookie will ensure that only 5% of the bags give rise to
complaints. This amounts to 43.68 raisins per pound of cookie dough.
3.5.6. 30 drops per square inch per minute amounts to 5 drops per square inch per 10-second
period. Thus, the chance of a particular square inch being missed by the rain in a given 10-second
period is e−5 = 0.0067. I am assuming that the rain is falling in accordance with a Poisson process,
and that the rate of rainfall is constant in time.
3.5.10. (a) E(3X + 5) = 3E(X) + 5 = 3λ + 5.
(b) Var(3X + 5) = Var(3X) = 9Var(X) = 9λ.
(c)
∞
∞
X
X
1 −λ λk
λk
−1
E[(1 + X) ] =
e
=
e−λ
k+1
k!
(k + 1)!
k=0
k=0
= λ−1
∞
X
e−λ
k=0
−1
=λ
∞
X
λ
λj
= λ−1
e−λ
(k + 1)!
(j)!
j=1
k+1
[1 − e−λ ]
3.5.11. (a) X + Y has the Poisson distribution with mean 2, so P (X + Y = 4) = e−2 24 /4! =
(2/3)e−2 = 0.09022.
2
(b) E (X + Y )2 = Var(X + Y ) + [E(X) + E(Y )] = Var(X) + Var(Y ) + 22 = 1 + 1 + 4 = 6.
(c) X + Y + Z has the Poisson distribution with mean 3, so P (X + Y + Z = 4) = e−3 34 /4! =
(27/8)e−3 = 0.1680.
4