Technique 2: We have that
5.1 BERNOULLI AND BINOMIAL RANDOM VARIABLES
Chapter 5
S pecial D iscrete
D istributions
q(y)= P (Y = y) = P
X= y− 1 2
=
4
y−1 2
(0.60)(y−1)/2(0.40)4−[(y−1)/2], y= 1, 3, 5, 7, 9.
8.
8 i=0
15 i
(0.8)i(0.2)15−i = 0.142.
9.
10 5
11 36
525 36
5
= 0.108.
10. (a) 1−
5 0
1 3
02 3
5
−
5 1
1 3
12 3
4
= 0.539. (b)
5 2
1 10
2 9 10
3
= 0.073.
11. We know that p(x) is maximum at[(n + 1)p]. If (n + 1)p is an integer, p(x) is maximum at [(n + 1)p] = np + p. But in such a case, some straightforward algebra shows that
n
np+ p
pnp+p(1− p)n−np−p =
n
np+ p − 1
pnp+p−1(1− p)n−np−p+1, implying that p(x) is also maximum at np+ p − 1.
12. The probability of royal or straight flush is 40
52 5
.If Ernie plays n games, he will get, on the average, n
40
52 5
royal or straight flushes. We want to have 40n
52 5
= 1; this gives n=
52 5
40= 64, 974.
13.
6 3
1 3
32 3
3
= 0.219.
14. 1− (999/1000)100= 0.095.
15. The maximum occurs at k= [11(0.45)] = 4. The maximum probability is
10 4
(0.45)4(0.55)6= 0.238.
16. Call the event of obtaining a full house success. X, the number of full houses is n independent poker hands is a binomial random variable with parameters (n, p), where p is the probability that a random poker hand is a full house. To calculate p, note that there are
52 5
possible poker hands and
4 3
4 2
13!
11! = 3744 full houses. Thus p = 374452 5
≈ 0.0014. Hence
Section 5.1 Bernoulli and Binomial Random Variables 89 E(X)= np ≈ 0.0014n and Var(X) = np(1−p) ≈ 0.00144n. Note that if n is approximately 715, then E(X) = 1. Thus we should expect to find, on the average, one full house in every 715 random poker hands.
17. 1−
6 6
1 4
63 4
0
−
6 5
1 4
53 4
≈ 0.995.
18. 1−
3000 0
(0.0005)0(0.9995)3000−
3000 1
(0.0005)(0.9995)2999≈ 0.442.
19. The expected value of the expenses if sent in one parcel is 45.20× 0.07 + 5.20 × 0.93 = 8.
The expected value of the expenses if sent in two parcels is (23.30× 2)(0.07)2+ (23.30 + 3.30)
2 1
(0.07)(0.93)+ (6.60)(0.93)2= 9.4.
Therefore, it is preferable to send in a single parcel.
20. Let n be the minimum number of children they should plan to have. Since the probability of all girls is (1/2)nand the probability of all boys is (1/2)n, we must have 1−(1/2)n−(1/2)n ≥ 0.95.
This gives (1/2)n−1 ≤ 0.05 or n − 1 ≥ ln 0.05
ln(0.5) = 4.32 or n ≥ 5.32. Therefore, n = 6.
21. (a) For this to happen, exactly one of the N stations has to attempt transmitting a message.
The probability of this is
N 1
p(1− p)N−1= Np(1 − p)N−1.
(b) Let f (p)= Np(1−p)N−1. The value of p which maximizes the probability of a message going through with no collision is the root of the equation f(p)= 0. Now
f(p)= N(1 − p)N−1− Np(N − 1)(1 − p)N−2 = 0.
Noting that p= 1, this equation gives p = 1/N. This answer makes a lot of sense because at every “suitable instance,” on average, Np= 1 station will transmit a message.
(c) By part (b), the maximum probability is f
1 N
= N1 N
1− 1
N N−1
= 1− 1
N N−1
.
As N → ∞, this probability approaches 1/e, showing that for large numbers of stations (in reality 20 or more), the probability of a successful transmission is approximately 1/e independently of the number of stations if p= 1/N.
22. The k students whose names have been called are not standing. Let A1, A2, . . . , An−kbe the students whose names have not been called. For i, 1≤ i ≤ n − k, call Ai a “success,” if he or she is standing; failure, otherwise. Therefore, whether Ai is standing or sitting is a Bernoulli trial, and hence the random variable X is the number of successes in n− k Bernoulli trials.
For X to be binomial, for i = j, the event that Ai is a success must be independent of the event that Aj is a success. Furthermore, the probability that Ai is a success must be the same for all i, 1≤ i ≤ n − k. The latter condition is satisfied since Ai is standing if and only if his original seat was among the first k. This happens with probability p = k/n regardless of i . However, the former condition is not valid. The relation
P
Aj is standing| Ai is standing
= k− 1 n ,
shows that given Aiis a success changes the probability that Ajis success. That is, Aibeing a success is not independent of Aj being a success. This shows that X is not a binomial random variable.
23. Let X be the number of undecided voters who will vote for abortion. The desired probability is
P
b+ (n − X) > a + X
= P
X < n+ (b − a) 2
=
[n+(b−a) 2 ]
i=0
n i
1 2
i1 2
n−i
=1 2
n[n+(b−a) 2 ]
i=0
n i
.
24. Let X be the net gain of the player per unit of stake. X is a discrete random variable with possible values−1, 1, 2, and 3. We have
P (X= −1) =
3 0
1 6
05 6
3
= 125 216, P (X= 1) =
3 1
1 6
5 6
2
= 75 216, P (X= 2) =
3 2
1 6
25 6
= 15 216, P (X= 3) =
3 3
1 6
35 6
0
= 1
216. Hence
E(X)= −1 ·125
216 + 1 · 75
216+ 2 · 15
216+ 3 · 1
216 ≈ −0.08.
Therefore, the player loses 0.08 per unit stake.
Section 5.1 Bernoulli and Binomial Random Variables 91 25.
E(X2)= n x=1
x2
n x
px(1− p)n−x = n x=1
(x2− x + x)
n x
px(1− p)n−x
= n x=1
x(x− 1)
n x
px(1− p)n−x+ n x=1
x
n x
px(1− p)n−x
= n x=2
n!
(x− 2)! (n − x)!px(1− p)n−x+ E(X)
= n(n − 1)p2 n x=2
n− 2 x− 2
px−2(1− p)n−x+ np
= n(n − 1)p2
p+ (1 − p)n−2
+ np = n2p2− np2+ np.
26. (a) A four-engine plane is preferable to a two-engine plane if and only if 1−
4 0
p0(1− p)4−
4 1
p(1− p)3>1−
2 0
p0(1− p)2.
This inequality gives p > 2/3. Hence a four-engine plane is preferable if and only if p > 2/3.
If p= 2/3, it makes no difference.
(b) A five-engine plane is preferable to a three-engine plane if and only if
5 5
p5(1− p)0+
5 4
p4(1− p) +
5 3
p3(1− p)2>
3 2
p2(1− p) + p3. Simplifying this inequality, we get 3(p− 1)2(2p− 1) ≥ 0 which implies that a five-engine plane is preferable if and only if 2p− 1 ≥ 0. That is, for p > 1/2, a five-engine plane is preferable; for p < 1/2, a three-engine plane is preferable; for p= 1/2 it makes no difference.
27. Clearly, 8 bits are transmitted. A parity check will not detect an error in the 7–bit character received erroneously if and only if the number of bits received incorrectly is even. Therefore, the desired probability is
4 n=1
8 2n
(1− 0.999)2n(0.999)8−2n = 0.000028.
28. The message is erroneously received but the errors are not detected by the parity-check if for 1≤ j ≤ 6, j of the characters are erroneously received but not detected by the parity–check, and the remaining 6−j characters are all transmitted correctly. By the solution of the previous exercise, the probability of this event is
6 j=1
(0.000028)j(0.999)8(6−j)= 0.000161.
29. The probability of a straight flush is 40
52 5
≈ 0.000015391. Hence we must have
1−
n 0
(0.000015391)0(1− 0.000015391)n ≥ 3 4. This gives
(1− 0.000015391)n≤ 1 4. So
n≥ log(1/4)
log(1− 0.000015391) ≈ 90071.06.
Therefore, n≈ 90, 072.
30. Let p, q, and r be the probabilities that a randomly selected offspring is AA, Aa, and aa, respectively. Note that both parents of the offspring are AA with probability (α/n)2, they are both Aa with probability
1− (α/n)2
, and the probability is 2(α/n)
1− (α/n)
that one parent is AA and the other is Aa. Therefore, by the law of total probability,
p= 1 ·α n
2
+1 4·
1−α n
2
+1 2 · 2α
n
1−α
n =1
4
α n
2
+1 2
α n
+1 4, q = 0 ·α
n 2
+1 2
1−α
n 2
+1 2· 2α
n
1−α
n = 1
2−1 2
α n
2
, r = 0 ·α
n 2
+1 4
1−α
n 2
+ 0 · 2α n
1−α
n =1
4
1−α
n 2
. The probability that at most two of the offspring are aa is
2 i=0
m i
ri(1− r)m−i.
The probability that exactly i of the offspring are AA and the remaining are all Aa is
m i
piqm−i.
31. The desired probability is the sum of three probabilities: probability of no customer served and two new arrivals, probability of one customer served and three new arrivals, and probability of two customers served and four new arrivals. These quantities, respectively, are (0.4)4·
4 2
(0.45)2(0.55)2,
4 1
(0.6)(0.4)3·
4 3
(0.45)3(0.55), and
4 2
(0.6)2(0.4)2· (0.45)4. The sum of these quantities, which is the answer, is 0.054.
Section 5.1 Bernoulli and Binomial Random Variables 93 32. (a) Let S be the event that the first trial is a success and E be the event that in n trials, the
number of successes is even. Then
P (E)= P (E|S)P (S) + P (E|Sc)P (Sc).
Thus
rn = (1 − rn−1)p+ rn−1(1− p).
Using this relation, induction, and r0= 1, we find that rn= 1
2
1+ (1 − 2p)n .
(b) The left sum is the probability of 0, 2, 4, . . . , or[n/2] successes. Thus it is the probability of an even number of successes in n Bernoulli trials and hence it is equal to rn.
33. For 0≤ i ≤ n, let Bibe the event that i of the balls are red. Let A be the event that in drawing kballs from the urn, successively, and with replacement, no red balls appear. Then
P (B0|A) = P (A|B0)P (B0) n
i=0
P (A|Bi)P (Bi)
= 1×1
2 n
n i=0
n− i n
k n i
1 2
n = 1
n i=0
n i
n− i n
k
.
34. Let E be the event that Albert’s statement is the truth and F be the event that Donna tells the truth. Since Rose agrees with Donna and Rose always tells the truth, Donna is telling the truth as well. Therefore, the desired probability is P (E | F ) = P (EF )/P (F ). To calculate P (F ), observe that for Rose to agree with Donna, none, two, or all four of Albert, Brenda, Charles, and Donna should have lied. Since these four people lie independently, this will happen with probability
1 3
4
+
4 2
2 3
21 3
2
+2 3
4
=41 81.
To calculate P (EF ), note that EF is the event that Albert tells the truth and Rose agrees with Donna. This happens if all of them tell the truth, or Albert tells the truth but exactly two of Brenda, Charles and Donna lie. Hence
P (EF )=1 3
4
+1 3·
3 2
2 3
21 3
= 13 81. Therefore,
P (E | F ) = P (EF )
P (F ) =13/81 41/81 = 13
41 = 0.317.