Conditional Probability and Independence
4.3 Independent Events
#51
26
$ = 43
459 ≈ 0.09368
4.3 Independent Events
4.43 a) By Definitions 4.2 and 4.3, if A and B are independent (with P (A) > 0), then P (B∩ A)
P (A) = P (B| A) = P (B), implying that P (B∩ A) = P (A)P (B).
b) If P (A) > 0 and P (A∩ B) = P (A)P (B), then P (B| A) = P (B∩ A)
P (A) = P (A)P (B)
P (A) = P (B).
4.44 a) No. P (C1) = 61.49.3 ̸= 25.81.3 = P (C1| S2), thus, C1 and S2 are not independent.
b) No. P (S1) = 35.661.4 ̸= 21.49.8 = P (S1| C2), thus, S1 and C2 are not independent.
4.45 Since P (A2) = 0.401, P (L1) = 0.126 and P (A2 ∩ L1) = 0.038 ̸= 0.401 × 0.126, then P (A2∩ L1)̸= P (A2)P (L1), implying that A2 and L1 are not independent. In other words, for a randomly selected U.S. citizen, 15 years of age or older, the event that the selected person is between 25 and 44 years old and the event that the person lives alone are not independent (thus, if one of the two events does in fact occur, then the probability of the other event changes).
4.46 a) 615 ≈ 0.0001286
b) 665 = 614 ≈ 0.0007716 c)
#6
2$ × 2 ×2!3!5!
65 = 300
65 ≈ 0.03858, since there are#6
2
$ways to pick which two of the six numbers are rolled, then there are two ways to decide which one of the two selected numbers will be doubled (and which one will be tripled). Next fix one such selection, say, three 1’s and two 2’s. Then there are 3!2!5! possible ordered arrangements of three 1’s and two 2’s in 5 positions.
Similar argument holds for any other such selection, and the required result follows.
4.47 a) Suppose first that P (A) = 0, then, for an arbitrary event B, we have A∩ B ⊂ A, implying that 0 ≤ P (A ∩ B) ≤ P (A) = 0, thus P (A ∩ B) = 0 = P (A)P (B), which by Definition 4.4 implies independence of A and B.
On the other hand, if P (A) = 1, then P (Ac) = 0, and, by the just proved statement, we have that Ac is independent of any other event. Then, by Proposition 4.4, A is independent of any other event.
b) If A and A are independent, then P (A) = P (A∩ A) = P (A)P (A), implying that either P (A) = 0 or P (A) = 1.
4.48 Yes, A and B are independent, since
P (A) = |[0.6, 0.7)|
|(0, 1)| = 0.1,
P (B) = +9
k=0
&
&
-0.04 + 10k, 0.05 +10k$&
&
|(0, 1)| = 10× 0.01 1 = 0.1, and
P (A∩ B) = |[0.64, 0.65)|
|(0, 1)| = 0.01 = 0.1× 0.1 = P (A)P (B).
4.49 False. The counter-example will vary. For instance, take an arbitrary event A with 0 < P (A) < 1 and let B = Ω and C = A. Then, by Exercise 4.47, A and B are independent, also B and C are independent, but A and C are not independent.
4.50 a) 0.255× 0.238 × 0.393 ≈ 0.02385 b) 3× 0.2552× 0.114 ≈ 0.0222
4.51 2n−1−n. For each m ∈ {2, · · · , n}, one needs to check that P (Ak1∩· · ·∩Akm) = .m i=1
P (Aki) for every (unordered) set {k1,· · · , km} of m distinct integers ki ∈ {1, · · · , n}, i = 1, · · · , m.
There are #n
m
$ such sets (for each m), therefore, one needs to check a total of +n m=2
#n
m
$ equations to check for mutual independence of n events. By Binomial theorem,
!n m=2
" n m
%
=
!n m=0
" n m
%
−"n 0
%
−"n 1
%
= (1 + 1)n− 1 − n = 2n− 1 − n.
4.52 a) By Example 4.15(d), assignments #1, #2 and #3 correspond to independent tosses of the coin. On the other hand, assignment #4 does not correspond to independent tosses since P (HTH) = 0.050 ̸= 0.220 = P (HHT), whereas, as established in Example 4.15(d), individual outcomes with the same number of heads must have the same probability.
b) 0.5, 1 and 0.2 for assignments #1, #2 and #3, respectively. If p is the probability of head on a given toss, then P (HHH) = p3, implying that p = (P (HHH))1/3, and the required answer easily follows in each case.
4.53 a) By the inclusion-exclusion principle (and after some elementary algebra),
P
b) For a series system,
P (system is functioning) = P
" n
For a parallel system,
P (system is functioning) = P
" n
where we applied the complementation rule, De Morgan’s law and Proposition 4.5.
4.54 244/495≈ 0.4929; Let X be the sum of scores on the two dice in the first roll. Then P (win) = P (X = 7) + P (X = 11) + ! which by (a) implies that A and B cannot be mutually exclusive.
c) The answers will vary.
4.56 Let Hi denote the event that the ith toss results in a head, i = 1, 2.
c) P (H1∩ H2) = 14 ·12+ p2·12 = 0.5(p2+ 0.25).
d) Events H1 and H2 are independent if and only if P (H1∩ H2) = P (H1)P (H2), which, by (a)–(c), holds if and only if
0.5(p2+ 0.25) = [(p + 0.5)/2]2,
which is equivalent to p2−p+0.25 = 0, or (p−0.5)2 = 0. Therefore, H1 and H2are independent if and only if p = 0.5, which means that the second coin has to be balanced in order for H1 and H2 to be independent.
4.57 a) Exactly two of the four children are boys means that either first two children born are boys and the 3rd and 4th are girls, or the 1st and 3rd are boys and the 2nd and 4th are girls, or the 1st and 4th are boys and the 2nd and 3rd are girls, or the 2nd and 3rd are boys and the 1st and 4th are girls, or the 2nd and 4th are boys and the 1st and 3rd are girls, or the first two are girls and the last two are boys.
b) P (S2) = 6p2(1− p)2. c) P (Sk) =#4
k
$pk(1− p)4−k, k = 0, 1, 2, 3, 4. If the nth child born is a boy, let us say that the nth position is occupied by a boy. Then, if exactly k of the four children born are boys, then there are #4
k
$ ways to pick a set of positions occupied by boys. Fix one such set of positions for boys, say, the first k children are boys and the last 4− k children born are girls. Probability of the latter event happening is pk(1− p)4−k. The same holds for every other fixed set of positions for boys. Thus, the required result follows.
Theory Exercises
4.58 Suppose A and B are independent, then
P (A∩ Bc) = P (A)− P (A ∩ B) = P (A) − P (A)P (B) = P (A)(1 − P (B)) = P (A)P (Bc), implying that A and Bc are independent. Interchanging the roles of A and B in the above argument leads to the conclusion that B and Ac are independent. Finally note that
P (Ac∩ Bc) = P ((A∪ B)c) = 1− P (A ∪ B) = 1 − (P (A) + P (B) − P (A ∩ B))
= 1− (P (A) + P (B) − P (A)P (B)) = (1 − P (A))(1 − P (B)) = P (Ac)P (Bc), implying that Ac and Bc are independent.
4.59 Suppose P (E) = p > 0. Let An be the event that E occurs for the first time on the nth try, n = 1, 2, . . . . Then
P (E eventually occurs) = P
"∞
)
n=1
An
%
=
!∞ n=1
P (An) =
!∞ n=1
(1− p)n−1p
= p
!∞ k=0
(1− p)k = p· 1
1− (1 − p) = 1, since 0≤ 1 − p < 1.
4.60 Let Enbe the event that A occurs for the first time on the nth try and B does not occur at all in the first n repetitions of the experiment. Since A and B are mutually exclusive, then the
probability that neither A nor B occurs on a given try is equal to 1−P (A∪B) = 1−P (A)−P (B),
b) False. Counterexamples may vary. For instance, consider rolling a pair of balanced dice, and let B be the event that the first die shows an odd number, and C be the event that the second die shows an odd number. Let A be the event that the sum of scores on the two dice is even. Then it’s easy to see that B and C are independent, but P (B| A) = P (C | A) = P (B ∩ C | A) = 0.5, implying that P (B∩ C | A) ̸= P (B | A)P (C | A).
c) False. Counterexamples may vary. For instance, consider events A, B, C such that B = A, C = Ac and 0 < P (A) < 1. Then
P (B∩ C |A) = P (B∩ C ∩ A)
P (A) = P (∅)
P (A) = 0 = 1× 0 = P (B | A)P (C | A), thus, B and C are conditionally independent given A, but
P (B∩ C) = P (∅) = 0 ̸= P (A)
i.e. B and C are not independent.
4.62 a) Since 1− x ≤ e−x, by independence of A1, A2. . . , for all n∈ N we have that
n=10 = 0. Therefore, by the De Morgan’s law,
b) Fix an event E such that 0 < P (E) < 1 and let An = E for all n ∈ N . Then, clearly, +∞
n=1P (An) = +∞
n=1P (E) = ∞, but P (A∗) = P (E) < 1. Thus, the second Borel-Cantelli lemma fails without the independence assumption.
4.63 a) If r ≤ 1/2, then the amoeba population eventually becomes extinct with probability 1; if r > 1/2, then the extinction probability equals to (1− r)/r. Let En be the event that the amoeba population is eventually extinct if the initial size of the population is n, and let pn= P (En). Then,
p1 = 1· P (1st amoeba dies) + P (E1| 1st amoeba splits)P (1st amoeba splits)
= (1− r) + p2r = (1− r) + p21r.
Note that the above quadratic equation rp21− p1+ (1− r) = 0 has the following roots: 1 and (1− r)/r. Thus, when r ≤ 1/2, (1 − r)/r ≥ 1, implying that the required extinction probability p1 equals to 1. On the other hand, if r > 1/2, then the required extinction probability equals to (1− r)/r.
b) For r = 1/3 and r = 1/2, the extinction probability is 1. For r = 3/4, the extinction probability equals to 1/3. For r = 9/10, the extinction probability equals to 1/9.
c) We use contextual independence to conclude that P (E1| 1st amoeba splits) = p2 = p21. 4.64 Note that in every repetition of the experiment, either event E or Ec occurs. Then {n(E) = k} = {E occurs in k repetitions and Ec occurs in n− k repetitions}. There are #n
k
$ ways to pick a set of those repetitions where E occurs. Therefore, using independence of the repetitions, we obtain that
P (n(E) = k) ="n k
%
(P (E))k(P (Ec))n−k="n k
%
pk(1− p)n−k, where k = 0, 1,· · · , n.
4.65 #2n
n
$/4n. Let X be the number of heads obtained by Jan in n independent tosses, and let Y be the number of heads obtained by Jean in n independent tosses. Then
P (X = Y ) =
!n j=0
P ({X = j} ∩ {Y = j}) =
!n j=0
P (X = j)P (Y = j),
where, by Exercise 4.64, P (X = j) = P (Y = j) ="n j
% 1
2n. Therefore,
P (X = Y ) =
!n j=0
"n j
%"n j
% 1 2n · 1
2n = 1 4n
!n j=0
"n j
%"
n n− j
%
= 1 4n
"2n n
% ,
since#n
j
$=# n
n−j
$ and the last equality holds by the Vandermonde’s identity (Exercise 3.48).