College aantekeningen
Kansrekening en statistiek Antwoorden
Lecture notes of 100 pages for the course Kansrekening en Statistiek at TU Delft
[Meer zien]Voorbeeld 3 van de 100 pagina's
Voorbeeld 3 van de 100 pagina's
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Voorbeeld van de inhoud
29A Modern Introduction to Probability and
Statistics
Full Solutions
February 24, 2006
©F.M. Dekking, C. Kraaikamp, H.P. Lopuhaä, L.E. Meester
,458 Full solutions from MIPS: DO NOT DISTRIBUTE
29.1 Full solutions
2.1 Using the relation P(A ∪ B) = P(A)+P(B)−P(A ∩ B), we obtain P(A ∪ B) =
2/3 + 1/6 − 1/9 = 13/18.
2.2 The event “at least one of E and F occurs” is the event E ∪ F . Using the
second DeMorgan’s law we obtain: P(E c ∩ F c ) = P((E ∪ F )c ) = 1 − P(E ∪ F ) =
1 − 3/4 = 1/4.
2.3 By additivity we have P(D) = P(C c ∩ D)+P(C ∩ D). Hence 0.4 = P(C c ∩ D)+
0.2. We see that P(C c ∩ D) = 0.2. (We did not need the knowledge P(C) = 0.3!)
2.4 The event “only A occurs and not B or C” is the event {A ∩ B c ∩ C c }. We
then have using DeMorgan’s law and additivity
P(A ∩ B c ∩ C c ) = P(A ∩ (B ∪ C)c ) = P(A ∪ B ∪ C) − P(B ∪ C) .
The answer is yes , because of P(B ∪ C) = P(B) + P(C) − P(B ∩ C)
2.5 The crux is that B ⊂ A implies P(A ∩ B) = P(B). Using additivity we obtain
P(A) = P(A ∩ B) + P(A ∩ B c ) = P(B) + P(A \ B). Hence P(A \ B) = P(A) − P(B).
2.6 a Using the relation P(A ∪ B) = P(A) + P(B) − P(A ∩ B), we obtain 3/4 =
1/3 + 1/2 − P(A ∩ B), yielding P(A ∩ B) = 4/12 + 6/12 − 9/12 = 1/12.
2.6 b Using DeMorgan’s laws we get P(Ac ∪ B c ) = P((A ∩ B)c ) = 1 − P(A ∩ B) =
11/12.
2.7 P((A ∪ B) ∩ (A ∩ B)c ) = 0.7.
2.8 From the rule for the probability of a union we obtain P(D1 ∪ D2 ) ≤ P(D1 ) +
P(D2 ) = 2 · 10−6 . Since D1 ∩ D2 is contained in both D1 and D2 , we obtain
P(D1 ∩ D2 ) ≤ min{P(D1 ) , P(D2 )} = 10−6 . Equality may hold in both cases: for
the union, take D1 and D2 disjoint, for the intersection, take D1 and D2 equal to
each other.
2.9 a Simply by inspection we find that
A = {T T H, T HT, HT T }, B = {T T H, T HT, HT T, T T T },
C = {HHH, HHT, HT H, HT T }, D = {T T T, T T H, T HT, T HH}.
2.9 b Here we find that Ac = {T T T, T HH, HT H, HHT, HHH},
A ∪ (C ∩ D) = A ∪ ∅ = A, A ∩ Dc = {HT T }.
2.10 Cf. Exercise 2.7: the event “A or B occurs, but not both” equals C = (A∪B)∩
(A ∩ B)c Rewriting this using DeMorgan’s laws (or paraphrasing “A or B occurs,
but not both” as “A occurs but not B or B occurs but not A”), we can also write
C = (A ∩ B c ) ∪ (B ∩ Ac ).
2.11 Let the two outcomes be called 1 and 2. Then Ω = {1, 2}, and P(1) = p, P(2) =
p2 . We must
√ have P(1) + P(2) = √P(Ω) = 1, so p + p2 = 1. This has two solutions:
p = (−1 + 5 )/2 and√ p = (−1 − 5 )/2. Since we must have 0 ≤ p ≤ 1 only one is
allowed: p = (−1 + 5 )/2.
2.12 a This is the same situation as with the three envelopes on the doormat, but
now with ten possibilities. Hence an outcome has probability 1/10! to occur.
2.12 b For the five envelopes labeled 1, 2, 3, 4, 5 there are 5! possible orders, and
for each of these there are 5! possible orders for the envelopes labeled 6, 7, 8, 9, 10.
Hence in total there are 5! · 5! outcomes.
, 29.1 Full solutions 459
2.12 c There are 32·5!·5! outcomes in the event “dream draw.” Hence the probability
is 32 · 5!5!/10! = 32 · 1 · 2 · 3 · 4 · 5/(6 · 7 · 8 · 9 · 10) = 8/63 =12.7 percent.
2.13 a The outcomes are pairs (x, y).
a b c d
The outcome (a, a) has probability 0 to
1 1 1
occur. The outcome (a, b) has probability a 0 12 12 12
1 1 1
1/4 × 1/3 = 1/12 to occur. b 12
0 12 12
1 1 1
The table becomes: c 12 12
0 12
1 1 1
d 12 12 12
0
2.13 b Let C be the event “c is one of the chosen possibilities”. Then C =
{(c, a), (c, b), (a, c), (b, c)}. Hence P(C) = 4/12 = 1/3.
2.14 a Since door a is never opened, P((a, a)) = P((b, a)) = P((c, a)) = 0. If the can-
didate chooses a (which happens with probability 1/3), then the quizmaster chooses
without preference from doors b and c. This yields that P((a, b)) = P((a, c)) = 1/6.
If the candidate chooses b (which happens with probability 1/3), then the quizmas-
ter can only open door c. Hence P((b, c)) = 1/3. Similarly, P((c, b)) = 1/3. Clearly,
P((b, b)) = P((c, c)) = 0.
2.14 b If the candidate chooses a then she or he wins; hence the corresponding
event is {(a, a), (a, b), (a, c)}, and its probability is 1/3.
2.14 c To end with a the candidate should have chosen b or c. So the event is
{(b, c), (c, b)} and P({(b, c), (c, b)}) = 2/3.
2.15 The rule is:
P(A ∪ B ∪ C) = P(A)+P(B)+P(C)−P(A ∩ B)−P(A ∩ C)−P(B ∩ C)+P(A ∩ B ∩ C) .
That this is true can be shown by applying the sum rule twice (and using the set
property (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)):
P(A ∪ B ∪ C) = P((A ∪ B) ∪ C) = P(A ∪ B) + P(C) − P((A ∪ B) ∩ C)
= P(A) + P(B) − P(A ∩ B) + P(C) − P((A ∩ C) ∪ (B ∩ C))
= s − P(A ∩ B) − P((A ∩ C)) − P((B ∩ C)) + P((A ∩ C) ∩ (B ∩ C))
= s − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C) .
Here we did put s := P(A) + P(B) + P(C) for typographical convenience.
2.16 Since E ∩ F ∩ G = ∅, the three sets E ∩ F , F ∩ G, and E ∩ G are disjoint.
Since each has probability 1/3, they have probability 1 together. From these two
facts one deduces P(E) = P(E ∩ F ) + P(E ∩ G) = 2/3 (make a diagram or use that
E = E ∩ (E ∩ F ) ∪ E ∩ (F ∩ G) ∪ E ∩ (E ∩ G)).
2.17 Since there are two queues we use pairs (i, j) of natural numbers to indicate
the number of customers i in the first queue, and the number j in the second queue.
Since we have no reasonable bound on the number of people that will queue, we
take Ω = {(i, j) : i = 0, 1, 2, . . . , j = 0, 1, 2, . . . }.
2.18 The probability r of no success at a certain day is equal to the probability
that both experiments fail, hence r = (1 − p)2 . The probability of success for the
first time on day n therefore equals rn−1 (1 − r). (Cf. Section2.5.)
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