程序代写案例-MAT 135 A

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MAT 135 A — Luis Rademacher
Midterm 1 — October 30th — 50 minutes
Open book and open notes. Every student must write his or her own solutions; looking for
solutions from external sources (the web, material from previous years, etc.) is prohibited.
Justify or prove all your answers.
1. (a) Sixty patients wish to enroll in a small study in which patients are divided into
four groups A,B,C,D (so, distinguishable groups) of 15 patients each. In how
many ways can this be done if every patient is assigned to exactly one group?
(b) Toss five fair coins and find the probability of three or more heads.
Solution: (a)
(
60
15
)(
45
15
)(
30
15
)
.
(b)
(53)+(
5
4)+(
5
5)
25
= 1/2.
2. Roll a six-sided die four times. Consider the events A = {all numbers are distinct},
B = {numbers are distinct and increasing}.
(a) Compute P (A), P (B), P (A | B), P (B | A).
(b) Are events A,B disjoint? Are events A,B independent? Justify your answer.
Solution: (a) 6 · 5 · 4 · 3/64. 6 · 5 · 4 · 3/(644!). 1. 1/4!
No. No.
3. An urn contains 6 balls labelled 1, 2, . . . , 6. Remove balls one by one at random (with-
out replacement) and stop when you extract the ball labelled 1.
(a) Determine the probability that you extract exactly 4 balls and the extracted balls
contain the balls labelled 2 and 3.
(b) Determine the probability that the extracted balls contain the balls labelled 2 and
3.
Solution: (a)
(31)
(53)
1
6
.
(b) Total probability, condition on the time 1 is extracted. 0 · 1
6
+ 0 · 1
6
+
(30)
(52)
1
6
+
(31)
(53)
1
6
+
(32)
(54)
1
6
+ 1 · 1
6
= 1/3.
Alternative solution: Continue removing all balls. We want the probability that 1 is
removed after 2,3. All permutations of 1,2,3 are equally likely so we want P(321 or
321), which is 1/3.
4. Let A,B,C be any three events in a probability space. Show:
(a) P (A ∪B ∪ C) ≤ P (A) + P (B) + P (C).
1
(b) P (A ∪B ∪ C) ≥ P (A) + P (B) + P (C)− P (A ∩B)− P (A ∩ C)− P (B ∩ C).
Solution: (a) P (A ∪B ∪ C) = P (A) + P (B \ A) + P (C \ (A ∪B)) ≤ . . ..
(b) P (A∪B∪C) = P (A) +P (B) +P (C)−P (A∩B)−P (A∩C)−P (B∩C) +P (A∩
B ∩ C) ≥ . . ..
5. Let X be a discrete random variable. Show that if Var(X) = 0 then there exists a ∈ R
such that P (X 6= a) = 0 (namely, X is a constant random variable).
Solution: If Var(X) = 0, then E((X − µ)2) = 0. Thus, ∑i(xi − µ)2P (X = xi) = 0,
which implies xi = µ for all i.
2

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