Download Probability and Statistics Exam Solutions - Vrije Universiteit Amsterdam and more Exams Probability and Statistics in PDF only on Docsity!
Final Exam Solutions: Probability
and Statistics (code 400178)
Vrije Universiteit Amsterdam
Note that there can be more than one correct way to solve a problem.
Problem 1.
a) The probability that (at least) two persons are drinking the same cock- tail is 1 − 100 × 10099 × 10 ... ×^91 ∼ 0 .37.
b) The probability that someone else is drinking the same cocktail as you is 1 − 100 ×^99 9 10010 ∼^0 .09.
Problem 2.
a) Disregarding the trivial cases p = 0 and p = 1, the correct answer is p = 1/3. One can see this for example by noting that P (C) = 3 p(1 − p)(1 − p) and P (C|A) = (1 − p)(1 − p). These two values are the same only if p = 1/3.
b) Although C can be independent of A and B, as shown in a), A, B and C are never mutually independent because P (A, B, C) = 0 while P (A)P (B)P (C) > 0 (again disregarding the trivial case p = 0).
Problem 3.
a) A and B are clearly independent. It is also the case that A is in- dependent of C and B is independent of C. To see this note that P (C|A) = P (C|B) = (^56 )^5 and P (C) = 6 × 16 × (^56 )^5 = (^56 )^5.
b) The three events are not mutually independent since P (A, B, C) = 0 while P (A)P (B)P (C) > 0.
Problem 4.
a)
(n k
(^12 )k(^12 )n−k^ =
(n k
(^12 )n
b)
∑n i=k
(n i
(^12 )n
c)
( (^) n n+ 2
(^12 )n
Problem 5.
a) F (x) =
∫ (^) x 0 λe
−λy (^) dy = 1 − e−λy
b) P (X < y) = F (y) = 1 − e−λy^ = 1/2 implies that y = ln 2 λ.
Problem 6.
a) (N − k)/
( N
k+
b) The maximum likelihood estimate for N is 2 since this value maximizes the probability of extracting the marked ball. In fact, for N = 2 the probability of extracting the marked ball is 1, while for every other value N > 2 it is strictly smaller than 1.
Problem 7.
a) The distribution of the temperatures expressed in Fahrenheit is Normal.
b) If Y = a+bX, then E(Y ) = a+bE(X) and V ar(Y ) = b^2 V ar(X). For a Normal distribution, E(X) = μ and V ar(X) = σ^2. Therefore the new expected value is 32 + 95 30 = 86 and the new variance 8125 × 4 = 12.96.
Problem 8. The probability that k customers arrive in the next minute is λk k! e
−λ, and the expected time between arrivals is 1/λ minutes. Here λ = 1.
We then obtain the following answers.
a) e−^1
b) 1 minute
c) 1 − e−^1 − e−^1 − e − 1 2 −^
e−^1 6 = 1^ −^
8 3 e
The probability that no customer arrives in the next T minutes is e−λT^ with λ = 1. Using this and the the memoryless property of the exponential distribution, we obtain the following answers.
d) e−^3
e) 1 − e−^2
