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Material Type: Exam; Professor: Valko; Class: Introduction to the Theory of Probability; Subject: MATHEMATICS; University: University of Wisconsin - Madison; Term: Spring 2009;
Typology: Exams
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(a) We have an urn with 5 red and 18 blue balls and we pick 4 balls without replacement. We denote the number of red balls in the sample by X. This is a hypergeometric random variable with parameters n = 4, N = 23 and m = 5. the probability mass function is
P(X = k) =
k
5 −k
4
) , 0 ≤ k ≤ 4
(b) We have an urn with 5 red and 18 blue balls and we pick 4 balls with replacement. We denote the number of red balls in the sample by X. This is a binomial random variable with parameters n = 4, p = 235. The probability mass function is P(X = k) =
k
)k ( 18 23
) 4 −k , 0 ≤ k ≤ 4
(c) We roll a fair die until we get a three or a four. X denotes the number of rolls needed. This is a geometric random variable with parameter p = 23. The probability mass function is P(X = k) =^2 3
3 k−^1
k ≥ 1
(a) What is the value of the cumulative distribution function of X at 0.3?
(b) Compute the expectation of X.
i=− 1
p(a)a = (−1)^1 5
(c) Compute the variance of X. We first compute the second moment:
i=− 1
a^2 p(a) = 1 · 1 5
which gives V ar[X] = E[X^2 ] − E[X]^2 =^7 5
f (x) =
x if 1^ ≤^ x^ ≤^3 0 otherwise Compute the expectation and the variance of X. We first compute C using the fact that
−∞ f^ (x)dx^ = 1. ∫ (^) ∞
−∞
f (x)dx =
1
x dx^ =^ C^ log 3 which gives C = (^) log 3^1. The expectation is E[X] =
−∞
xf (x)dx =
1
x C x
dx = 2C = 2 log 3
The second moment is
E[X^2 ] =
−∞
x^2 f (x)dx =
1
x^2
x dx^ =^
Cx^2 2
x= x=1 =^
log 3. Thus the variance is V ar[X] = E[X^2 ] − E[X]^2 = 4 log 3
(log 3)^2
P(1176 ≤ X ≤ 1236) = P( √^1176 −^ np np(1 − p)
√X^ −^ np np(1 − p)
√^1236 −^ np np(1 − p)
√X^ −^ np np(1 − p)
√X^ −^ np np(1 − p)