probability questions

2010-04-14 6:26 am

1. Let X be a random variable taking values in the positive integers,
with probablity mass function of the form:

px(k) = a.3^(-k)

a) Determine a.
b) Is X more likely to take even or odd values?
c) If they exist, calculate E(X) and V(X)

2. Ten computers are linked by only one line to the network. Each computer needs to log on to the network for 12 minutes an hour on average. Sessions initiated by different computers are independent of each other, and two computers cannot use the line simultaneously.

a) For k= 0,...10, what is the probablity that k computers simultaneously need the line? What is the most likely number of computers requiring the line at one time?

b) We add lines to the network, allowing simultaneous sessions. What is the minimum number c of lines that are reqiured so that the probablity of overload at some fixed time t is less than 0.007?

3. Let X be a continuose non negative random variable with distribution function F (x) and probablity density function f(x). Define the failure rate function as r(x) = f( x) / ( 1-F(x) )

and the survival function G(x) = 1 - F(x)

a) show that for a non negative continuous random variable X with failure rate function r, X has survival function
G(x)= exp ( - int ( r(u) du) )

b) Consider a certain type of ball-bearing that wears down gradually at a slow but steady rate. To model this, we suppose that its failure rate function is linearly increasing r(x) = ax for some a >0. It has been observed that the median lifetime is 3 years. What is the probability that such a ball-bearing lasts for more than 4 years?

c) Find the probablity density functions of non negative continous random variables with failure rate functions

i) r (x) = x^3
ii) r(x) = 1 = (1+x)

回答 (1)

匿名

2010-04-14 10:27 pm

✔ 最佳答案

1a: ∑px(k) = 1
1 = a * 3-1 + a * 3-2 + a * 3-3 + … …
1 = (a / 3) * [1 + (1 / 3) + (1 / 3)2 + (1 / 3)3 + … …]
1 = (a / 3) * {1 / (1 – (1 / 3)]}
a = 2

1b:
px(1) + px(3) + px(5) + … … = (2 / 3) * [1 + (1 / 9) + (1 / 9)2 + … …] = 0.75
px(2) + px(4) + px(6) + … … = (2 / 9) * [1 + (1 / 9) + (1 / 9)2 + … …] = 0.25
Hence, odd nos is more likely.

1c:
1 + (1 / 3) + (1 / 3)2 + (1 / 3)3 + … … = 1 / [1 – (1 / 3)] = 3 / 2 = 1.5
E(X) = ∑ [x * px(k)] = (2 / 3) * [1 + 2 * (1 / 3)1 + 3 * (1 / 3)2 + 4 * (1 / 3)3 + … …] = (2 / 3) * {[1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + (1 / 3)1 * [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + (1 / 3)2 * [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + (1 / 3)3 * [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + … …} = (2 / 3) * [(3 / 2) + (1 / 3)1 * (3 / 2) + (1 / 3)2 * (3 / 2) + (1 / 3)3 * (3 / 2) + … …] = (2 / 3) * (3 / 2) * [1 + (1 / 3) + (1 / 3)2 + (1 / 3)3 + … …] = 1.5
E(X2) = ∑ [x2 * px(k)] = (2 / 3) * [12 + 22 * (1 / 3)1 + 32 * (1 / 3)2 + 42 * (1 / 3)3 + … …] = (2 / 3) * {[1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + 3 * (1 / 3)1 * [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + 5 * (1 / 3)2 * [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + 7 * (1 / 3)3 * [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + … …} = (2 / 3) * [(3 / 2) + 3 * (1 / 3)1 * (3 / 2) + 5 * (1 / 3)2 * (3 / 2) + 7 * (1 / 3)3 * (3 / 2) + … …] = (2 / 3) * (3 / 2) * [1 + 3 * (1 / 3) + 5 * (1 / 3)2 + 7 * (1 / 3)3 + … …] = [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + 2 * (1 / 3)1 * [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + 2 * (1 / 3)2 * [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + 2 * (1 / 3)3 * [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] + … … = (3 / 2) + 2 * (1 / 3)1 * (3 / 2) + 2 * (1 / 3)2 * (3 / 2) + 2 * (1 / 3)3 * (3 / 2) + … … = (3 / 2) + 2 * (3 / 2) * (1 / 3) * [1 + (1 / 3)1 + (1 / 3)2 + (1 / 3)3 + … …] = (3 / 2) + (3 / 2) = 3
Var(X) = E(X2) – [E(X)]2 = (3 / 4) = 0.75

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