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Two software companies, MicroWare and BusiCorp, are preparing a new business package in time for a computer trade show 180 days in the future. They work independently.MicroWare has anticipated completion time, in days, exponential (1/150). BusiCorp has time to completion, in days, exponential (1/130). What is the probability both willcomplete on time; that at least one will complete on time; that neither will complete on time?
p1 = 1 - exp(-180/150)
p1 = 0.6988p2 = 1 - exp(-180/130)
p2 = 0.7496Pboth = p1*p2
Pboth = 0.5238Poneormore = 1 - (1 - p1)*(1 - p2) % 1 - Pneither
Poneormore = 0.9246Pneither = (1 - p1)*(1 - p2)
Pneither = 0.0754 Eight similar units are put into operation at a given time. The time to failure (in hours) of each unit is exponential (1/750). If the units failindependently, what is the probability that five or more units will be operating at the end of 500 hours?
p = exp(-500/750); % Probability any one will survive
P = cbinom(8,p,5) % Probability five or more will surviveP = 0.3930 The location of ten points along a line may be considered iid randomvariables with symmytric triangular distribution on . What is the probability that three or more will lie within distance 1/2 of the point ?
Geometrically,
, so that
P = cbinom(10,p,3) = 0.9996 .
A Christmas display has 200 lights. The times to failure are iid, exponential (1/10000). The display is on continuously for 750 hours (approximatelyone month). Determine the probability the number of lights which survive the entire period is at least 175, 180, 185, 190.
p = exp(-750/10000)
p = 0.9277k = 175:5:190;
P = cbinom(200,p,k);disp([k;P]')175.0000 0.9973
180.0000 0.9449185.0000 0.6263
190.0000 0.1381 A critical module in a network server has time to failure (in hours of machine time) exponential (1/3000). The machine operates continuously, except for brieftimes for maintenance or repair. The module is replaced routinely every 30 days (720 hours), unless failure occurs. If successive units fail independently, whatis the probability of no breakdown due to the module for one year?
p = exp(-720/3000)
p = 0.7866 % Probability any unit survivesP = p^12 % Probability all twelve survive (assuming 12 periods)
P = 0.056 Joan is trying to decide which of two sales opportunities to take.
Let X be the net profit on the first alternative and Y be the net gain on the second. Assume the pair is independent.
with . , where binomial (8, 0.57).
c = [570 525 465 0];pm = minprob([0.57 0.41 0.35]);canonic % Distribution for X
Enter row vector of coefficients cEnter row vector of minterm probabilities pm
Use row matrices X and PX for calculationsCall for XDBN to view the distribution
Y = 150*[0:8]; % Distribution for Y
PY = ibinom(8,0.57,0:8);icalc % Joint distribution
Enter row matrix of X-values XEnter row matrix of Y-values Y
Enter X probabilities PXEnter Y probabilities PY
Use array operations on matrices X, Y, PX, PY, t, u, and Pxmax = max(X)
xmax = 1560ymax = max(Y)
ymax = 1200k = [600 900 1000 1100];px = zeros(1,4);for i = 1:4
px(i) = (X>=k(i))*PX';
endpy = zeros(1,4);
for i = 1:4py(i) = (Y>=k(i))*PY';
enddisp([px;py]')0.4131 0.7765
0.4131 0.25600.3514 0.0784
0.0818 0.0111M = u>t;
PM = total(M.*P)PM = 0.5081 % P(Y>X) 
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