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A property is offered for sale. Experience indicates the number N of bids is a random variable having values 0 through 8, with respective probabilities

Number 0 1 2 3 4 5 6 7 8
Probability 0.05 0.15 0.15 0.20 0.15 0.10 0.10 0.05 0.05

The market is such that bids (in thousands of dollars) are iid symmetric triangular on [150 250].Determine the probability of at least one bid of $210,000 or more.

gN = 0.01*[5 15 15 20 15 10 10 5 5];PY = 0.5 + 0.5*(1 - (4/5)^2) PY = 0.6800>>PW = 1 - polyval(fliplr(gN),PY) PW = 0.6536%alternate gY = [0.68 0.32]; [D,PD]= gendf(gN,gY); P = (D>0)*PD' P = 0.6536
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Suppose N binomial ( 10 , 0 . 3 ) and the Y i are iid, uniform on [ 10 , 20 ] . Let V be the minimum of the N values of the Y i . Determine P ( V > t ) for integer values from 10 to 20.

gN = ibinom(10,0.3,0:10); t = 10:20;p = 0.1*(20 - t); P = polyval(fliplr(gN),p) - 0.7^10P = Columns 1 through 70.9718 0.7092 0.5104 0.3612 0.2503 0.1686 0.1092 Columns 8 through 110.0664 0.0360 0.0147 0 Pa = (0.7 + 0.3*p).^10 - 0.7^10 % Alternate form of gNPa = Columns 1 through 70.9718 0.7092 0.5104 0.3612 0.2503 0.1686 0.1092 Columns 8 through 110.0664 0.0360 0.0147 0
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Suppose a teacher is equally likely to have 0, 1, 2, 3 or 4 students come in during office hours on a given day. If the lengths of the individual visits, in minutes, areiid exponential (0.1), what is the probability that no visit will last more than 20 minutes.

gN = 0.2*ones(1,5); p = 1 - exp(-2);FW = polyval(fliplr(gN),p) FW = 0.7635gY = [p 1-p]; % Alternate[D,PD] = gendf(gN,gY);PW = (D==0)*PD' PW = 0.7635
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Twelve solid-state modules are installed in a control system. If the modules are not defective, they have practically unlimited life. However, with probability p = 0 . 05 any unit could have a defect which results in a lifetime (in hours) exponential (0.0025). Under the usual independence assumptions, what is the probability the unit does not failbecause of a defective module in the first 500 hours after installation?

p = 1 - exp(-0.0025*500); FW = (0.95 + 0.05*p)^12FW = 0.8410 gN = ibinom(12,0.05,0:12);gY = [p 1-p];[D,PD] = gendf(gN,gY);PW = (D==0)*PD' PW = 0.8410
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The number N of bids on a painting is binomial ( 10 , 0 . 3 ) . The bid amounts (in thousands of dollars) Y i form an iid class, with common density function f Y ( t ) = 0 . 005 ( 37 - 2 t ) 2 t 10 . What is the probability that the maximum amount bid is greater than $5,000?

P ( Y 5 ) = 0 . 005 2 5 ( 37 - 2 t ) d t = 0 . 45
p = 0.45; P = 1 - (0.7 + 0.3*p)^10P = 0.8352 gN = ibinom(10,0.3,0:10);gY = [p 1-p];[D,PD] = gendf(gN,gY); % D is number of "successes"Pa = (D>0)*PD' Pa = 0.8352
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A computer store offers each customer who makes a purchase of $500 or more a free chance at a drawing for a prize. The probability of winning on a draw is 0.05.Suppose the times, in hours, between sales qualifying for a drawing is exponential (4). Under the usual independence assumptions, what is the expected time between a winning draw?What is the probability of three or more winners in a ten hour day? Of five or more?

N t Poisson ( λ t ) , N D t Poisson ( λ p t ) , W D t exponential ( λ p ) .

p = 0.05; t = 10;lambda = 4; EW = 1/(lambda*p)EW = 5 PND10 = cpoisson(lambda*p*t,[3 5]) PND10 = 0.3233 0.0527
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Noise pulses arrrive on a data phone line according to an arrival process such that for each t > 0 the number N t of arrivals in time interval ( 0 , t ] , in hours, is Poisson ( 7 t ) . The i th pulse has an “intensity” Y i such that the class { Y i : 1 i } is iid, with the common distribution function F Y ( u ) = 1 - e - 2 u 2 for u 0 . Determine the probability that in an eight-hour day the intensity will not exceed two.

N 8 is Poisson (7*8 = 56) g N ( s ) = e 56 ( s - 1 ) .

t = 2; FW2 = exp(56*(1 - exp(-t^2) - 1))FW2 = 0.3586
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The number N of noise bursts on a data transmission line in a period ( 0 , t ] is Poisson ( μ t ) . The number of digit errors caused by the i th burst is Y i , with the class { Y i : 1 i } iid, Y i - 1 geometric ( p ) . An error correcting system is capable or correcting five or fewer errors in any burst. Suppose μ = 12 and p = 0 . 35 . What is the probability of no uncorrected error in two hours of operation?

F W ( k ) = g N [ P ( Y k ) ] P ( Y k ) - 1 - q k - 1 N t Poisson ( 12 t )

q = 1 - 0.35; k = 5;t = 2; mu = 12;FW = exp(mu*t*(1 - q^(k-1) - 1)) FW = 0.0138
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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