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A manufacturer seeks bids on a modification of one of his processing units. Twenty contractors are invited to bid. They bid with probability 0.3, so that the numberof bids binomial (20,0.3). Assume the bids Y i (in thousands of dollars) form an iid class. The market is such that the bids have a common distributionsymmetric triangular on (150,250). What is the probability of at least one bid no greater than 170, 180, 190, 200, 210? Note that no bid is not a low bid of zero, hence we must use the special case.
Solving graphically for , we get
Now . We use MATLAB to obtain
t = [170 180 190 200 210];p = [23/25 41/50 17/25 1/2 8/25];PV = 1 - (0.7 + 0.3*p).^20;
disp([t;p;PV]')
170.0000 0.9200 0.3848180.0000 0.8200 0.6705
190.0000 0.6800 0.8671200.0000 0.5000 0.9612
210.0000 0.3200 0.9896
Suppose the number of bids is 1, 2 or 3 with probabilities 0.3, 0.5, 0.2, respectively.
Determine in each case.
SOLUTION.
The minimum of the selected is no greater than t if and only if there is at least one Y less than or equal to t . We determine in each case probabilities for the number of bids satisfying . For each t , we are interested in the probability of one or more occurrences of the event . This is essentially the problem in Example 7 from "Random Selection", with probability .
t = [170 180 190 200 210];p = [23/25 41/50 17/25 1/2 8/25]; % Probabilities Y<= t are 1 - p
gN = [0 0.3 0.5 0.2]; % Zero for missing value
PV = zeros(1,length(t));for i=1:length(t)
gY = [p(i),1 - p(i)];
[d,pd]= gendf(gN,gY);
PV(i) = (d>0)*pd'; % Selects positions for d>0 and
end % adds corresponding probabilitiesdisp([t;PV]')170.0000 0.1451
180.0000 0.3075190.0000 0.5019
200.0000 0.7000210.0000 0.8462
[link] may be worked in this manner by using
gN = ibinom(20,0.3,0:20)
. The results, of course, are the same as in the previous solution.
The fact that the probabilities in this example are lower for each
t than in
[link] reflects the fact that
there are probably fewer bids in each case.
Electrical units from a production line are first inspected for operability. However, experience indicates that a fraction p of those passing the initial operability test are defective. All operable units are subsequenly tested in a batch under continuousoperation ( a “burn in” test). Statistical data indicate the defective units have times to failure Y i iid, exponential , whereas good units have very long life (infinite from the point of view of the test). A batch of n units is tested. Let V be the time of the first failure and N be the number of defective units in the batch. If the test goes t units of time with no failure (i.e., ), what is the probability of no defective units?
SOLUTION
Since no defective units implies no failures in any reasonable test time, we have
Since does not yield a minimum value, we have . Now under the condition above, the number of defective units binomial , so that . If N is large and p is reasonably small, N is approximately Poisson with and . Now ; for large n
For , and , MATLAB calculations give
t = 1:5;
n = 5000;p = 0.001;
lambda = 2;P = exp(-n*p*exp(-lambda*t));
disp([t;P]')
1.0000 0.50832.0000 0.9125
3.0000 0.98774.0000 0.9983
5.0000 0.9998
It appears that a test of three to five hours should give reliable results. In actually designing the test, one should probably make calculations with a number of differentassumptions on the fraction of defective units and the life duration of defective units. These calculations are relatively easy to make with MATLAB.
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