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Lowest bidder

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 N 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.

Solution

P ( V t ) = 1 - g N [ P ( Y > t ) ] = 1 - ( 0 . 7 + 0 . 3 p ) 20 where p = P ( Y > t )

Solving graphically for p = P ( V > t ) , we get

p = [ 23 / 25 41 / 50 17 / 25 1 / 2 8 / 25 ] for t = [ 170 180 190 200 210 ]

Now g N ( s ) = ( 0 . 7 + 0 . 3 s ) 20 . 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
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[link] With a general counting variable

Suppose the number of bids is 1, 2 or 3 with probabilities 0.3, 0.5, 0.2, respectively.

Determine P ( V t ) in each case.

SOLUTION.

The minimum of the selected Y ' s 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 Y t . For each t , we are interested in the probability of one or more occurrences of the event Y t . This is essentially the problem in Example 7 from "Random Selection", with probability p = P ( Y t ) .

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.

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Batch testing

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 ( λ ) , 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., V > t ), what is the probability of no defective units?

SOLUTION

Since no defective units implies no failures in any reasonable test time, we have

{ N = 0 } { V > t } so that P ( N = 0 | V > t ) = P ( N = 0 ) P ( V > t )

Since N = 0 does not yield a minimum value, we have P ( V > t ) = g N [ P ( Y > t ) ] . Now under the condition above, the number of defective units N binomial ( n , p ) , so that g N ( s ) = ( q + p s ) n . If N is large and p is reasonably small, N is approximately Poisson ( n p ) with g N ( s ) = e n p ( s - 1 ) and P ( N = 0 ) = e - n p . Now P ( Y > t ) = e - λ t ; for large n

P ( N = 0 | V > t ) = e - n p e n p [ P ( Y > t ) - 1 ] = e - n p P ( Y > t ) = e - n p e - λ t

For n = 5000 , p = 0 . 001 , λ = 2 , and t = 1 , 2 , 3 , 4 , 5 , 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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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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