# 15.1 Some random selection problems  (Page 3/6)

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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 $N\sim$ 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

$P\left(V\le t\right)=1-{g}_{N}\left[P\left(Y>t\right)\right]=1-{\left(0.7+0.3p\right)}^{20}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{where}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}p=P\left(Y>t\right)$

Solving graphically for $p=P\left(V>t\right)$ , we get

$p=\left[23/25\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}41/50\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}17/25\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}1/2\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}8/25\right]\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}t=\left[170\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}180\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}190\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}200\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}210\right]$

Now ${g}_{N}\left(s\right)={\left(0.7+0.3s\right)}^{20}$ . 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

## [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 $P\left(V\le t\right)$ in each case.

SOLUTION.

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

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.

## 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 $\left(\lambda \right)$ , 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

$\left\{N=0\right\}\subset \left\{V>t\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{so}\phantom{\rule{4.pt}{0ex}}\text{that}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(N=0|V>t\right)=\frac{P\left(N=0\right)}{P\left(V>t\right)}$

Since $N=0$ does not yield a minimum value, we have $P\left(V>t\right)={g}_{N}\left[P\left(Y>t\right)\right]$ . Now under the condition above, the number of defective units $N\sim$ binomial $\left(n,\phantom{\rule{0.166667em}{0ex}}p\right)$ , so that ${g}_{N}\left(s\right)={\left(q+ps\right)}^{n}$ . If N is large and p is reasonably small, N is approximately Poisson $\left(np\right)$ with ${g}_{N}\left(s\right)={e}^{np\left(s-1\right)}$ and $P\left(N=0\right)={e}^{-np}$ . Now $P\left(Y>t\right)={e}^{-\lambda t}$ ; for large n

$P\left(N=0|V>t\right)=\frac{{e}^{-np}}{{e}^{np\left[P\left(Y>t\right)-1\right]}}={e}^{-npP\left(Y>t\right)}={e}^{-np{e}^{-\lambda t}}$

For $n=5000,p=0.001,\lambda =2$ , 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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