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dipanjali gogoi wrote at 2014-02-19 09:51:36
For this problem we need to calculate the probability, using the Poisson distribution, of getting 0, 1, 2 and 3 defects per packet and then multiplying by 10,000 to get the expected value of the number of defective packets per consignment.

The probability of a defect per blade is p = 1/500 = 0.002. This means that for a packet of 10, the mean number of defects L = 10p = 0.02. The parameter L is used in the Poisson distribution to give the probability of the number of defects, n, in a packet of 10:

P(n) = exp(-L)･(L^n)/n!

For n = 0, I get P(0) = 0.9802, which means that the expected value for the number of packets with no defects is 9802. For n = 1, P(1) = 0.0002. You can do the arithmetic for the other values of n; the probabilities are very small.

I'm not sure if the problem asks for just the individual number for each n or if it wants the total sum, but you can just add them up yourself. This actually might cause some confusion because you might notice that the sum over all 11 possible values for n in a packet (including 0) using the Poisson distribution formula will not be quite 1; it should be exactly 1 of course since it is the sum over all possible outcomes. The error is really small, however. The sum is exactly one if the binomial distribution is used; the Poisson distribution is an approximation to the binomial distribution.

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