A computer factory received a batch of 1000 cheap chips. The
probability that a chip is defective is 0.001. Let X be the number of defective
chips in the batch.
(a) What kind of distribution does X have? What are the values of parameters
of this distribution?
I got the answer to part (a) ==> Binomial Distribution with $n=1000$, and $p=0.001$
(b) What is the probability that the batch contains no defective chips?
I thought of answering this as:
$P($No defective chip$) = 1 – P($All chips are defective$)$
$P($All chips are defective$) = 1000 * 0.001$
$P($No defective chip$) = 1 – 1 = 0$
Is this correct?
(c) One defective chip?
(d) More than two defective chips?
Please explain to me how this can be solved. Thank you
Best Answer
Your answer to part (b) is not correct, because the probability of no defective chip is not $1$ minus the probability that all chips are defective; it is $1$ minus the probability that at least one chip is defective. (That is to say, the complement of the event "no chips are defective" is "at least one chip is defective", not "every chip is defective".) Further, the probability that all chips are defective is not $1000 \times 0.001$, provided that chip failures are independent; it is just $0.001^{1000} = 10^{-3000}$, a very small number.
The probability that no chips are defective is the probability that chip $1$ is not defective ($0.999$), times the probability that chip $2$ is not defective ($0.999$), times the probability that chip $3$ is not defective ($0.999$), and so on, all the way to the probability that chip $1000$ is not defective ($0.999$). This long product can be written compactly as $0.999^{1000}$, a number that is very close to $1/e$, where $e \doteq 2.71828$ is the base of the natural logarithm.*
Part (b) can be seen as the special case of the binomial distribution probability distribution for this problem:
$$ P(\text{$k$ defective chips}) = \binom{1000}{k} 0.001^k 0.999^{1000-k} $$
This permits us to answer part (c) by just plugging in $k = 1$. Part (d) could ostensibly be answered by plugging in $k = 2$ through $k = 1000$, and summing the results, but it is answered most straightforwardly in the way that you tried to answer part (b): by observing that
$$ P(\text{$> 2$ defective chips}) = 1 - P(\text{$0$ defective chips}) - P(\text{$1$ defective chip}) $$
*Because one property of $e$ is
$$ 1/e = \lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n $$
and $n = 1000$ is pretty close to infinity for these purposes. :-)