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It is known that diskettes produced by a certain company will be defective with probability $0.0014$, independently of each other.*

--Find exactly-- and --compute approximately (?)-- the probability to find at least 2 diskettes to be defective , from a random sample of 1000 diskettes.

I'm new to this so here's my first thoughts, you don't have to use my notations:

*At this point we understand that 0.0014 is an independed event.

We should define the number of defective diskettes as $X$.

I guess, for starters, want to find $P(X \ge 2)$ which is equal to $$P(X=2)+P(X=3)+\dots+P(X=1000) = $$ $$\binom{2000}{2}(0.0014)^2(1-0.0014)^{2000-2}+\dots+\binom{2000}{2000}(0.0014)^{2000}(1-0.0014)^{0}$$

As ctlusto and Robert Israel stated the exact solution is right (and can be found with an easier way)


For the approximation solution:

So we get to the main questions (updated):

What does he mean by approximation?

How do you choose the right approximation (generally)?

Can someone help me solve it with a simple way, maybe some theory in between like in this question Probability about students and exams. would be appreciated.

By Robert Israel's guidance I think that by using Poisson distribution we have: $$1 - ({e^{-2.8}2.8^{0} \over 0!} + {e^{-2.8}2.8^{1} \over 1!})$$

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You correctly used the binomial distribution with $n= 2000$ and $p = .0014$ here (although, as ctlusto indicated, you would find it easier to compute $P(X < 2)$).

For the approximation, you want to use a Poisson distribution. When $n$ is large and $p$ is small, with $np$ not too big, the Poisson distribution with parameter $\lambda = np$ is a good approximation to the binomial distribution with parameters $n$ and $p$.

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  • $\begingroup$ Can you expand more about the approximation? $\endgroup$ Sep 8 '13 at 23:04
  • $\begingroup$ Look up "Poisson distribution" in your textbook. Almost certainly it will mention the approximation of the binomial distribution by the Poisson. Or look at en.wikipedia.org/wiki/Poisson_distribution#law_of_rare_events $\endgroup$ Sep 8 '13 at 23:42
  • $\begingroup$ Yes it does: $$ e^{-λ}λ^x \over x!$$. But how do we choose an approximation? And to solve this problem with Poisson would we do this: $$ 1 - ({e^{-2.8}2.8^{0} \over 0!} + {e^{-2.8}2.8^{1} \over 1!}) $$ $\endgroup$ Sep 8 '13 at 23:53
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    $\begingroup$ Yes, that's the correct approximation. $\endgroup$ Sep 9 '13 at 1:13
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    $\begingroup$ The "true" value (using the binomial distribution) is about $0.23086$; the "approximate" value (using the Poisson) is about $0.23108$. These are very close, so it's a very good approximation. $\endgroup$ Sep 9 '13 at 1:23
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While there's nothing technically wrong with your solution (you're thinking about it in the right way), there is a much simpler way to go about it.

Notice that the event "at least two defective diskettes" is the complement of the event "fewer than two defective diskettes." In other words, those two probabilities must sum to 1. Since the second event is much easier to deal with, you can compute that probability and then subtract it from 1 to find the answer you're interested in.

In other words, instead of computing $P(X \geq 2)$ directly, you should compute $1-P(X < 2)$. That's much simpler, since you only have to directly calculate $P(X=0)$ and $P(X=1)$. This is a handy technique to keep in mind as you learn about probability: sometimes it's much easier to calculate the complement of an event rather than the event itself.

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  • $\begingroup$ Nice way of thought but what about when they are asking for approximate solution? I don't know if i'm translating the term correctly (approximate solution)... $\endgroup$ Sep 8 '13 at 22:09

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