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The exercise:

A machine produces certain parts that are packaged in batches of 25 units. Usually produces 2% of defective parts.

Calculate the probability that there are two or more defective parts in a batch.

What I thought:

The $2$% of $25$ is $0.5$, then it produces $0.5 / 25 = 1/50 = p$ defective parts, and finally $25 = n$ parts must be analyzed for the probability of obtaining $X : "Number$ $of$ $defective$ $parts"$.

$P(X\geqslant2)=1-P(X<2)=1-P(X\leqslant 1) = 1-F_X(1)$

My doubts:

  • This is a $X$~$Bin(25,1/50)$?
  • The answer is $1-F_X(1)$?
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    I'd have used a Poisson model, personally, for simplicity. Of course the two answers are very close. To be clear: these are just models. The underlying process might be very different. – 2017-02-04
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    But my solution is correct? Or this is a $X$~$Bin(25,0.2)$? What about $P(X\geqslant2)=1-P(X<2)=1-P(1\leqslant X<2) = 1-F_X(1)$? Thanks. – 2017-02-04
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    It's a sensible model. "correct" seems too strong. Maybe the machine is usually perfect but once every $1250$ times it produces a batch of $25$ all of which are defective. That also gives a $2\%$ mean but it would yield a radically different answer to the question. – 2017-02-04
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    To be clear: you are assuming (with no evidence) that each unit has the same probability of being defective independent of the others in the batch. That doesn't seem crazy but neither does it seem inevitable. Maybe the defects cluster...maybe there is some external factor (like one operator is worse than the others). To answer the question you need to make some assumptions....just be clear which ones you are making. – 2017-02-04
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    To your detailed question...I'm not sure I follow. $P(X<2)=P(X=0)+P(X=1)$ regardless of the distribution we assume. You appear to be neglecting $P(0)$. – 2017-02-04
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    But if $F_X(x)$ is the cumulative probability to $x$, then consider the $P(0)$ as well. Or not? – 2017-02-04
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    You wrote $P(X<2)=P(1≤X<2)$. I don't understand that. – 2017-02-04
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    I mispronounced, what I wanted to put was that $P (X <2) = P (X≤1<2)$ – 2017-02-04
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    No, $P(X<2)=P(0≤X<2)$. – 2017-02-04
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    Ok, but $P(X<2)=F_X(1)$? – 2017-02-04
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    yes, that's right. – 2017-02-04

1 Answers 1

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Probability for no defectives is $(0.98)^{25}$

Probability for exact 1 defective piece is $\dbinom{25}{1}(0.02)^{1}(0.98)^{24}$

So, required probability is

$1-(0.98)^{25}-\dbinom{25}{1}(0.02)^{1}(0.98)^{24}$