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One percent of an amount of products are defect. How many products are required such that at least 1 product has the defect with probability 0.95?

What I came up with so far:

Declare $X: \text{the number of products which are defect.} $ so $X\sim Binomial(n,0.95)$

$p(X\geq1)=1-p(X=0)=1-0.05^n$

How do I proceed further?

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  • $\begingroup$ We have $\Pr(X\ge 1)=1-(0.99)^n$. $\endgroup$ – André Nicolas Jan 27 '16 at 17:43
  • $\begingroup$ Would you mind explaining why? $\endgroup$ – user122673 Jan 27 '16 at 17:48
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    $\begingroup$ One percent are defective. So the probability an item is good is $0.99$. The probability all $n$ items are good is therefore $(0.99)^n$. So the probability there is at least one bad is $1-(0.99)^n$. Now you will have to compute the suitable $n$ that gives probability at least one bad equal to roughly $0.95$. Logarithms will be involved. $\endgroup$ – André Nicolas Jan 27 '16 at 17:54
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For $n$ objects where each object has a defect with probability $d$ (and thus has no defects with probability $1 - d$), the probability that at least one has a defect is equal to $1$ minus the chance of there being no defects in every component, which gives us

$P(\text{chance of at least one defect}) = 1 - (1 - d)^{n}$.

We want to set this equal to $0.95$ and we have $d = 0.01$, which gives us the equation

$0.95 = 1 - (1 - 0.99)^{n}$

which you can solve for $n$.

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