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$2000$ cashew nuts are mixed thoroughly in flour. The entire mixture is divided into $1000$ equal parts and each part is used to make $1$ biscuit. Assume that no cashews are broken in the process. A biscuit is picked at random. The probability that it contains no cashew is

$1.$ Between $0$ and $0.1$
$2.$ Between $0.1$ and $0.2$
$3.$ Between $0.2$ and $0.3$
$4.$ Between $0.3$ and $0.4$

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    Sorry. Made the edit – 2017-02-03
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    Roughly Poisson with $\lambda=2$ (average of $2$ cashews per biscuit). So $P(0)\approx e^{-2}\approx 0.135$. – 2017-02-03

4 Answers 4

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(I'll assume each cashew has equal probability to appear in each biscuit.) For each cashew, in how many ways can it not appear in your biscuit? That is, in how many other biscuits can it appear? In how many biscuits can it appear total? The ratio of these values gives you the probability that a single cashew doesn't appear in your biscuit. Once you have that probability, you can raise it to $2000$ to get the probability that none of the $2000$ almonds appear in your biscuit.

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    please show us some numbers – 2017-02-03
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    @miracle173 When I see a question where the OP doesn't show any attempt in finding the answer themselves, I try to encourage them to use my answer to try a little harder, rather than spoon-feed them the solution. I think I gave a sufficient outline for an answer, and the OP only needs to answer some very basic questions to derive the solution. – 2017-02-03
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    I respect this but I posted this request for the numbers not before two other users have published two complete but contradicting answers. Particularly with regards to the contradicting answers the numbers would be interesting. – 2017-02-03
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    @miracle173 That's fair; I'll leave it as a comment to avoid bumping the question up, but my answer should get $(\frac{999}{1000})^{2000}\approx .135$. – 2017-02-03
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Distributing cashews in biscuits is the same as putting them randomly in numbered boxes. This again is the same as pulling a numbered ball from a bag and putting it back again.

The probability a certain number/biscuit isn't drawn is obviously $1 - \frac{1}{2000} = 0.9995$.

Repeating this 2000 times gives the probability a certain number wasn't drawn. Hence $(1 - \frac{1}{2000})^{2000} \approx 0.368$

I ignore the fact that it is very unlikely that a biscuit contains more than 3 or 4 cashews in practice

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    It's not unlikely at all that a biscuit contains more than 2 or 3 cashews. It's not d(biscuit) , biscuit has a finite area. – 2017-02-03
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    and the number of biscuits(1000) does not matter? – 2017-02-03
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    1000 biscuits means the probability should be 0.999 – 2017-02-03
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    The calculation remains the same. The original post said 2000. it was ambiguous because the header stated 1000. – 2017-02-03
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Assuming that whether a cashew is in a particular biscuit is independent of other cashews in that biscuit (not true if cashews can crowd each other out) then

  • probability a particular cashew is in a particular biscuit is $\frac1{1000}=0.001$
  • probability a particular cashew is not in a particular biscuit is $1-\frac1{1000}=0.999$
  • probability no cashew is in a particular biscuit is $\left(1-\frac1{1000}\right)^{2000}\approx 0.1351999$

This last number is close to mjqxxxx's approximation of $e^{-2}\approx 0.1353353$

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    The "last number" isn't only close to mjqxxxx's approximation of $e^{−2}$ but it is close to $e^{−2}$ itself. – 2017-02-03
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    @miracle173 When I said "mjqxxxx's approximation of $e^-2$" I meant mjqxxxx saying $P(0)\approx e^-2$ – 2017-02-03
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For rough calculations, suitable to multiple-choice decisions, note that if there are $2000$ biscuits, your chance of not getting a cashew is roughly $\frac 1e \approx 0.37$. So given that each actual biscuit is like two of those smaller biscuits, you would have to have two "fails" i.e $0.37^2 \approx 0.14$. So option (2).