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Suppose in a class of $20$ students $25$% know Java .If we take a random sample of $5$ students,then What is the probability that exactly $2$ of them know Java ?

Well, I thought Binomial distribution is perfect for this.

$C(5,2) * (0.25)^2 * (0.75)^3= 0.263$

Even, I thought to use Hypergeometric distribution here,

$\frac{C(5,2) * C(15,3)}{C(20,5)} = 0.29$

Why is there such a difference ?

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    The hypergeometric distribution is the correct one to use here. The binomial distribution assumes that you are picking students independently each time and seeing if they know Java, which is not the case here. – 2017-02-02
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    @ParclyTaxel You mean picking one by one ? – 2017-02-02
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    Suppose that students numbered 1 to 5 know Java and those numbered 6 to 20 don't. The binomial model allows you to pick the same number multiple times. Sampling, though, forbids you from picking two with the same number. – 2017-02-02

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Hypergeometric distribution assumes that students are picked without replacement whereas Binomial distribution assumes that students are picked with replacement

Hypergeometric distribution is the correct one to use here

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    sorry, yes it is a typo. corrected now. – 2017-02-02