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I have the following problem:

Place $K$: where $35\%$ of students live
Place $N$: where $45\%$ of students live
Place $H$: where $20\%$ of students live

$4$ students are randomly selected
What is the probability that $2$ students are from $K$, $1$ from $N$ and $1$ from $H$

My way of solving:
Multinomial term: $(x^2)yz => (K^2)NH$
Coefficient: $\dbinom42\times\dbinom21\times\dbinom11 = 12$

=> $P(2K,1N,1H) = 12 * (0.35)^2 * (0.45) * (0.2) = 0.1323$

So, I thought about how to solve this in another way:

Sample space = $\dbinom{100}4$
$2$ students from $K = \dbinom{35}2$
$1$ student from $N = \dbinom{45}1$
$1$ student from $H = \dbinom{20}1$

$P(2K,1N,1H) = \dfrac{\left[ \dbinom{35}2\dbinom{45}1\dbinom{20}1 \right]}{ \dbinom{100}4} = 0.1366$

I have problems understanding why the second way does not give the same result.

JMoravitz
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jhuk
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1 Answers1

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It doesn't specify that there are a total of 100 students, only specifies percentages.

You will find that if you increase N to 200, say, the figure will change to 0.1344,

and if you increase it to 2000, it will become 0.1325.

(i.e. come nearer and nearer to the correct figure by the first approach)

Note

To be a bit more technical, the hypergeometric distribution converges to the binomial (or multinomial) as $N$ tends to infinity. See a proof here

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true blue anil
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