Question

Suppose N= 15 and r=4. What is the probability of x = 3 for n = 10 (to 4 decimals)?

Answer 1

Using the Hypergeometric distribution principle, the probability of X = 3 in the function is 0.4396

Defining an hypergeometric distribution function :

$P(X = k) = \frac{ \binom{r}{k} \binom{N - r}{n - k} }{ \binom{N}{n} }$

The parameters given :

• N = 15
• r = 4
• X = 3
• n = 10

$P(X = 3) = \frac{ \binom{4}{3} \binom{15 - 4}{10 - 3} }{ \binom{15}{10} } = \frac{ \binom{4}{3} \binom{11}{7} }{ \binom{15}{10} }$

Using Combination :

Recall : nCr = n! ÷ (n-r)!r!

Using a calculator :

$P(X = 3) = \frac{ \binom{4}{3} \binom{15 - 4}{10 - 3} }{ \binom{15}{10} } = \frac{ \binom{4}{3} \binom{11}{7} }{ \binom{15}{10}} = \frac{(4 \times 330)}{3003} = \frac{1320}{3003} = 0.4396$

Therefore, the probability of X = 3 in the hypergeometric function is 0.4396 (rounded to 4 decimal places).

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