Hypergeometric Distribution is a concept of statistics. It defines the chances that a specific number of successes would be attained when a certain number of draws are done. After withdrawals, replacements are not made. In terms of the formula used. “K” is the number of successes that have to be attained. Along with that, “N” is the total number of draws which have to be done. If you have a look at the concept of hypergeometric distribution, it is very similar to the binomial theorem.
The formula of hypergeometric distribution is given as follows.
\( P(X=k) = \dfrac{\dbinom{K}{k} \space \dbinom{N-K}{n-k}}{\dbinom{N}{n}} \)
Where:
\(K\) defines the number of successes in the population
\(k\) is the number of observed successes
\(N\) is the population size
\(n\) is the total number of draws
Consider that you have a bag of balls. In the bag, there are 12 green balls and 8 red balls. If you want to draw 5 balls from it out of which exactly 4 should be green. This means that one ball would be red. No replacements would be made after the draw.
When you apply the formula listed above and use the given values, the following interpretations would be made.
\( P(X=k) = \dfrac{(\mathrm{C}_{4}^{12})(\mathrm{C}_{1}^{8})}{(\mathrm{C}_{5}^{20})} \)
\( P ( X=k ) = 495 \times \dfrac {8}{15504} \)
\( P(X=k) = 0.25 \)
The Hypergeometric function is used for a hypergeometric series of numbers. Here when you talk about the hypergeometric function, it can be a Gaussian as well as the ordinary one.
What is probability? It defines the chances that the desired outcome would be obtained. However, when you are selecting between replacing and not replacing, the sample size changes. In other words, the probability value is affected.
Consider that we have a bag of glasses. There are fifteen glasses in total out of which 6 are green and 9 are yellow.
Consider that you want to draw a ball randomly. What is the probability that it would be green? There are 6 green balls and the total count is 15. Hence, the probability would be given.
The probability of getting a green ball will be.
\( P(Green) = \dfrac{6}{15} \)
Similarly, what is the probability that a randomly selected ball is yellow? There are 9 yellow balls and the total sample size is 15. Hence probability would be given as.
\( P(Yellow) = \dfrac{9}{15} \)
In both these cases, the same size remained the same because each time, the withdrawn ball was kept back.
Consider that you want to draw a random ball, what is the probability of getting a green one.
\( P(Green) = \dfrac{6}{15} \)
Now consider that you get a green ball and do not put it back. Here, the sample size would become 14. Hence, the probability of getting a yellow ball would be given.
\( P(Yellow) = \dfrac{9}{14} \)
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