Using the hypergeometric distribution, it is found that there is a 0.0232 = 2.32% probability of getting exactly two winning numbers with one ticket.
<h3>What is the hypergeometric distribution formula?</h3>
The formula is:
![P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7DC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
For this problem, the parameters are given as follows:
N =A + B = 54, k = 4, n = 4.
The probability of getting exactly two winning numbers with one ticket is P(X = 2), hence:
![P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7DC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
![P(X = 2) = h(2,54,4,4) = \frac{C_{4,2}C_{50,2}}{C_{54,4}} = 0.0232](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20h%282%2C54%2C4%2C4%29%20%3D%20%5Cfrac%7BC_%7B4%2C2%7DC_%7B50%2C2%7D%7D%7BC_%7B54%2C4%7D%7D%20%3D%200.0232)
There is a 0.0232 = 2.32% probability of getting exactly two winning numbers with one ticket.
More can be learned about the hypergeometric distribution at brainly.com/question/24826394
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