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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,54,4,4) = \frac{C_{4,2}C_{50,2}}{C_{54,4}} = 0.0232$

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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