If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has A objects of one type, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and nminusx objects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects four numbers from 1 to 55 (without repetition), and a winning four-number combination is later randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use Aequals4, Bequals51, nequals4, and xequals2.) P(x)equalsStartFraction A exclamation mark Over left parenthesis Upper A minus x right parenthesis exclamation mark x exclamation mark EndFraction times StartFraction B exclamation mark Over left parenthesis Upper B minus n plus x right parenthesis exclamation mark left parenthesis n minus x right parenthesis exclamation mark EndFraction divided by StartFraction left parenthesis Upper A plus Upper B right parenthesis exclamation mark Over left parenthesis Upper A plus Upper B minus n right parenthesis exclamation mark n exclamation mark EndFraction

where A is the number of objects of type A, B is the number of objects of type B, N is the population size, n is the sample size, and x is the number of successes.

for this problem, N is 55. A is the number of objects pulled, which is 4. B is the remaining objects, which is 55. n is the sample size, which is 4, and x is 2:

[tex]\frac{(_4C_2)(_{51}C_2)}{_{55}C_4}\\\\=\frac{6(1275)}{341055}\\\\=\frac{7650}{341055} \approx 0.0224[/tex]