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 n−x objects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects six numbers from 1 to 60 (without repetition), and a winning six-number combination is later randomly selected. Find the probabilities of getting exactly four winning numbers with one ticket. (Hint: Use A=6, B=54, n=6, and x=4.)

Question

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 n−x objects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects six numbers from 1 to 60 (without repetition), and a winning six-number combination is later randomly selected. Find the probabilities of getting exactly four winning numbers with one ticket. (Hint: Use A=6, B=54, n=6, and x=4.)

P(4)= ??

Answer 1

To find the probability of getting exactly four winning numbers with one ticket in the lottery game, we can use the hypergeometric distribution formula.

The formula for the hypergeometric distribution is:

P(x) = (A C x) * (B C (n - x)) / (A + B C n)

where:
P(x) is the probability of getting x objects of type A and n - x objects of type B,
(A C x) is the number of combinations of A objects taken x at a time,
(B C (n - x)) is the number of combinations of B objects taken (n - x) at a time, and
(A + B C n) is the number of combinations of the total population taken n at a time.

In this case, we have A = 6 (the number of possible winning numbers), B = 54 (the number of non-winning numbers), n = 6 (the number of numbers selected), and x = 4 (the desired number of winning numbers).

Plugging these values into the formula, we get:

P(4) = (6 C 4) * (54 C (6 - 4)) / (6 + 54 C 6)

To calculate the combinations, we use the formula:

n C r = n! / (r! * (n - r)!)

where n! is the factorial of n.

Calculating the combinations, we have:

(6 C 4) = 6! / (4! * (6 - 4)!) = 6! / (4! * 2!) = (6 * 5 * 4 * 3) / (4 * 3 * 2 * 1) = 15
(54 C (6 - 4)) = 54! / ((6 - 4)! * (54 - (6 - 4))!) = 54! / (2! * 50!) = (54 * 53) / (2 * 1) = 1,431
(6 + 54 C 6) = (6 + 54)! / (6! * 54!) = 60! / (6! * 54!) = (60 * 59 * 58 * 57 * 56 * 55) / (6 * 5 * 4 * 3 * 2 * 1) = 2,413,210

Plugging these values into the formula, we get:

P(4) = 15 * 1,431 / 2,413,210 ≈ 0.00888

Therefore, the probability of getting exactly four winning numbers with one ticket is approximately 0.00888.