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 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 is P(x) = A!/(A - x)!x! middot B!/(B - n + x)!(n - x)! + (A + B)!/(A + B - n)!n! In a relatively easy lottery, a bettor selects 5 numbers from 1 to 13 (without repetition), and a winning 5-number combination is later randomly selected. What is the probability of getting all 5 winning numbers? Round your answer to four decimal places.

Answer 1

Answer:

0.0008

Step-by-step explanation:

The formula for a hypergeometric probability is

$\frac{(_AC_x)(_BC_{n-x})}{_NC_n}$

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

In this problem, N is 13.  A is 5, since there are 5 objects drawn, and B is 8, since there are 8 remaining objects.

The sample size, n, is 5, and x is 5:

$\frac{(_5C_5)(_8C_0)}{_{13}C_5}\\\\=\frac{1(1)}{1287}=\frac{1}{1287}\approx 0.0008$