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 minus ?x objects of type B under the hypergeometric distribution is given by the following formula. In a lottery? game, a bettor selects four four numbers from 1 to 49 49 ?(without repetition), and a winning four four?-number combination is later randomly selected. Find the probabilities of getting exactly two two winning numbers with one ticket.? (Hint: Use A equals = 4 4?, B equals = 45 45?, n equals = 4 4?, and x equals = 2 2?.)

Answer 1

Suppose a population consists of N items, k of which are successes. And a random sample drawn from that population consists of n items, x of which are successes. Then the hypergeometric probability is:

h(x; N, n, k) = [ ^kC_x ] [ ^N-kC_n-x ] / [ ^NC_n ]

For the given problem, N = 49, k = 4, (population consists of 49 numbers, 4 of which are successes/winning numbers)

n = 4, x = 2 (random sample drawn from that population consists of 4 items, 2 of which are successes/winning numbers)

So, the probabilities of getting exactly two two winning numbers with one ticket is,

h(x=2; N=49, n=4, k=4) = [ ⁴C₂ ] [ ^49-4C_4-2 ] / [ ⁴⁹C₄ ] =  [ ⁴C₂ ] [ ⁴⁵C₂ ] / [ ⁴⁹C₄ ]

= [ (4 * 3)/2 ] [ 45 * 44 / 2] / [(49 * 48 * 47 * 46) / (4 * 3 * 2)]

= 6 * 990 / 211876

= 0.02803527