MATH SOLVE

4 months ago

Q:
# 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 six numbers from 1 to 56 (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 Aequals6, Bequals50, nequals6, and xequals4.)

Accepted Solution

A:

Answer:5/4324 = 0.001156337Step-by-step explanation:To better understand the hyper-geometric distribution consider the following example:
There are 100 senators in the US Congress, and suppose 60 of them are republicans so 100 - 60 = 40 are democrats).
We extract a random sample of 30 senators and we want to answer this question:
What is the probability that 10 senators in the sample are republicans (and of course, 30 - 10 = 20 democrats)?
The answer using the h-g distribution is:
[tex]\large \frac{\binom{60}{10}\binom{100-60}{30-10}}{\binom{100}{30}}=\frac{\binom{60}{10}\binom{40}{20}}{\binom{100}{30}}[/tex]
Now, imagine there are 56 senators (56 lottery numbers), 6 are republicans (6 winning numbers and 50 losers), we extract a sample of 6 senators (the bettor selects 6 numbers). What is the probability that 4 senators are republicans? (What is the probability that 4 numbers are winners?).
As we see, the situation is exactly the same, but changing the numbers. So the answer would be
[tex]\large \frac{\binom{6}{4}\binom{56-6}{6-4}}{\binom{56}{6}}=\frac{\binom{6}{4}\binom{50}{2}}{\binom{56}{6}}[/tex]
Now compute each combination separately:
[tex]\large \binom{6}{4}=\frac{6!}{4!2!}=15\\\\\binom{50}{2}=\frac{50!}{2!48!}=1225\\\\\binom{50}{6}=\frac{50!}{6!44!}=15890700[/tex]
and now replace the values:
[tex]\large \frac{\binom{6}{4}\binom{50}{2}}{\binom{56}{6}}=\frac{15*1225}{15890700}=\frac{18375}{15890700}=\frac{5}{4324}[/tex]
and that is it.
If the decimal expression is preferred then divide the fractions to get 0.001156337