Find probability of determinate data on a sample of a finite population

[JKAW]

I would like to know if my answer is correct

Question: If a random sample of size n = 3 is drawn from a finite population of size N = 50, what is the probability that a particular element of the population will be included in the sample?

Answer: ((49C2) x 3) / (50C3) = 0.18 = 18% I calculated the number of possible groups of 2 of 49 people and then multiplied by 3 because you can have the wanted person in every of the 3 positions and then divided by all the possible groups of 50 people taken by 3

 

[JKAW]

Solution: Hypergeometric distribution P(X=1) = 3/50

 

[pka]

Lets say that \(X\) is the particular element . There are \(\dbinom{50}{3}=19600\) ways to choose three from fifty.
Of those there are \(\dbinom{49}{2}=1176\) that contain \(X\). So now what?
SEE HERE

 

[JeffM]

There is a reason you are taught urn problems.

You have an urn with 1 red ball and 49 black balls.

You draw three balls without replacement. What is the probability that one of the three is red.

Two ways to solve. You can add up the probabilities of selecting the red ball on the first or second or third draw.

[MATH]\dfrac{1}{50} + \dfrac{\cancel {49}}{50}* \dfrac{1}{\cancel {49}}+ \dfrac{\cancel {49}}{50} * \dfrac{\cancel {48}}{\cancel {49}} * \dfrac{1}{\cancel {48}} = \dfrac{1}{50} + \dfrac{1}{50} + \dfrac{1}{50} = \dfrac{3}{50}.[/MATH]
Or you can divide the number of ways to select 1 red ball and 2 black balls by the number of ways to select 3 balls out of 50.

[MATH]\dbinom{1}{1} * \dbinom{49}{2} \div \dbinom{50}{3} = \dfrac{49!}{2! * \cancel {47!}} * \dfrac{3! * \cancel {47!}}{50!} = \dfrac{3 * \cancel 2 * \cancel {49!}}{50 * \cancel 2 * \cancel{49!}} = \dfrac{3}{50}.[/MATH]