Answer to Question #127044 in Statistics and Probability for Izzeddin

Question

For the scenarios below, answer the following:
Is the situation Binomial or Hypergeometric?
Referring to your criteria, how do you know which it is?
Change the situation so that it could be calculated using the other distribution.

Scenario 2:
You draw 5 cards from a standard deck of 52 cards. If X is a random variable that represents the number of hearts you get, what is the probability of 3 or more hearts?

Scenario 3:
Your manager at work is giving away tickets for the local Lacrosse game. She has 10 tickets: 4 in the blue section and 6 in the Green section. She gives you two at random. What is the probability that they are both in the blue section?

Scenario 4:
40% of the music you listen to is by a Canadian artist. If you listen to 10 songs, what is the probability that 5 of them are by Canadian artists?

Answer 1

Scenario 2:

The random variable $X$ has a hypergeometric distribution with parameters 52, 13, 5 (see https://en.wikipedia.org/wiki/Hypergeometric_distribution). We have 52 cards and 13 hearts among them. We make a set of five cards and looking for the probability of 3 or more hearts. It fits into the scheme for a hypergeometric distribution. The aim is to find:

$P(X\geq3)=P(X=3)+P(X=4)+P(X=5)=\frac{C_{13}^3C_{39}^2}{C_{52}^5}+\frac{C_{13}^4C_{39}^1}{C_{52}^5}+\frac{C_{13}^5}{C_{52}^5}$

The binomial coefficients have the form:

$C_{13}^3=\frac{13!}{10!3!}=\frac{13\cdot12\cdot11}{6}=286$ ; $C_{39}^2=\frac{39!}{37!2!}=\frac{39\cdot38}{2}=741$;

$C_{13}^4=\frac{13!}{9!4!}=\frac{13\cdot12\cdot11\cdot10}{24}=715$ ; $C_{39}^1=\frac{39!}{38!1!}=39$ ; $C_{13}^5=\frac{13!}{8!5!}=\frac{13\cdot12\cdot11\cdot10\cdot9}{120}=1287$ ;

$C_{52}^5=\frac{52!}{47!5!}=\frac{52\cdot51\cdot50\cdot49\cdot48}{120}=2598960$ ;

We receive:

$P(X\geq3)=\frac{286\cdot741+715\cdot39+1287}{2598960}\approx0.0928$ (it is rounded to 4 decimal places)

In order to change the distribution, we make another formulation: There is a big amount of cards, 25% are hearts. We draw 5 cards. What is a probability that 3 or more of them are hearts?

Scenario 3:

The situation is similar to the previous one. $X$ (presents a number of tickets in a blue section) has a hypergeometric distribution with parameters 10, 4, 2.

$P(X=2)=\frac{C_{4}^2}{C_{10}^2}=\frac{6}{45}\approx0.1333$ (it is rounded to 4 decimal places)

In order to change the distribution, we make another formulation: There is a big amount of tickets. 40% are in the blue section. The manager gives two of them. What is a probability that they are in the blue section?

Scenario 4:

Suppose that $X$ is a random variable that presents a number of songs by Canadian artists. It has a binomial distribution with parameters 10 and 0.4. We have:

$P(X=5)=C_{10}^5(0.4)^5(0.6)^5=\frac{10\cdot9\cdot8\cdot7\cdot6}{5!}(0.24)^5\approx0.2007$ (it is rounded to 4 decimal places)

In order to change the distribution, we make anothe formulation: We take 10 songs from the collection of 100 songs that contains 40 songs by Canadian artists. What is the probability that 5 of them are by Canadian artists?

Answers: Scenario 2: 0.0928; Scenario 3: 0.1333; Scenario 4: 0.2007 (the answers are rounded to 4 decimal places)