Probability Question About Probabilities with "And", "Or", "Exactly", etc in it.?

Probability Question About Probabilities with "And", "Or", "Exactly", etc in it.?



In a survey of a recent Statistics class at a commuter college in Florida, students were asked if they owned their own home, 
rented an apartment or house, or lived with their parents. The table below shows the responses: 

1. What is the probability of selecting one student from this class and getting a male who owns his own home? 
2. What is the probability of selecting one student from this class and getting a female? 
3. What is the probability of selecting one student from this class and getting a male or someone who is renting a house or 
apartment? 
4. What is the probability that one student is selected and that student is a renter given that the student is known to be a male? 
5. What is the probability of selecting one student from this class and getting a male or a female? 
6. What is the probability of selecting two students from this class (without replacement) and getting two females? 

Can someone please help me and show your work? That would really help me learn this material.‚Äč





Posted Answers:

1st total members of the class = 3+12+5 +2+ 7+9 = 38 
2nd total number of males = 3+12+5 = 20 
3rd total number of females = 2+7 +9 = 18 
Probability are based = number of who met criteria/total number in group 

In several of the questions , the "and" means nothing. 
You could use two separate independent statement. 
1. What is the probability of selecting one student from this class and getting a male who owns his own home? 

There are 3 male who own a home out of 38 students 
P= 3/38 

2. What is the probability of selecting one student from this class and getting a female? 
There are 18 females in the class out of 38 students 
p = 18/38 = 9/19 

3. What is the probability of selecting one student from this class and getting a male or someone who is renting a house or 
apartment? 
P (male or a renter or apartment dweller) 
there are 20 males who are in the class 
additional there are 2 females who rent and 7 females who live in apartments. 
so the total meeting these requirement are 
20 + 2 + 7 = 29 
p = 29/38 
4. What is the probability that one student is selected and that student is a renter given that the student is known to be a male? 
Two way to determine this common sense 
P= number of male students who rent/ total number of male student. 
By definition = 12/20 = 3/5 
P( A | B) = P( A and (intersection) B) / P(B) 
Let A = student is a renter 
B = Student is male 
P(A intersection B) = P( student is a male and a renter) = 12/38 
P(B) = P(Student is a male) = 20/38 
P( A | B) = (12/38) / (20/38) = 12/20 = 3/5 
Here since it is simple conditional probability, you can do it either way. 
However, in more complicated situation, the definition can be helpful. 

5. What is the probability of selecting one student from this class and getting a male or a female? 
P( getting a male or female ) 
there are 20 males and 18 females out of class of 38 
P( getting a male or female ) = (20+18)/38 =38/38 = 1 
P(getting a male or female ) = 1 

6. What is the probability of selecting two students from this class (without replacement) and getting two females? 
Probability of 1st pick being females = 18/40 
for the second chose , they are only 17 females left of the 39 remaining students. 
since you need both event to happen 
P(getting two females) = 18/40*17/39 = 9/20*(17/39) =153/780