Answer:
1) [tex]\displaystyle\frac{5}{18}\approx27.78\%[/tex]
2) [tex]\displaystyle\frac{4}{7}\approx57.14\%[/tex]
3) [tex]\displaystyle \frac{13}{16}=81.25\%[/tex]
Step-by-step explanation:
We are given a two-way frequency table. Using the table, we will determine the probability for each case.
Question 1)
P(A Student With A Part Time Job Without A Car)
Using the first column, the total number of students that have a part time job is 78+30=108.
Likewise, using the first column, we can see that out of those 108 students, 30 do not have a car.
Hence, the probability that a student with a part time job without a car is:
[tex]\displaystyle P=\frac{30}{108}=\frac{5}{18}\approx27.78\%[/tex]
Question 2)
P(No Car | Does Not Have A Part Time Job)
Remember that the vertical line means conditional probability.
So, we want the probability of a student having no car given that they do not have a part time job.
Using the second column, we can see that the total number of students that do not have a part time job is 18+24=42.
Likewise, using the second column, 24 do not have a car.
Hence, the probability that a student with a part time job without a car is:
[tex]\displaystyle P=\frac{24}{42}=\frac{4}{7}\approx57.14\%[/tex]
Question 3)
P(Part Time Job | Car)
So, we want to probability of a student having a part time job given that they have a car.
Using the first row, the total students that have a car is 78+18=96.
And of those 96 students, 78 have a part time job.
Hence, the probability that a student with a car has a part time job is given by:
[tex]\displaystyle P=\frac{78}{96}=\frac{13}{16}=81.25\%[/tex]
