Question

I'm giving brainliest to the most helpful answer. For 50 points:

Answer 1

Answer:

1) $\displaystyle\frac{5}{18}\approx27.78\%$

2) $\displaystyle\frac{4}{7}\approx57.14\%$

3) $\displaystyle \frac{13}{16}=81.25\%$

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:

$\displaystyle P=\frac{30}{108}=\frac{5}{18}\approx27.78\%$

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:

$\displaystyle P=\frac{24}{42}=\frac{4}{7}\approx57.14\%$

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:

$\displaystyle P=\frac{78}{96}=\frac{13}{16}=81.25\%$