Question

from 6 - 8 in a group of 100 students 50 are eating whataburger (w) and 60 are drinking dr.pepper (d) and 40 are eating and drinking both.

1. find ( n(wcap d^{c}) ), the number of students only eating whataburger.

a. 10
b. 20
c. 30
d. 40
e. none of the above

1. find ( p(wcap d) ), the probability a student eats whataburger and drinks dr.pepper.

a. ( \frac{10}{100} )
b. ( \frac{20}{100} )
c. ( \frac{30}{100} )
d. ( \frac{40}{100} )
e. none of the above

1. find ( p(w|d) ), the conditional probability a student eats whataburger, given that a student drinks dr.pepper.

a. ( \frac{20}{100} )
b. ( \frac{20}{60} )
c. ( \frac{40}{60} )
d. ( \frac{40}{100} )
e. none of the above

Explanation:

6.

Step1: Use the formula for \(n(A\cap B^{c})\)

We know that \(n(W\cap D^{c})=n(W)-n(W\cap D)\).

Step2: Substitute the values

Given \(n(W) = 50\) and \(n(W\cap D)=40\). Then \(n(W\cap D^{c})=50 - 40=10\).

7.

Step1: Use the probability formula \(P(A\cap B)\)

The formula for probability \(P(A\cap B)=\frac{n(A\cap B)}{n(S)}\), where \(n(S) = 100\) (total number of students) and \(n(W\cap D)=40\).

Step2: Calculate the probability

\(P(W\cap D)=\frac{40}{100}\)

8.

Step1: Use the conditional - probability formula \(P(A|B)\)

The formula for conditional probability is \(P(W|D)=\frac{P(W\cap D)}{P(D)}\). Also, \(P(W\cap D)=\frac{n(W\cap D)}{n(S)}\) and \(P(D)=\frac{n(D)}{n(S)}\), so \(P(W|D)=\frac{n(W\cap D)}{n(D)}\) (since \(n(S)\) cancels out).

Step2: Substitute the values

Given \(n(W\cap D) = 40\) and \(n(D)=60\). Then \(P(W|D)=\frac{40}{60}\)

Answer:

1. A. 10
2. D. $\frac{40}{100}$
3. C. $\frac{40}{60}$