Question

A certain university has 8 vehicles available for use by faculty and staff. Six of these are vans and 2 are cars. On a particular day, only two requests for vehicles have been made. Suppose that the two vehicles to be assigned are chosen at random from the 8 vehicles available. (Enter your answers as fractions.)
a.) Let E denote the event that the first vehicle assigned is a van. What is P(E) ?
b.) Let F denote the probability that the second vehicle assigned is a van. What is P(F|E)?
c.) Use the results of parts(a) and (b) to calculate P(E and F)

Answer 1

Answer:

a) $P(E) = \frac{6}{8}$

b) $P(F|E) = \frac{5}{7}$

c) $P(E \cap F) = \frac{15}{28}$

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

We have that:

8 vehicles, of which 6 are vans.

a.) Let E denote the event that the first vehicle assigned is a van. What is P(E) ?

8 vehicles, of which 6 are vans.

So

$P(E) = \frac{6}{8}$

b.) Let F denote the probability that the second vehicle assigned is a van. What is P(F|E)?

P(F|E) is the probability that the second vehicle assigned is a van, given that the first one was.

In this case, there are 7 vehicles, of which 5 are vans. So

$P(F|E) = \frac{5}{7}$

c.) Use the results of parts(a) and (b) to calculate P(E and F)

$P(F|E) = \frac{P(E \cap F)}{P(E)}$

$P(E \cap F) = P(F|E)P(E)$

$P(E \cap F) = \frac{6}{8}\frac{5}{7}$

$P(E \cap F) = \frac{15}{28}$