Question

A certain university has 18 vehicles available for use by faculty and staff. Five of these are vans and 13 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 18 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 ∩ F).

Answer 1

Answer:

a

 $P(E) = 0.278$

b

  $P(F |E) = 0.2353$

c

$P(E and F) = 0.0654$

Step-by-step explanation:

The number of vehicles available is $N = 18$

The number of vans is k = 5

The number of cars is n = 13

Generally P(E) is mathematically represented as

$P(E) = \frac{k}{N}$

=>    $P(E) =  \frac{5}{18}$

=>    $P(E) = 0.278$

Generally P(F| E)  means the probability that the second vehicle assigned is  a van given that the first one selected is a van this mathematically calculated as

      $P(F |E) =  \frac{5 -1}{18-1}$

        $P(F |E) =  \frac{4}{17}$

        $P(F |E) = 0.2353$

Generally  $P(F | E)$ is also mathematically represented as

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

=>   $P(E and  F) =P(E) * P(F | E)$

=>   $P(E and  F) = 0.278  * 0.2353$

=>$P(E and  F) = 0.0654$