A certain university has 7 vehicles available for use by faculty and staff. Five 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 in a completely random fashion from among the 7.
(a) Let E denote the event that the first vehicle assigned is a van. What is P(E)?
I got the answer for this being .714
(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) (Hint: Use the definition of P(F | E).)
To answer these questions, we will use some basic principles of probability. The event E represents the first vehicle assigned being a van, and event F represents the second vehicle assigned being a van.
a) To find P(E), we need to determine the probability of the first vehicle being a van. There are a total of 7 vehicles, 5 of which are vans. Therefore, the probability of selecting a van as the first vehicle is 5/7.
P(E) = 5/7 ≈ 0.714
b) To find P(F|E), we need to calculate the probability of the second vehicle being a van given that the first vehicle assigned is a van. Since the first vehicle is a van, there are now 6 vehicles left, of which 4 vans remain. Therefore, the probability of selecting a van as the second vehicle, given that the first vehicle was a van, is 4/6.
P(F|E) = 4/6 = 2/3
c) To calculate P(E and F), we can use the definition of conditional probability:
P(E and F) = P(E) * P(F|E)
From parts (a) and (b) above:
P(E and F) = P(E) * P(F|E) = (5/7) * (2/3) = 10/21 ≈ 0.476
So, the probability that both the first and second assigned vehicles are vans is approximately 0.476.