Bunuel
A rental firm has only two distinct types of vehicles: minivans and sedans. Each vehicle runs on exactly one fuel type, either gasoline or electricity. If a vehicle is selected at random, is the probability that it is a sedan running on electricity greater than the probability that it is a minivan running on gasoline?
(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
Assume
MG = minivan running on gasoline
ME = minivan running on electricity
SG = sedan running on gasoline
SE = sedan running on electricity
We know that
P(MG) + P(ME) + P(SG) + P(SE) = 1
Question: Is P(SE) > P(MG)?
Statement (1): The chance that a vehicle is either a minivan OR runs on gasoline is 2/5.
The only type of vehicle that is neither a minivan nor runs on gasoline is a sedan that runs on electricity. Therefore,
P(SE) = 1 - 2/5 = 3/5
Now, since the combined probability of all the other three categories (MG, ME, SG) is 2/5, it means
P(MG) + P(ME) + P(SG) = 2/5
So P(MG) is part of this 2/5, meaning it must be less than or equal to 2/5.
Hence, we can conclude that 3/5 is definitely greater than 2/5.
Statement (2): The chance that a vehicle is either a sedan OR runs on electricity is 5/8.
Now the only kind of vehicle that is not a sedan and doesn’t run on electricity is a minivan running on gasoline. Therefore,
P(MG) = 1 - 5/8 = 3/8
So now we know the probability of MG.
Is P(SE) > 3/8?
But we don’t know anything about P(SE) from this statement. It could be smaller than 3/8 or larger.
Hence, the statement alone is not sufficient.
Option A