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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 \(\frac{2}{5}\).
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is \(\frac{5}{8}\).
(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is \(\frac{2}{5}\).
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is \(\frac{5}{8}\).
07 Aug 2025, 04:03
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Official Solution:
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?
Firstly, to simplify, let's assume the total number of cars is 40, which is the LCM of the denominators of the probabilities given in the statements.
Next, notice that we have four distinct, non-overlapping types of cars: \(M_G\) (minivans on gasoline), \(M_E\) (minivans on electricity), \(S_G\) (sedans on gasoline), and \(S_E\) (sedans on electricity), and their total is \(M_G + M_E + S_G + S_E = 40\). We’re asked to find whether \(S_E > M_G\).
(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is \(\frac{2}{5}\).
This implies that \(M_G + M_E + S_G = \frac{2}{5} * 40 = 16\), so \(S_E = 40 - 16 = 24\). Thus, \(M_G\), which is less than or equal to 16, cannot be more than \(S_E = 24\). Sufficient.
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is \(\frac{5}{8}\).
This implies that \(S_E + S_G + M_E = \frac{5}{8} * 40 = 25\), so \(M_G = 40 - 25 = 15\). Since \(S_E\) is less than or equal to 25, it may or may not be greater than \(M_G = 15\). Not sufficient.
Answer: A
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?
Firstly, to simplify, let's assume the total number of cars is 40, which is the LCM of the denominators of the probabilities given in the statements.
Next, notice that we have four distinct, non-overlapping types of cars: \(M_G\) (minivans on gasoline), \(M_E\) (minivans on electricity), \(S_G\) (sedans on gasoline), and \(S_E\) (sedans on electricity), and their total is \(M_G + M_E + S_G + S_E = 40\). We’re asked to find whether \(S_E > M_G\).
(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is \(\frac{2}{5}\).
This implies that \(M_G + M_E + S_G = \frac{2}{5} * 40 = 16\), so \(S_E = 40 - 16 = 24\). Thus, \(M_G\), which is less than or equal to 16, cannot be more than \(S_E = 24\). Sufficient.
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is \(\frac{5}{8}\).
This implies that \(S_E + S_G + M_E = \frac{5}{8} * 40 = 25\), so \(M_G = 40 - 25 = 15\). Since \(S_E\) is less than or equal to 25, it may or may not be greater than \(M_G = 15\). Not sufficient.
Answer: A
