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08 Jan 2025, 04:44
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If a car is randomly selected from a factory's production line, what is the probability that the selected car is electric?
(1) The probability that a car selected at random is neither red nor electric is \(\frac{1}{3}\).
(2) The probability that a car selected at random is red but not electric is \(\frac{2}{15}\).
(1) The probability that a car selected at random is neither red nor electric is \(\frac{1}{3}\).
(2) The probability that a car selected at random is red but not electric is \(\frac{2}{15}\).
08 Jan 2025, 04:44
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Official Solution:
If a car is randomly selected from a factory's production line, what is the probability that the selected car is electric?
This question is essentially about overlapping groups framed as a probability problem. We use the formula:
Total = Electric + Red - Both + Neither
Our goal is to calculate Electric/Total * 100. To make calculations easier, let's assume Total = 45 (the least common multiple of the denominators from the statements).
(1) The probability that a car selected at random is neither red nor electric is \(\frac{1}{3}\).
45 = Electric + Red - Both + 15
Not sufficient.
(2) The probability that a car selected at random is red but not electric is \(\frac{2}{15}\).
The above implies:
Red - Both = 6.
Not sufficient.
(1) + (2) Substituting Red - Both = 6 into 45 = Electric + Red - Both + 15, we get 45 = Electric + 6 + 15, which simplifies to Electric = 24. Therefore, Electric/Total * 100 = 24/45 * 100. Sufficient.
Answer: C
If a car is randomly selected from a factory's production line, what is the probability that the selected car is electric?
This question is essentially about overlapping groups framed as a probability problem. We use the formula:
Total = Electric + Red - Both + Neither
Our goal is to calculate Electric/Total * 100. To make calculations easier, let's assume Total = 45 (the least common multiple of the denominators from the statements).
(1) The probability that a car selected at random is neither red nor electric is \(\frac{1}{3}\).
45 = Electric + Red - Both + 15
Not sufficient.
(2) The probability that a car selected at random is red but not electric is \(\frac{2}{15}\).
The above implies:
Red - Both = 6.
Not sufficient.
(1) + (2) Substituting Red - Both = 6 into 45 = Electric + Red - Both + 15, we get 45 = Electric + 6 + 15, which simplifies to Electric = 24. Therefore, Electric/Total * 100 = 24/45 * 100. Sufficient.
Answer: C
13 Jan 2025, 10:01
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I don’t quite agree with the solution. How do we know that there are only red and electric cars at the factory? Couldn't it be true that there are blue cars?
