One week a certain vehicle rental outlet had a total of 40 cars, 12 trucks, 28 vans, and 20 SUVs available. Andre and Barbara went to the vehicle rental outlet and chose 2 vehicles at random, with the condition that Andre and Barbara would not select two of the same type of vehicle (in other words, if one of them has an SUV, the other won't take an SUV, so the second person doesn't even consider the SUVs).
What is the probability that, of the two vehicles, one of them is a car or a van?
Total vehicles = 40 + 12 + 28 + 20 = 100
If we exclude the probability of selecting a truck and a SUV, then we get the probability that, of the two vehicles, one of them is a car or a van.
The probability of selecting a truck first and then a SUV = 12/100 * 20/88 = 3/25 * 5/22 = 3/110
The probability of selecting a SUV first and then a truck = 20/100 * 12/80 = 1/5 * 3/20 = 3/100
The probability of selecting a SUV first and then a truck in either order = 3/110 + 3/100 = 63/1100
The probability that, of the two vehicles, one of them is a car or a van = 1 - 63/1100 = 1037/1100
IMO D