MGMAT Word Translations Probability Strategy Question #11

[jaredroc]

The question reads as follows:

One week a certain vehicle rental outlet had a total of 40 cars, 12 trucks, 28 vans, and 20 SUV's 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 SUV's). What is the probability that, of the two vehicles, one of them is a car or a van?

As an answer, the commentary states to take the complement of the various combinations that will NOT include either a car or a van. Thus taking the complement of the probability that Andre will pick a truck and Barbara an SUV and vice versa. This equals 1037/1100. This makes since but I don't understand why we don't have to subtract all of the chances that both Barbara and Andre will pick the same vehicle besides the truck or SUV (e.g. Barbara picks a car and Andre picks a car). Doesn't the 1037/1100 account for all of possibilities besides the SUV/Truck probability, including the times that they will pick the same vehicle?

Thanks in advance for your help! Hope my question is clear.

-Jared

[StaceyKoprince]

No, the 1100 possibilities do not include any instances in which they pick the same vehicle. Take a look at the chart in the explanation in the book. Note the parts of the problem when you set up the math:

(T=truck, S=SUV)

-----Andre--------Barbara
T (12/100)------S (20/88)
S (20/100)------T (12/80)

Notice that Andre can choose from among all 100 vehicles, but Barbara cannot (the denominators of the fractions), nor can she choose from among 99 (since Andre has already chosen 1). Instead, she can only choose from 88 in the first scenario (because Andre chooses one truck and then we remove the other 11 trucks because we know the constraint that she can't pick what he picks). Then, in the second scenario, Barbara only gets to choose from among 80 vehicles).

In other words, we've already removed from the problem the possibility that they pick the same vehicle, so that we don't have to deal with it later on. If you hadn't done that, then, yes, you'd have to subtract later on (and that would make the problem a lot more annoying).

[shaji]

The correct answer is 136/221. The 'MGMAT' answer is flawed. Please do an audit of all possibilities and U can spot the flaw.

Please provide all answer choices and I will list the complete audit for all to see.

No, the 1100 possibilities do not include any instances in which they pick the same vehicle. Take a look at the chart in the explanation in the book. Note the parts of the problem when you set up the math:

(T=truck, S=SUV)

-----Andre--------Barba//ra
T (12/100)------S (20/88)
S (20/100)------T (12/80)

Notice that Andre can choose from among all 100 vehicles, but Barbara cannot (the denominators of the fractions), nor can she choose from among 99 (since Andre has already chosen 1). Instead, she can only choose from 88 in the first scenario (because Andre chooses one truck and then we remove the other 11 trucks because we know the constraint that she can't pick what he picks). Then, in the second scenario, Barbara only gets to choose from among 80 vehicles).

In other words, we've already removed from the problem the possibility that they pick the same vehicle, so that we don't have to deal with it later on. If you hadn't done that, then, yes, you'd have to subtract later on (and that would make the problem a lot more annoying).

[StaceyKoprince]

There are no answer choices for this problem - this is not a GMAT-type question; it's just a pure math question.

[Shaji]

There are no answer choices for this problem - this is not a GMAT-type question; it's just a pure math question.

The total audit of all possblities.
There R 100 cars. The total number of possibilities are 100*99=9900 for selecting any two vehicles.

Consider all possibilities where the two vehicles R of diff type.

Car-Truck=40*12*2=960;
Car- Van=40*28*2=2240;
--
--
Likewise, the total=7072.
Now consider, the two vehicles are same.

Car=40*39=1560
Trucks=12*11=132
--
--
Lilke wise total is 2828

Now 2828+7072=9900. Complete audit

Now the required prob=4352*/7072=136/221

*7072-2240-480(deduct the car van & truck SUV possiblities

[StaceyKoprince]

Ah, I see. The language in the question is imprecise enough to allow two interpretations. Our solution assumes that the "car AND van" option should be included in the numerator and your solution assumes that it shouldn't. The question asks for the probability that "one of them is a car or a van" - which techically means: pick one and is that one a car or a van? (Without commentary on what the other one might be.) But you could also read that to mean: either a car or a van, but not both. I'll pass this along to our curriculum director for a look. Thanks!