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Probability question

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Probability question

by jnellaz » Mon Dec 15, 2008 8:24 am
One week a certain vehicle rental outlet has 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 Barara would not select two of the same type of vehicle (in other words, if one of them has chosen an SUV, the other won't take an SUV. an SUV would not even be considered). What is the probability that of the two vehicles chosen, one of them is a car or a van?

[spoiler]1037/1100[/spoiler][spoiler]
[/spoiler]
Last edited by jnellaz on Mon Dec 15, 2008 8:59 am, edited 1 time in total.
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Re: Probability question

by tritrantran » Mon Dec 15, 2008 8:56 am
jnellaz wrote:One week a certain vehicle rental outlet has 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 Barara would not select two of the same type of vehicle (in other words, if one of them has chosen an SUV, the other won't take an SUV. an SUV would not even be consider). What is the probability that of the two vehicles chosen, one of them is a car or a van?
Prob(Car) = Car/Total = 40/(40+12+28+20) = 40/100 = 2/5

Prob(Van) = VanTotal = 28/100 = 7/25

Prob(Car or Van) = Prob(Car) + Prob(Van) - Prob(Car and Van)

= (2/5) + (7/25) - 0 = 17/25

I think this may be wrong...I'm not sure.
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by jnellaz » Mon Dec 15, 2008 8:59 am
Sorry. I added the spoiler!
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Re: Probability question

by logitech » Mon Dec 15, 2008 9:49 am
https://www.manhattangmat.com/forums/mgm ... t1810.html
Last edited by logitech on Mon Dec 15, 2008 11:33 am, edited 1 time in total.
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by mowie » Mon Dec 15, 2008 9:50 am
One week a certain vehicle rental outlet has 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 Barara would not select two of the same type of vehicle (in other words, if one of them has chosen an SUV, the other won't take an SUV. an SUV would not even be considered). What is the probability that of the two vehicles chosen, one of them is a car or a van?


moment please..

first c or v:

68/100 * 1 (already fulfilled criteria, second choice does not matter)

first truck(afterwards only 99cars and 11 trucks left)

12/100 * 68/88

88=99-11=100-12

first suv

20/100 * 68/80

sum: 68 * (1/100 + 12/(100*88) + 20/(100*80))

~ 68/1000 * ( 10 + 1,4 + 2,5)
~ 68 * 14 / 1000
~ 952 / 1000

and time for poe now


I explained the exact result just a few posts below this one.
Last edited by mowie on Mon Dec 15, 2008 11:59 am, edited 1 time in total.
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by jnellaz » Mon Dec 15, 2008 10:16 am
I think the trick to this problem is that when you are calculating the probability of the first choice, say it is an SUV, per the condition - the second choice can not be an SUV so you have to subtract the remaining SUVs from the total number of vehicles. So for example:

Probability:
Andre chooses an SUV = 20/100
Barbara chooses a Truck = 12/(100 minus remaining SUVs) or 81
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by logitech » Mon Dec 15, 2008 11:08 am
jnellaz wrote:I think the trick to this problem is that when you are calculating the probability of the first choice, say it is an SUV, per the condition - the second choice can not be an SUV so you have to subtract the remaining SUVs from the total number of vehicles. So for example:

Probability:
Andre chooses an SUV = 20/100
Barbara chooses a Truck = 12/(100 minus remaining SUVs) or 81

I like where you going with this..let me edit my post
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by mowie » Mon Dec 15, 2008 11:55 am
Thats exactly the way I solved it.
But you need to be careful.
If one SUV is chosen there are not only 19 SUVs left, but also only 99 cars total.
I just was too lazy to calculate the result exactly.
I recognized that it is GREAT to approximate the calculation. So you should look at the answer choices first.

So, lets have a try to solve it exactly:
68 * (1/100 + 12/(100*88) + 20/(100*80))

68 * (1/100 + 12/8800 + 20/8000)
68 * (88/8800 + 12/8800 + 22/8800)
68 * (88+12+22)/8800
68 * 122 / 8800
2*2*17*2*61 / 8800
17 * 61 / 1100
(60 * 17 + 17 )/1100
(600 + 420 + 17)/1100
1037/1100
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by mrsmarthi » Mon Dec 15, 2008 6:05 pm
Here is another way arriving at the answer.

The question asked to find the probability that atleast a car or van is selected. To get this, we can find the probability of not selecting a car and Van.

Probability of not selecting the car and Van is equal to the probability of selecting an SUV and a Truck.

Hence P(Sub and Truck) = 20 / 100 * 12/80 + 12/100*20/88

Solving the above we get 63/1100.

Now P(Car or Van) = 1 - P(Not Car and Van)

==> 1-63/1100 = 1037/1100
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by tritrantran » Mon Dec 15, 2008 7:27 pm
mrsmarthi wrote:Here is another way arriving at the answer.

The question asked to find the probability that atleast a car or van is selected. To get this, we can find the probability of not selecting a car and Van.

Probability of not selecting the car and Van is equal to the probability of selecting an SUV and a Truck.

Hence P(Sub and Truck) = 20 / 100 * 12/80 + 12/100*20/88

Solving the above we get 63/1100.

Now P(Car or Van) = 1 - P(Not Car and Van)

==> 1-63/1100 = 1037/1100
I like this, so much simpler!
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