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There are 24 cars in the parking lot outside of my building. They are all either 2-door or 4-door, are either white, black, or red, and are either Ford, Honda, Cadillac, or Volvo. No two cars are the same with respect to these properties (in other words, for example, there are not two 4-door red Fords). If 2 cars are chosen at random, then what is the probability that they are different in all three of these properties? (For example, a 2-door white Volvo is different from a 4-door red Cadillac, but not from a 4-door white Honda).

Guest Apr 3, 2015

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 #1
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+5

 

 

I assume you're sayng that all the cars differ in at least one aspect......so...there are 24 "different" cars, thusly:

(H, W,2)          (V, W,2)            (F, W, 2)          (C, W,2) 

(H, W,4)          (V, W,4)            (F, W, 4)          (C, W,4)                   

(H, R, 2)          (V, R,2)             (F,  R,  2)         (C, R, 2)

(H, R, 4)          (V, R 4)             (F, R,  4)          (C, R, 4)  

 

Notice that there are 4 ways  to  pair the  Hondas with a "different" Volvo, "different" Ford  or "different" Cadillac.  So, 4 + 4 + 4 = 12

And there are 4 ways  to pair the Volvos wth a "different" Ford or "different' Cadillac. {We've already paired each with a "different" Honda.} So 4 + 4 = 8

And since we've already paired the Fords with a "different" Honda or Volvo, we have 4 ways that we can pair them with a "different" Cadillac.

So there are 12 + 8 + 4 = 24 "different" pairings

And the total number of possible pairs made by selecting any 2 cars from 24  = C(24,2) = 276

So 24 / 276 = 2/23 = about an 8.7% chance of pairing two "different" cars......

 

  

CPhill  Apr 3, 2015
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4+0 Answers

 #1
avatar+85644 
+5
Best Answer

 

 

I assume you're sayng that all the cars differ in at least one aspect......so...there are 24 "different" cars, thusly:

(H, W,2)          (V, W,2)            (F, W, 2)          (C, W,2) 

(H, W,4)          (V, W,4)            (F, W, 4)          (C, W,4)                   

(H, R, 2)          (V, R,2)             (F,  R,  2)         (C, R, 2)

(H, R, 4)          (V, R 4)             (F, R,  4)          (C, R, 4)  

 

Notice that there are 4 ways  to  pair the  Hondas with a "different" Volvo, "different" Ford  or "different" Cadillac.  So, 4 + 4 + 4 = 12

And there are 4 ways  to pair the Volvos wth a "different" Ford or "different' Cadillac. {We've already paired each with a "different" Honda.} So 4 + 4 = 8

And since we've already paired the Fords with a "different" Honda or Volvo, we have 4 ways that we can pair them with a "different" Cadillac.

So there are 12 + 8 + 4 = 24 "different" pairings

And the total number of possible pairs made by selecting any 2 cars from 24  = C(24,2) = 276

So 24 / 276 = 2/23 = about an 8.7% chance of pairing two "different" cars......

 

  

CPhill  Apr 3, 2015
 #2
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hmm 2/23 didnt work for me

Guest Apr 7, 2016
 #3
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Since there are  possible combinations of the above properties, each possible type of car is represented exactly once. Once we choose the first car, the second car has  possibility for the doors to be different,  possibilities for the color to be different, and  possibilities for the brand to be different. Thus there are  cars that have no properties in common with the first car. We could have chosen this second car in 23 equally-likely ways, and hence our answer is 6/23.

 

 

So the answer is 6/23 :)

Guest Apr 17, 2016
edited by Guest  Apr 17, 2016
 #4
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don't cheat on aops problems guys

Guest Aug 2, 2016

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