A famous car production company produces 3 different models of cars, available in 5 colors and 3 possible CC(1900 cc,2000 cc and 2200 cc). suppose that every combination of model, color and CC has equal probability to be selected by a client:

  • If in one week 6 cars are sold, what is the probability that at least two cars are equal(same color,cc, and model)

My answer: Total possible combinations are 45^6, instead of finding "at least" two cars i would find 1-(Probability to get zero equal cars + probability to get one equal car) :

P(get 0 equal cars)= 44^6
P(Get 1 equal cars) 44^5

So the answer is: $1-(44^6+44^5/45^6)= 1-0.893= 0.107$

Is my answer correct?


  • $\begingroup$ One good way to approach problems with largish but arbitrary numbers is to make them smaller so you can see what's going on. Suppose there were only two kinds of car, chosen with equal probability. It looks like your formula in that case would be $1-(1^6 + 1^5/2^6),$ but that's a negative number (and obviously wrong). Could you solve this simpler two-car problem? If not, change $6$ to $2$ and try the two-car problem: it's easy to write down all the possibilities and their probabilities and looking at them should reveal how to proceed. $\endgroup$ – whuber Nov 20 '13 at 17:39
  • $\begingroup$ Why would i get a negative number with the example you gave? I would get $1-(2/2^6) = 1 - 0.03125= 0.6875$ $\endgroup$ – tutak Nov 20 '13 at 19:41
  • $\begingroup$ I see--you forgot some parentheses. Unfortunately, when they are inserted your formula is still incorrect. This is the old socks-in-a-drawer problem: suppose your drawer has two (or $45$) colors of socks. You are getting dressed in the dark, so you will take some socks out of the drawer and carry them into the light. How many must you take in order to be sure of having at least one pair of matching socks? The answer is three (or $(46)$). If you take more than that--such as six--you will still have a matching pair. Your formula therefore should give a value of $1$ in this case, not $0.6875.$ $\endgroup$ – whuber Nov 20 '13 at 19:56

What i wrote is not correct, it should have been: $1-((45!/39!)/45^6)$ where the nominator $(45!/39!)/45^6$ represents the probability to get zero equal cars. Having 6 cars, i have 45 choices for the 1st car, 44 for the 2nd, 43 for the 3d, 42 for the 4th, 41 for the 5th and 40 for the 6th.

Question answered by slash^ from ##statistics

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