1
$\begingroup$

I'm having trouble with understanding a combinatorics question, and wanted to develop some intuition for these kinds of problems. The question is as follows:

There is a basket of 10 oranges and 9 apples. You randomly choose 6 fruits (without replacement). What is the probability that you choose at least 4 oranges?

Solution: Let Q=size of sample space, and P=size of event space. Q = The total number of choices you can make = ${19}\choose{6}$. P = The number of ways you can choose at least 5 oranges, which is ${{10}\choose{4} } {9\choose 2}+ {10\choose5}{9\choose1} + {10\choose6}{9\choose0} $. The probability is P/Q.

I understand this solution and can see why it is the right answer. However, I had an alternative way of thinking about this question: First you choose 4 oranges from a pool of 10, giving you ${10\choose 4} $. Then you choose your remaining 2 fruits from the fifteen fruits you have left, i.e. ${15\choose2} $.

By this logic, P= ${10\choose4}{15\choose2} $

Now, I know that this definitely is a wrong answer, but I don't know why that is the case. Why is this an invalid way to think about the problem? I know that the answer is too large, but I can't come up with an argument that disproves my own faulty thinking.

Any help in understanding this would be appreciated.

Thanks!

$\endgroup$
  • 1
    $\begingroup$ The difficulty here is that you are "marking" the first four oranges as special, but the oranges are indistinguishable. So, for instance, say that you wrote the number 1-10 on the oranges. The case where you use oranges 1-5 is counted 5 times: once where you first choose 1-4 and then choose 5, once where you choose 1,2,3,5 first and then choose 4, once when... $\endgroup$ – Nick Peterson Feb 2 '17 at 21:44
  • $\begingroup$ Thanks, I see what the problem is now. Then is it true that in general, you can only mutiply two combinations together when the results are distinguishable? $\endgroup$ – SharKCS11 Feb 2 '17 at 21:55
  • $\begingroup$ I wouldn't quite put it that way. Instead, I'd say that you can only multiply two counts when it comes to making decisions that don't impact each other. In this case, your choice of the first four oranges impacts which oranges are available for the second choice. $\endgroup$ – Nick Peterson Feb 2 '17 at 21:57
1
$\begingroup$

You are overcounting. If you choose oranges $1,2,3,4$ in the first step then $5,6$ in the second, those are the same six fruit as if you had chosen oranges $1,2,3,6$ in the first step then $4,5$ in the second. Your solution, however, counts them as different choices.

In other words, in your solution, the order matters.

$\endgroup$
  • $\begingroup$ Ok, I think I see what the problem is. Thanks. $\endgroup$ – SharKCS11 Feb 2 '17 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.