0
$\begingroup$

For a set that contains $=\{1,2,3,4,5,6,7,8,9\}$

How many times minimally do I have to randomly pick a number from the set to ensure the sum is even?

$\endgroup$
  • $\begingroup$ any thought about the question? $\endgroup$ – Siong Thye Goh Nov 1 '18 at 15:12
  • 2
    $\begingroup$ Hint: for a bad scenario, you must first draw an odd number and thereafter draw even numbers. $\endgroup$ – lulu Nov 1 '18 at 15:17
  • $\begingroup$ Is there any general formula to use? $\endgroup$ – user10456099 Nov 1 '18 at 15:20
  • 1
    $\begingroup$ Please show some efforts. If you can solve the problem in this case, I think you will see how to do it in general. $\endgroup$ – lulu Nov 1 '18 at 15:21
  • $\begingroup$ Do refer to my working below,i am not sure if it is the correct answer. $\endgroup$ – user10456099 Nov 1 '18 at 17:26
1
$\begingroup$

Remember: $$even+even=even$$ $$odd+odd=even$$ $$odd+even=odd$$

If you want to ensure the sum is even, you have to find the maximum number of picks. $$odd+even+even+even+even=odd$$ is the maximum number of picks to get odd.

So minimum of picks to ensure you have an even sum is six.

$\endgroup$
  • $\begingroup$ @user10456099 so? $\endgroup$ – Math Girl Nov 1 '18 at 15:37
  • $\begingroup$ even+odd+odd/even = even OR odd+even+odd/even = even $\endgroup$ – user10456099 Nov 1 '18 at 15:43
  • $\begingroup$ So,3 maximum picks if same number cannot be pick again? $\endgroup$ – user10456099 Nov 1 '18 at 15:43
  • $\begingroup$ Yes if you cannot have even-even or odd-odd after each other, than the it must be 3, but that was not the question right? $\endgroup$ – Math Girl Nov 1 '18 at 15:51
  • $\begingroup$ No,My question was if i randomly picked number 2,i can no longer able to pick number 2 again.I may randomly picked any number other than 2 in the set. $\endgroup$ – user10456099 Nov 1 '18 at 15:53

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.