Question number 18. In a bag there are 3 green apples, 5 yellow apples, 7 green pears and 2 yellow pears. Simon randomly takes fruit out of the bag one by one. How many pieces of fruit must he take out in order to be sure that he has at least one apple and one pear of the same color? Of course if we consider Simon to be very lucky, it would be possible for him to first pick out a green apple and then on the second try to pick out a green pear. And then he would be done after two tries. Or he could first pick out a yellow apple and then pick out a yellow pear and also be done after just two tries. But what we're going to have to consider here is in order for Simon to be absolutely sure that no matter how many tries he has made, he has the fruit he needs. We will have to account for the possibility that he has the worst possible luck. So in the worst case, what happens is Simon selects first all 7 of the green pears. Just one by one he picks out 7 fruit. They're all green pears. Then the next worst thing that can happen to him is he makes 5 more selections and they're all yellow apples. And after having made 12 picks from the bag, he still does not have the fruit he needs. The next pick, the 13th choice, will be either one of the remaining 3 green apples or one of the remaining 2 yellow pears. And since already in his possession he has a lot of yellow apples and a lot of green pears, this next choice will guarantee that he has one apple and one pear of the same color. So 13 is the minimum number of fruit Simon needs to pick if he is having a really unlucky day. But that, on the other hand, will guarantee in his possession at least one apple and one pear of the same color. So E here is the answer, 13.

fruit selection

apples and pears

worst-case scenario

color matching

minimum picks