SBD98 wrote:
A box contains 10 tablets of Medicine A and 15 tablets of Medicine B. What is the minimum number of tablets that need to be removed from the box to guarantee that at least two tablets of each type are among the ones extracted?
A. 12
B. 15
C. 17
D. 19
E. 21
The worst-case scenario would occur if we remove all 15 tablets of medicine B first. In this scenario, we would still not have any tablets of medicine A. However, the next two tablets we remove must be of Medicine A. Therefore, to guarantee that at least two tablets of each type will be taken, we should remove a minimum of \(15+2=17\) tablets.
Answer: C
Hi Bunuel,
I have a confusion. For example: if I pick all 10 tablets of A first, then pick 2 tablets of B from the box, the minimum number of tablets removed would be 12. In this case, I follow the condition of picking at least 2 tablets of each type of medicine. So, why 12 isn't the answer?
The reason 12 isn't the correct answer is that it depends on a specific order of removal (all 10 of A first, then 2 of B). The question asks for a number that
guarantees at least two of each type, regardless of the order. This doubt has been addressed in the thread, and for further practice, you can refer to other
Worst Case Scenario Questions from our
Special Questions Directory to practice.
To elaborate further, it's important to recognize that the tablets can be removed in many different orders. Ideally, the best case scenario to meet the requirement of removing at least two tablets of each type would involve immediately removing 2 tablets of Medicine A and 2 tablets of Medicine B. However, this specific order of removal is not guaranteed. What we need to determine is a number of draws that, regardless of the order of removal, will always ensure that at least two tablets of each type are extracted. Hence, to guarantee that at least two tablets of each type are removed under any sequence, it becomes necessary to consider the worst-case scenario, which would be most unfavorable removing order.