M21-24

Question

If a box contains 10 red pills, 5 blue pills, and 12 yellow pills, what is the minimum number of pills one must remove from the box to guarantee that at least three pills of each color are among those removed?

A. 12
B. 17
C. 18
D. 23
E. 25

Show Answer

Official Answer

E

A. 12
B. 17
C. 18
D. 23
E. 25

Bunuel · Accepted Answer

Official Solution:

If a box contains 10 red pills, 5 blue pills, and 12 yellow pills, what is the minimum number of pills one must remove from the box to guarantee that at least three pills of each color are among those removed?

A. 12
B. 17
C. 18
D. 23
E. 25

Show Answer

Official Answer

E

A. 12
B. 17
C. 18
D. 23
E. 25

In the worst-case scenario, one could draw all 10 red pills and all 12 yellow pills, resulting in a total of 22 pills without obtaining a single blue pill. In this case, we would have at least three red pills and at least three yellow pills, but no blue pills at all. To ensure that there are at least 3 blue pills among the extracted ones, one must additionally draw 3 more pills, which would definitely be blue pills. Therefore, a total of 25 pills must be removed to guarantee that at least three pills of each color are among those taken out.

Answer: E

luisnavarro · Answer

Hi,

I think the anser is pretty logic, simple and fast. But I have one doubt... what if I have to solve with combinatorics formulas "C", wich would be the right solution?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

Bunuel · Answer

This is not combinations question, it's a so called worst case scenario, min/max, question.

Check other  Worst Case Scenario Questions  from our  Special Questions Directory  to practice.

Hope it helps.

shalini14 · Answer

Hi Bunuel,

The question has asked for the least number of balls that can be extracted.

For that reason I chose 18 as an answer. Can you please explain the solution in detail and why 18 isn't right.
15 balls (without any yellow balls), then 3 more to include at least 3 yellow and that makes the total as 18.

Please correct me what is wrong with my logic.

Thanks!

Bunuel · Answer

We want to  ensure  that at least three pills of each color are among those extracted. 18 is not correct answer because we can have 10 red pills and then 8 yellow pills and we still won't have three pills of 3 different colors.

For more check other  Worst Case Scenario Questions  from our  Special Questions Directory  to practice.

Hope it helps.

JIAA · Answer

Hi  Xylan

Can you please see what am i missing here??

The question asks for the least number of balls that can be extracted.

10 red balls
5 blue balls
and 3 yellow balls to include  "at least three pills of each color" as the Q stem states

This makes a total of 18 balls !!

would appreciate your help!
Thanks

XyLan · Answer

JIAA It's one of the coolest Qs that work on Logic rather than on Combinatorics/formulae . It's okay to get it wrong in the learning stage . However, Do understand the finer nuances and what EXACTLY the Q is seeking from the test-taker. Given: 10 R ed pills | 5 B lue pills | 12 Y ellow Pills - Total pills: 10 + 5 + 12 = 27 Q-statement: What is the least number of pills one must extract from the box to ensure that at least three pills of each color are among those extracted? Q-meaning: The LEAST number of pills one should pick such that the picker MUST have ALL the three pills of EACH color - focus on the capital case-words; they ARE contextually vital. Let's solve it in the reverse-way to help you better understand: Let's say you pick all the 27 pills: Now, you would definitely have ALL the pills of EACH color, let alone at least 3 colors of each pill. Left-out pills: 0 - The pills which ARE NOT picked Now, Decrease the picks such that, you are still FOR SURE left with ALL the three pills of EACH color. Let's say you pick all the 26 pills: Left out pill: 1 - Can be anything from R, B, and Y. Worst case scenario: If the 1 left-out pill is B, you are STILL left with a minimum of 4 pills of B Now, this part is where it becomes interesting : Let's say you pick all the 25 pills: Left out pill: 2 - Can be anything from R, B, and Y. Worst case scenario: If the 2 left-out pills are B, you are STILL left with a minimum of 3 pills of B Another decreased-pick, i.e., Picking 24 pills MAY result in being having LESS THAN 3 pills of B, i.e., 2 pills of B in the picked-pill selection , If the 3 left-out pills are of B-color. Since we need to BE SURE of at least 3 pills of EACH color, Picking >= 25 pills is a SAFE bet. Thus, if we PICK 25 pills, we can SURE that we will have AT LEAST 3 pills of EACH color: worst case scenario would be that 10R, 3B, and 12Y are picked: 10 + 3 + 12 = 25 - The least pick count Another approach: Let's say, there is a highly unfortunate picker in the Casino of Las-Vegas, and he needs to be SURE of the minimum picks such that at least 3 pills of EACH color are present in order to win the GRAND prize. - A made-up story, Yet Contextually striking!

The unfortunate picker in order to CONFIRM his chances of winning thought of the worst case scenario, i.e., what-if all the initial picks gave him pills of identical color. Thus, the case where all his initial picks resulted in picking 12 Yellow ones Now, being aware of the unfortunateness , he continued to pick the other balls of the identical color, i.e., Next pick of 10 Red ones. The interesting scenario - After picking all the pills of R and Y, the picker is NOW sure that the next pick will DEFINITELY give him the blue ones. Now , knowing this, he can simply add 3 more to the initial picks to confirm the grand prize!

12 + 10 + 3 = 25 - The least pick count Picking 18 does NOT guarantee that one would have at least 3 of each color. 18 may result from (12Y & 6R) OR (9Y & 9R). - Hence, incorrect .

arun8sharma · Answer

Bunuel  or  chetan ,

Just for my understanding what would be the best-case scenario? 18

Bunuel · Answer

Well, the best case scenario would be if you pick 3 red pills, 3 blue pill, and 3 yellow pill right away. So, 9 picks.

deep31993 · Answer

Why is the answer not 9. I am not able to understand any of the explanations given here on this post.

Bunuel · Answer

We want to  ensure  that at least three pills of each color are among those extracted. 9 is not correct answer because we can have 9 red pills and we still won't have three pills of 3 different colors.

For more check other  Worst Case Scenario Questions  from our  Special Questions Directory  to practice.

Hope it helps.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

JAIDEEP27 · Answer

The question is asking for the minimum number of pills that must be removed. What if we remove all Red and Blue first and then 3 yellow, in this case we just need to remove 18 to get atleast 3 pills of each color. What am i missing here?

Bunuel · Answer

Your approach of removing 18 does not necessarily  guarantee  that you will end up with a minimum of three pills of each color. For instance, you could remove 12 yellow and 6 red pills, and you would still not have three pills of each color. However, with 25 attempts,  you can be certain  to have at least three pills of each color, irrespective of the combination you pick. This is because in fewer than 25 picks, there's always a possibility of a scenario where you do not get the minimum of three pills of each color, such as in the 18 picks scenario you provided.

I recommend visiting this link  Worst Case Scenario Questions  and practicing similar questions to enhance your understanding.

Hope it helps.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

Rave25 · Answer

I think this is a  high-quality  question and I agree with explanation.

gagan03 · Answer

If we draw all the red pills, all the blue pills, and just 3 yellow pills — making a total of 18 pills — wouldn't that be the minimum number needed?

Bunuel · Answer

Try checking the discussion: https://gmatclub.com/forum/m21-184278.html#p3237157

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