JIAA wrote:
Bunuel wrote:
Official Solution:
If a box contains 10 red pills, 5 blue pills, and 12 yellow pills, 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?
A. 12
B. 17
C. 18
D. 23
E. 25
In the worst possible scenario, one will draw 22 pills without drawing a single blue pill. To make sure that there will be 3 blue pills among the extracted, one has to draw 3 more than 22 (i.e. 25).
Answer: E
Hi
XylanCan you please see what am i missing here??
The question asks for the least number of pills that can be extracted.
10 red pills
5 blue pills
and 3 yellow pills to include
"at least three pills of each color" as the Q stem states This makes a total of 18 pills !!
would appreciate your help!
Thanks
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 Red pills | 5 Blue pills | 12 Yellow 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.
_________________