testcracker wrote:
chetan2u wrote:
Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10
B. 16
C. 23
D. 42
E. 43
In other words, the worst case is when the last 4 are all of different colour, and the one you pick will result in picking all of one type, so ans = 12+10+15+9-4+1=43
hi
do you think 38 can also be an answer, although the choice is not in the answer choice ?
say first we pick all 15 green balls, then all 12 red balls, and then all 10 blue balls, so lastly we end up with only black balls remaining
so 15 + 12 + 10 +1 = 38
so, where am I getting wrong ?
thanks
The problem is with understanding the question, which can become quite perplexed.
The question says the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color.
The catch is ASSURED. I can pick up just 9 all and all 9 can be blacks and that will be the best scenario.
But the word ASSURED means that the probability of one colour has to be 100%. These 9 could actually be combinations of all 4 colours.
This means that I pick up some number blindly and then be able to claim that I have picked up atleast one set of colours, and here we will consider the worst scenario.
That is I have picked up all but 4, and these 4 are one of each colour.
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