Answer to the Hidden balls problem
The answer is 1/3.
The technical way to go about the answer is thru conditional probability. But that is the laborious way and CAT is not forgiving on those who take time to solve the problems :)
The key basic idea to learn here is that when you dont know what balls were taken out, it does not change the probability. Hence the probability would remain the same 1/3 even if Ashok had removed 0, 1, 2 or 3 balls.
If this feels unintutive, just think that if Sunita had put here hand inside the bag and randomly put aside two balls without looking and then choosen the 1 ball from the rest, the ptobability would still have been the same. What Ashok has done is equivalent to this action
Now think about a small variations.If the probability does not change when Ashok removes 0, 1, 2, 3.. balls, does it change if he removes all the balls. If so what is wrong in our thinking
Click here for the question
Gyan-ee 2:49 AM
3 Comments:
Hi...Though its difficult to digest what u said but i still pondering over the concept....R u sure on that??
What is my understanding from ur Answer is that u have taken same colored balls as packet and irrespective of their number ,u want to say that we want to select one packet out of three thats 1/3...Is that correct??? or should it be 1/4 i.e 1 out of 4 green colored ...Please explain a little more...
Regarding U said if he removes all non green then probability will definitely be affected because Number of balls removed is greter then Number of balls of one particular color..The funda u mentioned is only applicable till number of balls removed is less then Number of balls of any particular color AM i right?
Subrosa,
The funda is applicable as long as he does not remove all the balls. Even if he leaves only one ball, the probability is 1/3.
The reason the probability changes once all the balls are removed is because, then the colours of all the balls removed is known. AS long as we dont have that knowledge, the probability does not change.
As I said, Ashok removing the balls is equivalent to Sunita putting aside the balls inside the bag before choosing from the second set.
Let me try and give another anology. There is a pack of cards in which you have to pick an Ace. The probability is 1/52. Now if you divide the cards into 2 sections before choosing the card, the probability will remain the same (unless you know what cards are on which side). The probability will remain the same even if you make 3 to 52 different sets of the cards
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