### Problem 2 - The hidden balls

For the last problem 72 people visited the blog, but only 2 posted the answers. Dont be shy in putting the answers, you can even do it anonymously or with a pseudoname. Only then can you gain confidence

Here comes the second problem. And as i have said, it is simple yet you need to have your basics perfect to answer these.

"There are 3 black, 5 red and 4 green balls in a bag. Ashok comes in, takes two balls at random and walks away with them. Sunita comes and picks a ball from the bag. What is the probability that the ball Sunita picked is green"

Think carefully before answering.

Click here for the answer

Here comes the second problem. And as i have said, it is simple yet you need to have your basics perfect to answer these.

"There are 3 black, 5 red and 4 green balls in a bag. Ashok comes in, takes two balls at random and walks away with them. Sunita comes and picks a ball from the bag. What is the probability that the ball Sunita picked is green"

Think carefully before answering.

Click here for the answer

## 8 Comments:

Guys,

The answer is 1/3.

The technical way to go about this is thru conditional probability. But that is also 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

the answer would definitely become 0 in that case , the flaw is if the balls picked up by ashok were green the probability would not remain the same infact that is what i didnt understand even in ur line of thought , even if sunita puts her hand inside the bag and just moves the 2 balls aside , the probability would depend on the balls she moved aside ( which i presume she cant pick up) ??

Anonymous (Can you leave a short name which I can use to address :),

When you dont know what balls have been removed, the probability does not change. It changes only when you know that Ashok has removed green balls. Or red for that matter. Check this by applying conditional probability for all scenarios and you will arrive at the same answer.

The other way to think about probability is betting. If you had placed bets on Sunita removing green before ashok removed the ball, it would have been at 1/3. It will change only if you know what has been removed, assuming Ashok is doing it at random and not to influence the outcome.

The approach using conditional probability is given below

There are three scenarios, Ashok removes 2 non green balls, Ashok removes one non green ball and one green ball and Ashok removes 2 green balls.

The probability in the three conditions are as follows,

(8C2/ 12C2) * 4/10 = 224/1320.

(8C1*4C1/12C2) * 3/10 = 192/1320.

(4C2/ 12C2) * 2/10 = 24/1320.

Hence answer is the sum of the three conditional probabilities (224+192+24)/1320 = 1/3

This comment has been removed by the author.

Thanks! Great blog and interesting insights into a range of topics. Btw you can address me by this name.

hi d ans is 1/3

it wud hav 3 cases

Case 1) when both d balls withdrawn are green n finally a green ball is taken out

4C2/12C2 . 2C1/10C1

Case2)when 1 ball is green n other is non green n then finally a green ball is withdrawn

4C1.8C1/12C2 . 3C1/10C1

Case3)when both are non green n finally a green ball is withdrawn

8C2/12C2 . 4C1/10C1

so addin 3 cases we get 1/3

Amazing question i must say...

great to clear the basics... and feels good to realize that am not that rusted yet. ;-)

Post a Comment

<< Home