# How to solve below question on Probability?

You have a bag with 347 black balls and 278 white balls. Without looking, you pick up two balls from the bag and apply the following rule. If both balls are of the same colour, you throw them both away. Otherwise, you throw away the black ball and return the white ball to the bag. You keep repeating this process. If at some stage there is exactly one ball left in the bag, which of the following is true?

A) The ball in the bag is definitely white. B) The ball in the bag is definitely black. C)Both colours are possible, but the probability of it being white is greater. D) Both colours are possible, but the probability of it being black is greater.

the sample space is

1. Picking both black balls .
2. Picking both white balls .
3. Picking a black and a white ball

So probablity that a black ball will be thrown is 2/3 becoz in two scenarios it will be thrown , while a white ball will be thrown in only one case so accordingly option C goes , please rectify my approach .

Let us simulate your question where the total number of balls in a bag are 9 in which 5 are black and 4 are white. The probability of picking white or black ball out of the nine balls will remain the same as in the original problem.The probability of picking black ball in the original question is $$\frac{347}{625}=0.5552$$, whereas in this simulation it is $$\frac59=0.5555$$

Now the probability of picking randomly 2 balls out of 9 balls in which both have the same colour is $$(0.5555)^2+ (0.4444)^2=0.506072.$$...(1)

And the probability of picking randomly 2 balls out of 9 balls in which both balls are of different colours is $$2*(0.5555)(0.4444)=0.493817$$...(2)

So the probability of (1)is more than (2).So let us assume you pick 2 balls of same colour i-e black and you remove these 2 balls from 9 balls. So now you have total 7 balls in which 4 are white and 3 are black.

Lastly in any scenario, when there are 3 balls left,1 of them will be Black and other 2s are white.if you pick any 2 balls of white colour,ball remained in the bag is black. And if you pick any 2 balls of different colours,one black ball will be thrown out and another white ball will be put inside the bag. So there are 2 white balls in the bag,when you pick them,both balls will be thrown out.So there will be nothing in the bag.So the probability of black ball is 100% when there is only 1 ball remained in the bag.

Hint: the parity of white balls can never change.

EDIT: Further explanation:

The reason for this is this: The only case in which you throw out white balls is when you pick two of the same kind. And then you throw out two of them, so the parity stays the same.

How this can be used to solve the problem: The problem is about an event when there is just one ball left. You start with even number of white balls. So can the last (one) ball left be white?

• Sorry , couldn't get it . – Radha Gogia Nov 11 '18 at 13:36
• @RadhaGogia This answer is a tremendous hint. You start with $278$ white balls. Can there ever be $277$ whilte balls? $276$? $275$? Investigate what the first few steps of the procedure possibly can be (don't worry about the exact likelihood yet) and what the number of white balls is after each step. Look for a pattern. If you don't see it right away, maybe you will see it after trying more examples. But if you don't see it yet and you don't try you will not see it. When you see the pattern, think about how your observations would extend to smaller numbers of white balls. – David K Nov 11 '18 at 14:57
• @RadhaGogia Are you puzzled about why this holds, or how it solves the problem? David's comment already should pretty much clarify them but I edited some further explanations also to the answer. – ploosu2 Nov 12 '18 at 8:07
• @RadhaGogia, There is 100% probability that one ball remained in the bag will be a black ball.For clarification, read out my answer. – Dhamnekar Winod Nov 12 '18 at 14:15