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
- Picking both black balls .
- Picking both white balls .
- 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 .
Best Answer
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?