newton9 wrote:Hii Stuart,
I have a question here.
I agree with your explanation when the balls are replaced.
But here the balls are not replaced. Does it still hold good??
Say for example in first 3 attemps, I got black balls.
So there are 2 black balls and 3 Yellow balls left. How can the probability be 5/8 in this case? Is it not 2/5.
Thanks in advance.
Hi,
replacement is irrelevant; the same logic applies.
If the answer is less than 5/8, that means that the chance of selecting a yellow ball 4th is greater than 3/8 (it has to be either black or yellow, right?); if the answer is greater than 5/8, that means that the chance of selecting a yellow ball 4th is less than 3/8. There's no logical reason why the probabilities would chance from the original chances.
It's only if we actually know the previous results that the probability would change. For example, if the question had been:
A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. If the first 3 balls chosen are black, what's the probability that the fourth ball selected is black?
In the question at hand, we have to consider every possible case that includes the black ball as the #4 selection; when we add those up, we'll get 5/8 for black and 3/8 for yellow.
