@Suman - I see what you are saying, but actually you are making a slight error.
How did you get 10*9 from Becky's formula? Remember -she has not made some general formula here, but simply split up the two events of picking the marbles up. The denominator of her expression (8+y)*(7+y) is a simplified expression and does not denote the total number of selections as you seem to have assumed.
Coming back to your first post, why did you divide the denominator by 2?
When you say that you are picking both reds "at the same time" you are forgetting that this is not practically different from picking them up one at a time.
CASE 1: There are 10 balls. I pick 1. There are 9 left. I pick another out of these 9.
CASE 2: There are 10 balls. I take 2 out.
In CASE 2, is it possible that I can take the same ball out twice? No! So even though I am "lifting" both balls out simultaneously, I have actually made a decision to choose 1, and then choose the other.
I am not sure if this makes sense but it is a common mistake, and I completely agree with Stuart.
This can be proven by testing the following case.
For y = 2. With Becky's formula of 56/(8+y)(7+y), the probability of 2 red balls is simply 28/45
Using combinatorics:
Selections with red balls = 8C2 = 28
Total selections = 10C2 = 45
So Probability = 28/45
Same answer. So whether or not you treat the 2 balls as together or one at a time, real world tells us even if the event is happening at the same time you are still choosing 1 at a time since you cannot choose the same ball twice.
For more about this you can search the topic of "Dependent events in probability"
Hope this clarifies your doubt!
~Abhay
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
