Probability math problem?

Question

It's been a long time since I did any real math, and it's catching up to me. I can't figure out how to do this one, and I need the answer:
There's 10 beads in a bag. One of them is red.
Person A takes 7 beads from the bag.
Person B takes 5 beads from the bag.
What's the probability of neither of them having a red bead?

Answer 1

Well, the question seems incomplete, but let me tell you some assumptions to help us: Since there are only 10 beads, Person A takes 7, then returns it just before Person B picks his 5. Also, I am assuming that the colors of the other 9 beads do not matter, as long as the tenth bead is the only red bead.

The probability of Person A not picking any red bead would be 9C7/10C7 (these are combination symbols), which simplify to 36/120 or 3/10. The probability of Person B not picking any red bead would be 9C5/10C5, which simplifies to 126/252 or 1/2.

The probability that neither picks red would be the product of the 2 probabilities, which is:

3/10 * 1/2 or 3/20

Answer 2

9 of the beads are not red.
For person A there are 9!/(7! * 2!) possible ways to not take the red bead 9!/(7!/2!) = 36. The number of possibleways to take 7 beads are 10!/(7!*3!) = 120. 36/120 = 0.3.

For person B there are 9!/(5! * 4!) possible ways to not take the red bead 9!/(5! * 4!) = 126. And the number of combinations is 10!/(5! * 5!) = 252. The probability is (126/252) = 1/2.

Answer 3

1/9*1/8*1/7*1/6*1/5*1/4*1/3*
1/9*1/8*1/7*1/6*1/5
2!*4!/9!*9!