Five card poker hand probability

[F.B]

Sorry to post so many questions.
But i need help.

1. A five card poker hand is dealt and the last card is turned face up. Determine the probability that you have been dealt four aces, given that the card turned over is an ace.

I think its a conditional probability problem but here's what i did instead.

Probability of the first to be an ace is: 4/52
Second card: 3/51
Third: 2/50
Fourth to not be an ace: 48/49
Fifth:1

I multiply all those together but i don't get the right answer.

2. In how many ways can a group of ten people be chosen from 6 adults and 8 children if the group must contain at least 2 adults.

I did this.

C(14,10) - C(6,0) x C(8,10) - C(6,1) x C(8,9)
This doesn't work out for some reason. I thinks how you suppose to do it but it won't work.

3. There are 25 students in a data management class. Determine the probability that at least two share the same birthday. (Assume that every year has 365 days)

This one i have no clue how to do it.

 

[NewScientist]

Sorry to post so many questions.
But i need help.
1. A five card poker hand is dealt and the last card is turned face up. Determine the probability that you have been dealt four aces, given that the card turned over is an ace.
I think its a conditional probability problem but here's what i did instead.
Probability of the first to be an ace is: 4/52
Second card: 3/51
Third: 2/50
Fourth to not be an ace: 48/49
Fifth:1
I multiply all those together but i don't get the right answer.

Is there only 1 deck or not? Secondly, the probability is given by

(3 / 51) * (2 / 51) * (1/50)

 

[Tide]

3. HINT: Calculate the probability that none of the 25 students share a birthday.

 

[NewScientist]

Part 3 in as a common question - the answer is near a 1/2

 

[HallsofIvy]

A point about problem 1: suppose the first card were turned face up and was an ace. How many cards are left in the deck? How many of them are aces? Can you see that the fact that it was the last card turned face up isn't important? You still know that particular ace can't be one of the face down cards. What is now the probability of getting "A, A, A, not an ace". And be careful to remember that that particular order is not important. In how many ways can you order "AAAN"?

2. In how many ways can a group of ten people be chosen from 6 adults and 8 children if the group must contain at least 2 adults.

I did this.

C(14,10) - C(6,0) x C(8,10) - C(6,1) x C(8,9)
This doesn't work out for some reason. I thinks how you suppose to do it but it won't work.

I have no idea why you think you should do it that way since you haven't given your reasoning. Here's how I would do it:
First select the two adults for the group- how many ways can you select 2 out of the 6 adults? That leaves 4 adults. Together with the 8 children, you have 12 people left to select from to get the remaining 8 people. How many ways can you select 8 people out of 12?