Math question. Please help? 10 points best answer!

In a pack of ten cards, exactly four are known to be red. If three cards are selected at random:
(a) What is the probability that all three are red
(b) What is the most likely number of red cards

Please guys help me as i have been stuck on this question and i cant figure it out. thanks =]

3 black, 0 red
(6*5*4) / 10*9*8 = 120 / 720 = 1/6
each time you pick card, it reduces number of that color by one (hence 6,5,4)
each time you pick card, it reduces number of total cards by one (hence 10,9,8)

2 black, 1 red
3 * (6*5)*(4) / 10*9*8 = 360 / 720 = 1/2
multiply by 3 because the one red card could be 1st, 2nd. or 3rd picked

1 black, 2 red
3 * (6)*(4*3) / 10*9*8 = 216 / 720 = 3/10
multiply by 3 because the one black card could be 1st, 2nd. or 3rd picked

0 black, 3 red
(4*3*2) / 10*9*8 = 24 / 720 = 1/30

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ANSWERS
(a) prob(3 red) = 1/30
(b) prob(1 red) = 1/2
1 red is most likely number of red cards

a)
First card: 4 reds in 10 cards, so 4/10 or 2/5 chance
Second card: 3 reds in 9 cards, so 3/9 or 1/3 chance
Third card: 2 reds in 8 cards, so 2/8 or 1/4 chance

2/5 * 1/3 * 1/4 = 2/30 = 1/30

b) your possible answers are 0, 1, 2 or 3

To get 0, you never pick a red card & there are 6 black cards, so probability of selecting black on all cards is 6/10 * 6/9 * 6/8 = 3/5 * 2/3 * 3/4 = 18/60 = 3/10

To get 1, it has to be selected on the first, second or third card, with the other two being black
Selecting red on first card only: 4/10 (odds of getting red on first) + 6/9 (odds of getting black on second) + 5/8 (odds of getting black on third) = 2/5 * 2/3 * 5/8 = 20/120 = 1/6
Selecting red on second card only: 6/10 (odds of getting black on first) + 4/9 (odds of getting red on second) + 5/8 (odds of getting black on third) = 3/5 * 4/9 * 5/8 = 60/360 = 1/6
By similar equations we find odds of getting red on third card only = 1/6, so probability of getting 1 red is these three scenarios added together, or 1/6 + 1/6 + 1/6 = 3/6 = 1/2

Since the probability of getting one red is 1/2 & there are three other options, we know none of them can be more likely than 1/2 so we don't have to figure out the probability of getting 2 reds, the most likely scenario is one red.

chance first one red is 4/10
chance second red is 4/10 or 3/9 if card not put back in pack
chance 3d card red is 4/10 or 2/8 if card not put back in pack

so chance 3 reds in a row is (4/10)^3 or (4/10)*(3/9)*(2/8) if card not put back in pack

P(first card red ) = 4/10 = 2/5

P(2nd card red) = 3/9 = 1/3

P(3rd card red) = 2/8 = 1/4

a) P(all red) = (2/5)(1/3)(1/4) = 1/30

A. 4/10 × 4/10 ×4/10 = (just enter in calculator)
B. Not sure

a) 30%
b) 2