Calculating Winning Probability in Horse Racing and Sock Selection

[six789]

this is the problem...
1. nine horses are entered in a horse race. If you "box" three horses (three are chosen and they can finish in any of the first 3 positions in the race), determine the probability that you will hold the winning ticket.
-->i tried this one... (3C1*3C2*3C3)/9C3 but my answer is 3/28, which is wrong, it should be 1/14
2. A drawer contains four red socks and five blue socks. Three socks are drawn one at a time and then put back before the next selection. Determine the probability that
i. exactly red socks are selected
--> i did this... [(4C2*5C3) + (4C1*5C4) + (4C0* 5C5)]/9C3 = 27/28, but it is wrong since the answer in the book is 304/729
ii. at least two red socks are selected
--> i don't get this one..

 

[Tide]

There are 6 ways of holding the winning ticket and C(9,3) total ways of selecting any three horses for win, place and show.

 

[Ajoo]

For the Horse problem the answer is 3P3/9C3
(not sure if it is P cause in portugal we use A but in my calculator it says P it means the number of possible combinations of 3 horses in 3 different places where order counts (3!/(3-3)! = 3!)

The second problem we can deal with it with binommial distribution:
being x the number of consecutive red socks u take, u have
n=3; p=4/9
if u want to know the porbability of getting 1 red sock:
p(x=1)=3C1*(4/9)^1*(5/9)^2=300/729

for at least 2 red socks
p(x>1)=p(x=2)+p(x=3)=3C2*(4/9)^2*(5/9)+3C3*(4/9)^3=304/729

 

[Ajoo]

Sorry, my previous solution on the socks problem is wrong. I'm rly sry but i was really sleepy. Binomial distribution doesn't apply cause each act of choosing a sock is not indepandent from the others (Choosing 1 sock makes the probability of the next sock different). I also don't get the first question: do u want the probability of how many red socks?