Calculating Winning Probability in Horse Racing and Sock Selection

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Homework Help Overview

The discussion revolves around calculating probabilities in two distinct scenarios: horse racing and sock selection. The first problem involves determining the probability of holding a winning ticket when boxing three horses in a race with nine entries. The second problem focuses on drawing socks from a drawer containing red and blue socks, specifically calculating the probability of selecting a certain number of red socks.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore different methods for calculating probabilities, including combinations and permutations for the horse racing problem. There is also discussion about using binomial distribution for the sock selection problem, though some participants question its applicability due to dependency in the selection process.

Discussion Status

Multiple interpretations of the horse racing problem are being explored, with some participants suggesting different approaches to calculating the winning probability. In the sock selection problem, there is acknowledgment of potential errors in previous calculations, and participants are seeking clarification on the requirements of the questions posed.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is also uncertainty regarding the correct application of probability concepts in the sock problem due to the dependency of selections.

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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..
 
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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.
 
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
 
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?