Permutations/Combinations Probabiltiy Problem

  • Thread starter kwikness
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In summary, the conversation discusses a complaint of discrimination that was voted on by an advisory board. The vote was split along gender lines, with 5 women voting for the complaint and 3 men voting against it. The company's attorney appealed the decision, claiming sex bias. The probability of the vote splitting along gender lines is 1 out of 256 possible combinations, which is equivalent to a probability of 0.0039. Another possible approach to solving the problem is to calculate the probability of choosing 5 black balls to say 'yes' out of 5 total black balls AND 3 white balls to say 'no' out of 3 total white balls, which would result in a probability of 1 out of 56 combinations
  • #1
kwikness
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Homework Statement


The 8 member Advisory Board considered the complaint of a woman who claimed discrimination. The board, composed of 5 women and 3 men voted 5-3 (respectively). The company's attorney appealed the decision based on proposed sex bias. If there was no sex bias it might be reasonable to assume any group of 5 members would be as likely to vote for the complaintant as any other 5. If this were the case, what is the probability that the vote would split along sex lines as it did (five women for, 3 men against)?


The Attempt at a Solution


5C5/8C4
 
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  • #2
With problems like these it is easier to break it down into smaller cases. Try drawing 1 black ball and 1 white ball on a piece of paper, and listing out all of their possible decisions as a 'yes' or a 'no'.

So you would have white = yes black = no, white = no black = yes.. and so on. In this case there are 4 ( total cases = [tex]2^2[/tex] (w = no b = yes, w = no b = no, w = yes b = no, w = yes b = yes). What is the probability of w = no and b = yes?

An equivalent question, how many ways can you choose 1 black ball to say yes out of 1 total black balls AND 1 white ball to say 'no' out of 1 total white balls. The answer here is 1. What is the probability of this? Divide by the total number of 'decisions' (or cases) The answer is: [tex]\frac{1}{4}[/tex]

Now extend this example to the full problem. Count the total number of possible 'decisions' ([tex]2^8[/tex]) and now ask yourself, how many ways can you choose 5 black balls to say 'yes' out of 5 black balls, AND (note the emphasis on 'AND', think multiplication) 3 white balls to say 'no' out of 3 total white balls, then divide by the total number of 'decisions'.
 
  • #3
Worst case scenario, draw out every single case.
 
  • #4
I really don't understand.

I can only see one possible combination where all males would vote no and all females would vote yes. That would make 1 out of 256 combinations-- a probability of 0.0039

I also tried figuring it out your way, but to no avail:
how many ways can you choose 5 black balls to say 'yes' out of 5 black balls
1
3 white balls to say 'no' out of 3 total white balls
1
(note the emphasis on 'AND', think multiplication)
1*1=1
divide by the total number of 'decisions'
1/256

I also thought you might have meant for me to calculate them separately,
i.e. (3 black balls saying yes while white balls say anything) * (5 white balls saying no while black balls say anything) = 15
Then dividing 15 by 256 and getting 0.0586, but this is also incorrect.

Did I misunderstand the way you explained it?

(The correct answer is 1/56 but I can't figure out how to get there)
 
  • #5
Well I suppose I don't understand the question then, because there is only 1 out of those 256 possible ways for all the women to say 'yes' and all the men to say 'no'.
 

1. What is the difference between permutations and combinations?

Permutations are ordered arrangements of a set of objects, while combinations are unordered selections of objects from a set.

2. How do I calculate the number of possible permutations or combinations?

The number of permutations can be calculated using the formula n!/(n-r)!, where n is the total number of objects and r is the number of objects in each permutation. The number of combinations can be calculated using the formula n!/(r!(n-r)!), where n is the total number of objects and r is the number of objects in each combination.

3. Can permutations and combinations be used in probability problems?

Yes, permutations and combinations can be used to calculate the probability of certain events occurring. The number of total outcomes is used as the denominator, while the number of desired outcomes is used as the numerator in the probability formula.

4. How can I determine if a problem requires the use of permutations or combinations?

If the problem involves arranging objects in a specific order, permutations should be used. If the problem involves selecting objects without regard to order, combinations should be used.

5. Are there any real-world applications of permutations and combinations?

Yes, permutations and combinations are used in many fields such as genetics, computer science, and statistics. For example, in genetics, combinations are used to determine the likelihood of inheriting certain traits from parents. In computer science, permutations are used in algorithms for sorting and searching data. In statistics, combinations are used in sampling methods for surveys and experiments.

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