Confusing Probability and Stats assignment :(

I'm doing an assignment for my stats class. It's about probability and combinations / permutations. Here's a question I'm having trouble with.

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The executive of an organization is to be selected from a group of 8 males and 7 females.

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a) How many ways are there to choose the president and secretary?

i) 105

ii) 56

iii) 210

iv) 28

15 choose 2 = 105 ?

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b) How many ways are there to have a female president and a male secretary?

i) 105

ii) 56

iii) 210

iv) 28

Not sure at all how to approach this problem. Would 7 choose 1 * 8 choose 1 be valid?

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c) What is the probability that a randomly chosen president is female and a randomly chosen secretary is male?

i) .50

ii) .53

iii) .13

iv) .27

7/15 * 8/14 = .27 ?

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d) How may ways are there to choose a president, secretary and treasurer?

i) 455

ii) 448

iii) 2730

iv) 392

15 choose 3 = 455 ?

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e) How many ways are there to choose a president, secretary and treasurer so that two are male and one is female?

i) 1176

ii) 392

iii) 98

iv) 196

No clue XD

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f) What is the probability that a randomly chosen president, secretary and treasurer include two males and one female?

i) .43

ii) .07

iii) .14

iv) .21

Not sure what to do here, either :(

Re: Confusing Probability and Stats assignment :(

Hey RubberDucky.

For 1) you are correct if there are no conditions on the executive or secretary being male/female (i.e. anything goes).

For 2) Consider the two choices separately (female president, male secretary) and remember that the choices are independent of each other (which means you can use the multiplication or product rule). Your answer is in-line with this (hopefully you can understand the answer based on what I've written).

For 3) You have to specify how they have been selected: you will get different answe rs based on how you select them (Q1 vs Q2)

For 4) What conditions does the treasurer have? (Are they male/female/both?)

For 5) List all of the possible combinations by hand - MMF MFM FMM and now add up the combinations

For 6) Probability = number of combinations for that event / total number of possibilities. You will need to answer 5 to get to 6.

Show us how you go with these hints and if/where you get stuck.

Re: Confusing Probability and Stats assignment :(

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Originally Posted by

**chiro** Hey RubberDucky.

For 2) Consider the two choices separately (female president, male secretary) and remember that the choices are independent of each other (which means you can use the multiplication or product rule). Your answer is in-line with this (hopefully you can understand the answer based on what I've written).

So the multiplication rule can be used when doing combinations as well?

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For 3) You have to specify how they have been selected: you will get different answers based on how you select them (Q1 vs Q2)

I copied the question exactly as it appears on my assignment. I had to assume they would be selected one and then the other, because if they weren't, I would be 7/15 * 8/15 = 0.25, which is not an option.

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For 4) What conditions does the treasurer have? (Are they male/female/both?)

This is why I find this assignment frustrating - I know that if it weren't multiple choice, I would end up doing it differently. From the information provided (which is all in my original post), there are no conditions on any of the positions.

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For 5) List all of the possible combinations by hand - MMF MFM FMM and now add up the combinations

Sorry, I'm not sure what you mean by add up the combinations :/

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For 6) Probability = number of combinations for that event / total number of possibilities. You will need to answer 5 to get to 6.

Re: Confusing Probability and Stats assignment :(

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Originally Posted by

**RubberDucky**

I disagree with that answer. We can elect Amy & Allen as officers.

But electing Amy as president and Allen as secretary is different from electing Allen as president and Amy as secretary .

Thus this is a permutation so $\displaystyle _{15}\mathcal{P}_2=(15)(14)$.

Re: Confusing Probability and Stats assignment :(

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Originally Posted by

**Plato** I disagree with that answer. We can elect Amy & Allen as officers.

But electing Amy as president and Allen as secretary is different from electing Allen as president and Amy as secretary .

Thus this is a permutation so $\displaystyle _{15}\mathcal{P}_2=(15)(14)$.

So you propose 210 as the correct answer?

Also, do you have any thoughts about the last 3 questions?

Re: Confusing Probability and Stats assignment :(

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Originally Posted by

**RubberDucky** So you propose 210 as the correct answer? It is correct.

Also, do you have any thoughts about the last 3 questions?

f) What is the probability that a randomly chosen president, secretary and treasurer include two males and one female?

There are $\displaystyle \binom{8}{2}=28$ ways to choose two males and $\displaystyle 7$ to choose a female.

There are $\displaystyle 6$ ways to assign the offices any group of three.

So?

Re: Confusing Probability and Stats assignment :(

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d) How may ways are there to choose a president, secretary and treasurer?

Since the positions don't have any constraints, I get 15 options for first position * 14 people left for second position * 13 = 2730? Or equivalently, as Plato showed, order is important when the positions are being chosen, and 15P3 = 2730.

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e) How many ways are there to choose a president, secretary and treasurer so that two are male and one is female?

As Plato said, there should be 8C2 ways to have two males, and 7C1 to have a single female. I get 28 * 7 = 196 ?

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f) What is the probability that a randomly chosen president, secretary and treasurer include two males and one female?

There are 196 ways for this event to happen, out of a possible 2730 events, as seen in d & e respectively. 196/2730 = 0.07 ?

Thankfully all of these answers are among the options, but I'd imagine the options are composed in such a way that if I made any common mistakes, those wrong answers would be options. Does everything look okay?

Re: Confusing Probability and Stats assignment :(

Quote:

Originally Posted by

**RubberDucky** As Plato said, there should be 8C2 ways to have two males, and 7C1 to have a single female. I get 28 * 7 = 196 ?

That is not all I said.

The answer is $\displaystyle \binom{8}{2}(7)(6)$.

There are 196 ways to choose two males and one female.

But any group of three can be assigned to three offices is six ways.