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consider an experiment that consists of determining the type of job - either blue-collar or white collar- and the political affiliation -republicans, democratic or independent - of the 15 members of an adult soccer team. how many outcomes are in the sample space?

I calculated it by doing the following calculation:

15x2x3=90 outcomes

However the answer is 6^15. How did they calculate it like this and why is my calculation wrong?

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    $\begingroup$ Could you explain how you got to $15 \times 2 \times 3$? $\endgroup$ Jun 22 at 2:57
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    $\begingroup$ Yes, I had 15 people, and for each 15 I had two possibilities for their occupations being either blue collar or white collar, and then three possibilities for their political affiliations. So I multiplied the possibilities and came up with 15x2x3. $\endgroup$
    – user112167
    Jun 22 at 3:11
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for every player, the total number of pairings (job,political affiliation) are $\binom{2}{1}\times \binom{3}{1} = 6$. Assuming that pairings of any 2 soccer players are independent, each player has 6 options of the pairings. thus it would be $ \underbrace{6 \times 6 \times 6 \cdots 6}_{\textrm{15 times}} = 6^{15}$.

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    $\begingroup$ Thanks, I understand now how they arrived at this calculation. However, I am still not sure why the calculation I derived was incorrect? $\endgroup$
    – user112167
    Jun 22 at 3:12
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Using the basic counting principle of multiplication, it must be $2^{15}\times 3^{15}=6^{15}$.

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  • $\begingroup$ Thanks this makes it clear how it was solved. $\endgroup$
    – user112167
    Jun 23 at 14:04
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$ \underbrace{6 \times 6 \times 6 \cdots 6}_{\textrm{15 times}} = 6^{15}$

This is because of the fundamental principle of product that states that if a job is doable in $m$ ways and another job is doable in $n$ ways, the number of ways in which both the jobs are doable is $m\times n$.

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Since there are $6$ choices in each of the $15$ cases, it is $6^{15}$.

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Yes, it is an application of the basic principle of counting.

$$2^{15}\times 3^{15}=6^{15}$$

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