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In a business school at ABC University, 1/6 of the enrollment are freshmen, 3/10 are sophomores, 1/3 are juniors and 1/5 are seniors. They compete with several other schools within their region in an annual financial case competition.

Three students were randomly selected to form a team for this year’s competition.

Calculate the probability that the juniors would make up the majority of the team.

I thought that this probability is equal to the probability that exactly 2 out 3 selected are juniors plus the probability that all 3 are juniors. Therefore, I computed the following:

$$P(exactly~2~juniors) = \frac13 \times \frac13 \times \frac23 = \frac2{27}$$ $$P(3~juniors) = (\frac13)^3$$ Therefore, $$P(juniors~make~up~the~majority)= \frac1{27} + \frac2{27} = \frac3{27}$$

My answer is wrong as I have only partially computed the probability. I would appreciate some explanation as what I am missing as I am having trouble figuring it out. Thank you

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Your formula $\frac{1}{3}\cdot\frac{1}{3}\cdot \frac{2}{3}$ to pick the two juniors is the probability of first selecting two juniors, and then the non-junior. You can also first pick a junior, then a non-junior, and then a junior again, the probability of which is: $\frac{1}{3}\cdot\frac{2}{3}\cdot \frac{1}{3}$. And, of course, you can first pick the non-junior, and then the two juniors, with a probability of $\frac{2}{3}\cdot\frac{1}{3}\cdot \frac{1}{3}$.

All these give you 2 juniors and 1 non-junior, so the probability of picking exactly 2 juniors is:

$\frac{1}{3}\cdot\frac{1}{3}\cdot \frac{2}{3}+\frac{1}{3}\cdot\frac{2}{3}\cdot \frac{1}{3}+\frac{2}{3}\cdot\frac{1}{3}\cdot \frac{1}{3} = \frac{6}{27}=\frac{2}{9}$

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  • $\begingroup$ I thought that the order in which I choose the students doesn't matter. It obviously does per your solution (thank you!). $\endgroup$ – the boy 88 Jul 13 '17 at 20:59
  • $\begingroup$ Can I think of my sample space as all combinations of 3-student tuples? And the event I want is choosing exactly 2 juniors and 1 non-junior so the outcomes that satisfy the event are { JJSo, JJF, and JJSe}? So = sophomore, Se = senior, F= freshman. $\endgroup$ – the boy 88 Jul 13 '17 at 21:07
  • $\begingroup$ @Kovs95 Yes, you can. And you can see them as unordered triples (basically sets with 3 elements), or as ordered triples. If seeing them as unordered, you need to multiply by 3 which is what sds did.. if you see them as ordered you get the formula I got. Same result of course! Now, there is really no need to distinguish between freshmen and sophomores and senior ... you just need to distinguish between junior (1/3 chance) and non-juniors (2/3 chance) $\endgroup$ – Bram28 Jul 13 '17 at 21:22
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You can select the non-junior in 3 different ways (1st, 2nd, 3rd), so

$$P(exactly~2~juniors) = \frac13 \times \frac13 \times \frac23 \times 3 = \frac29$$

Thus

$$P(juniors~ make~ up~ the~ majority)=\frac29+\frac1{27}=\frac{7}{27}$$

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  • $\begingroup$ yes sorry I fixed the 2/3 probability of not selecting a junior. $\endgroup$ – the boy 88 Jul 13 '17 at 20:43
  • $\begingroup$ yes, but this is not the only problem - see "times 3"! $\endgroup$ – sds Jul 13 '17 at 20:44

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