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Question 3.27 from the book Openintro statistics confuses me. The question states:

In a classroom with 24 students, 7 students are wearing jeans, 4 are wearing shorts, 8 are wearing skirts, and the rest are wearing leggings(5). If we randomly select 3 students without replacement, what is the probability that one of the selected students is wearing leggings and the other two are wearing jeans? Note that these are mutually exclusive clothing options.

I tought the answer would be $\frac{5}{24} \times \frac{7}{23} \times \frac{6}{22} = 0.0173$ However, the answer states:

$\frac{5}{24} \times \frac{7}{23} \times \frac{6}{22} = 0.0173$. However, the person with leggings could have come 2nd or 3rd, and these each have this same probability, so $ 3 \times 0.0173 = 0.0519$

This confuses me greatly. How would you for example compare this to the probabilty of drawing 3 hearts from a shuffled deck of cards, assuming no replacement. This would be $\frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} = 0.0129$ But here we don't multiply by 3, why? What about the probability of drawing 2 hearts and 1 diamond?

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The answer is a probability with the number of 'favorable' outcomes in the numerator and the number of 'all possible' outcomes in the denominator. In many cases, you have your choice whether 'outcomes' are ordered (order matters) or unordered (doesn't matter). However, once you make your choice, you need to be consistent, using either ordered or unordered outcomes both in numerator and denominator.

A method with unordered outcomes is as follows:

$$P(\mathrm{2\, Jeans\, and\, 1\, Leggings}) = \frac{{7\choose 2}{5\choose 1}{12\choose 0}}{{24\choose 3}} = 0.05187747.$$

Computation in R:

choose(7,2)*choose(5,1)/choose(24,3)
[1] 0.05187747

Note: By general agreement, for most card games, outcomes are regarded is unordered. (The order in which cards are dealt has no effect on the value of the hand.) So a straightforward combinatorial solution uses unordered outcomes in numerator and denominator.

Thus, getting three hearts (taken as indistinguishable) when dealing three cards from a well-shuffled deck is ${13\choose 3}{39\choose 0}/{52\choose 3}= 0.01294118.$

choose(13,3)/choose(52,3)
[1] 0.01294118
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  • $\begingroup$ Thank you for your answer. I have never seen probabilities being calculated this way. It will probably take me a couple of days to get to that section. Do you mind adding a part where you explain using my way of calculating probabilities? $\endgroup$
    – Gertjan Brouwer
    Commented Sep 25, 2020 at 23:06
  • $\begingroup$ As soon as you write $\frac{5}{24} \times \frac{7}{23} \times \frac{6}{22} = 0.0173$ you are looking at ordered samples. That's why you need the modification of the answer book. For the the card example you're treating all Hearts as the same, so there is no worrying about order. // By your approach you'd need to consider order for the probability of two Hearts and a Diamond. $\endgroup$
    – BruceET
    Commented Sep 25, 2020 at 23:25
  • $\begingroup$ Two Hearts and a Diamond: Unordered, choose(13,2)*13/choose(52,3) returns $0.04588235.$ Ordered, (13*12*13)/(52*51*50) returns $0.01529412,$ which needs to be multiplied by $3$ to arrange the cards HHD once you've got them. // If you write ${13 \choose 2}$ as $13!/(11!\cdot 2!),$ etc. maybe you can track the difference between the two methods. $\endgroup$
    – BruceET
    Commented Sep 25, 2020 at 23:36
  • $\begingroup$ I know I need to multiply by 3, because of it being ordered. I however still don't see the difference between the two examples. I'll work through the book to get to your way of calculating probabilities and see if I understand at that point. I'll get back to you on this. Thanks a lot $\endgroup$
    – Gertjan Brouwer
    Commented Sep 26, 2020 at 11:30

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