0
$\begingroup$

Case 1: We choose $3$ people from the same group, without replacement or caring about the order. The number of ways you could do this is $n\choose3$, if $n$ is the number of people in the group. So the probability of choosing a specific $3$ people out of all the people in the group is ${n\choose3}$, I think.

Case 2: We choose $3$ people again, except each is chosen from their own groups of size $n$. So we have $n \choose 1$ $=n$ ways of choosing each of the three people. But how many ways in total are there for choosing the $3$ people?

I'm not sure how to go from the number of ways to choose each person to the number of ways to choose all three people. Do I add to get $3$$n\choose 1$, or multiply to get $n\choose 1$$^3$? Furthermore, what would the probability then be of choosing those three specific people out of all the ones we could've chosen?

$\endgroup$
  • 1
    $\begingroup$ So in the latter case you mean there are $300$ people split into $3$ groups? $\endgroup$ – Parcly Taxel Oct 30 '19 at 1:59
  • 1
    $\begingroup$ $|\{a,b\}\cup \{x,y,z\}| = |\{a,b\}|+|\{x,y,z\}| = 2+3=5$. On the other hand $|\{a,b\}\times \{x,y,z\}| = |\{(a,x),(a,y),(a,z),(b,x),(b,y),(b,z)\}| = 2\times 3 = 6$. You add if it is "I want to choose one from $a,b$ or from $x,y,z$ but only one of them total." You multiply if what you want is "I want to choose one from $a,b$ and one from $x,y,z$ simultaneously" $\endgroup$ – JMoravitz Oct 30 '19 at 2:09
  • $\begingroup$ @JMoravitz Thank you, framing this in terms of set notation helps! Your last statement makes sense, but could you expand on what you mean by "I want to choose one from $a, b,$ or from $x,y,z$, but only one of them total"? Does that mean we're only choosing one object, while in your last statement you're talking about choosing two objects? $\endgroup$ – James Ronald Oct 30 '19 at 23:30
  • $\begingroup$ @ParclyTaxel Well, I suppose you could though that's not how I initially thought of it. There are simply 3 groups of $100$: there wasn't necessarily a "splitting up" process carried out at any point $\endgroup$ – James Ronald Oct 30 '19 at 23:31
0
$\begingroup$

Use a simpler example: suppose you have two groups of $3$ people each (say, the groups are $\{ A,B,C \}, and ${ D,E,F }$. In how many ways can you choose two people, one from each group?

Well, you can pick $A$ and $D$, or $A$ and $B$, or $A$ and $F$, or .... and I think you will quickly see that there will be $9$ possible combinations. Indeed, for combinations, you multiply, not add.

$\endgroup$
  • $\begingroup$ Thank you for the help! So in the latter case in my title, I'd have $100$ choices for each person and thus $100^3$ choices in total for all three people? Which means that the probability of choosing three specific people would be $1/100^3$? $\endgroup$ – James Ronald Oct 30 '19 at 23:32
  • $\begingroup$ @JamesRonald That's exactly correct! $\endgroup$ – Bram28 Oct 31 '19 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.