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$18$ people in a class. Teacher chooses $3$ students. What is probability that teacher chooses Jason, Kim and Ellie? The problem wants the answer without using combinations (nCr). Lost.

I would think the answer would be $(1/18) \cdot (1/17) \cdot (1/16)$, but that doesn't seem to be the way to do it with combinations.

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  • $\begingroup$ I would think the answer would be (1/18) * (1/17) * (1/16), but that doesn't seem to be the way to do it with combinations. $\endgroup$ – user163862 Jan 9 '17 at 5:50
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The chance that the first choice is one of Jason, Kim and Ellie is $\dfrac{3}{18}$

Given that the first choice is one of Jason, Kim and Ellie, that leaves $2$ desirable students out of $17$ so the conditional probability that the second choice is also is also one of Jason, Kim and Ellie is $\dfrac{2}{17}$

Given that the first and second choices are both from Jason, Kim and Ellie, that leaves $1$ desirable students out of $16$ so the conditional probability that the third choice is also is also one of Jason, Kim and Ellie is $\dfrac{1}{16}$

So the probability that all three choices are Jason, Kim and Ellie in any order is $\dfrac{3}{18} \times\dfrac{2}{17} \times\dfrac{1}{16} = \dfrac{1}{816}$

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You essentially need to derive the combinations answer. Your comment is almost right, but would require the children to be picked in one order. Multiply by the number of orders you can pick them in and you are there.

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  • $\begingroup$ that makes sense. so 3! times (1/18)*(1/17) * (1/16) $\endgroup$ – user163862 Jan 9 '17 at 6:01
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Split it into disjoint events, and then add up their probabilities:

  • Event #$1$: choosing Jason, then Kim, then Ellie
  • Event #$2$: choosing Jason, then Ellie, then Kim
  • Event #$3$: choosing Kim, then Jason, then Ellie
  • Event #$4$: choosing Kim, then Ellie, then Jason
  • Event #$5$: choosing Ellie, then Jason, then Kim
  • Event #$6$: choosing Ellie, then Kim, then Jason

As you already calculated, the probability of each event is $\frac{1}{18\cdot17\cdot16}$

Therefore, the probability of either one of these events is $\frac{6}{18\cdot17\cdot16}$

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