18 boys and 2 girls are made to stand in a line in a random order.Let $X$ be the number of boys standing in between the girls .Find $P[X=5]$ and $E[X]$. I proceed in this way: Note that $$P[X=0]=\frac{19.18!.2!}{20!}=\frac{19-0}{20\choose2}$$ $$P[X=1]=\frac{18.18!.2!}{20!}=\frac{19-1}{20\choose2}$$$$P[X=2]=\frac{17.18!.2!}{20!}=\frac{19-2}{20\choose2}$$So, $$P[X=x]=\frac{19-x}{20\choose2}$$,$x=0(1)18.$ so.$$P[X=5]=\frac{19-5}{20\choose2}=0.0737$$ and $$E[X]=\sum_{x=0}^{18}x.\frac{19-x}{20\choose2}=6$$
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closed as not constructive by cardinal, William, Noah Snyder, Norbert, rschwieb Oct 15 '12 at 18:28
As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or specific expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, see the FAQ for guidance.
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You got it all correct. Just to add a little bit of proof here, imagine the boys as black dots and the two girls as red dots. The configuration $X=5$ implies that the distance between the two red dots is $7$ dots. The number of working configurations is thus $20-7+1=14$ (just translate the fixed line between the red dots). Total configurations correspond to the choice of the position of red dots, which is $$\binom{20}{2} = 190$$ Thus $$ P(X=5) = \frac{14}{190} = 0.0737$$ You got the rest correct. |
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