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Can we generate event with probabilities (a) $6.75\space p^2q$, (b) $20\space p^3q^2$, (c) $3.9\space pq$? If yes, then how? If no, then why?

Here, in a coin toss, let P(occurring head)$=p=1-q$

I can easily generate events for $3\space pq$(with three trials which is just $3\space p^2q+3\space pq^2$), $10\space p^3q^2$ (with five trials) and $3\space p^2q$ (with three trials), but could not find anything in these cases. Thanks for any help!

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For (b), for $0\le k\le4$ choose $20\binom4k$ patterns with $3+k$ heads and $2+(4-k)$ tails, and thus each with probability $p^{3+k}q^{2+(4-k)}$. This is possible because $\binom9{3+k}\ge20\binom4k$ for all $k$. The sum of the probabilities is

$$ \sum_{k=0}^420\binom4kp^{3+k}q^{2+(4-k)}=20p^3q^2(p+q)^4=20p^3q^2\;. $$

You can't do the other two for general $p$. Consider $p$ the reciprocal of any odd prime. Then the denominator of any probability in any finite number of trials is always a power of that odd prime and can never contain a factor of $2$, whereas the denominators of the two probabilities in $(a)$ and $(c)$ do.

This of course assumes that you were implying that the problem should be solved with a finite number of trials. It's easy to achieve any desired probability with infinitely many trials.

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  • $\begingroup$ but the first calculation gives $10p^3q^2$ $\endgroup$ – Stat_prob_001 Jul 16 '18 at 17:47
  • $\begingroup$ @Stat_prob_001: Sorry! I replaced it by a slightly more involved solution that works (unless I've made another mistake...). $\endgroup$ – joriki Jul 16 '18 at 19:41
  • $\begingroup$ @Stat_prob_001: I posted and answered a more general form of this question here: math.stackexchange.com/questions/2853884. $\endgroup$ – joriki Jul 16 '18 at 21:39

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