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I have $6$ random numbers and the probability to pick them are the following:

$p_1 = p_2 = p_3$

$p_4 = p_5 = p_6 = 2p_1$

The first $3$ have the probability and the other $3$ have double the probability of the first ones.

How can I find a number representation for them? In my book, it's written that $p_1 = p_2 = p_3 = 1/9$, and $p_4 = p_5 = p_6 = 2/9$, but there's no explanation from where do we get those numbers.

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    $\begingroup$ Surely, you do not mean that $p_4 = p_5 = p_6 = 2\pi > 1$. Did you intend to indicate that $p_4 = p_5 = p_6 = 2p_1 = 2p_2 = 2p_3$? $\endgroup$ Commented Apr 2, 2016 at 10:23
  • $\begingroup$ I meant the second one, p4=p5=p6=2pi=2p2=2p3. 2/9=2/9=2/9=2*1/9=2*1/9=2*1/9. Sorry about that. $\endgroup$
    – user327949
    Commented Apr 2, 2016 at 10:24
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    $\begingroup$ Use the fact that (if I understand the meaning) the probabilities $p_1,\ldots, p_6$ add up to $1$. $\endgroup$
    – hardmath
    Commented Apr 2, 2016 at 10:25
  • $\begingroup$ Here is a tutorial on how to typeset mathematics on this site. $\endgroup$ Commented Apr 2, 2016 at 10:28
  • $\begingroup$ That's what I tried to do, make an equation a+2b=1. Then b=(1-a)/2, but I don't know how to proceed and find a real value. $\endgroup$
    – user327949
    Commented Apr 2, 2016 at 10:30

1 Answer 1

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Since there are six possible events in the sample space, their probabilities must add to $1$. $$p_1 + p_2 + p_3 + p_4 + p_5 + p_6 = 1$$ Moreover, you know that $p_1 = p_2 = p_3$ and that $p_4 = p_5 = p_6 = 2p_1$. Now express $p_2$, $p_3$, $p_4$, $p_5$, and $p_6$ in terms of $p_1$, then solve for $p_1$. Substitute to find the other probabilities.

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    $\begingroup$ I get it now. I think I found the solution. 9p1 = 1, hence p1 = 1/9. Thanks a lot! $\endgroup$
    – user327949
    Commented Apr 2, 2016 at 10:33

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