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Let $E_1,E_2,E_3,E_4,E_5$ exchangeable events. We have the following probabilities: $$ P(E_2)=\frac{1}{2} $$ $$ P(E_3\wedge E_5)=\frac{1}{4} $$ $$ P(E_1\wedge E_2^C \wedge E_3 \wedge E_4^C \wedge E_5)=P(E_1\wedge E_2^C \wedge E_3^C \wedge E_4^C \wedge E_5^C)=P(E_1\wedge E_2 \wedge E_3 \wedge E_4 \wedge E_5)=\frac{1}{30} $$ Compute:

  1. $P(E_2\wedge E_3\wedge E_4)$
  2. $P(E_1\wedge E_2\wedge E_3\wedge E_4)$
  3. $P(E_1\wedge E_2\wedge E_3^C\wedge E_4^C\wedge E_5^C)$

So I know that $\omega_1=\frac{1}{2}$, $\omega_2=\frac{1}{4}$; first question is about computing $$ P(E_2\wedge E_3\wedge E_4)=\omega_3 $$ I have reached this point: $$ \omega_3=\frac{\omega_3^5}{10}+\frac{4\omega_4^5}{10}+\omega_5=\frac{1}{300}+\frac{4\omega_4^5}{10}+\frac{1}{30} $$

but I am stuck on computing $\omega_4^5$, how can I compute it? Thank you!

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  • $\begingroup$ Check for language translation errors. What is an "exchangeable event"? $\endgroup$
    – JMoravitz
    Mar 15, 2022 at 0:48
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    $\begingroup$ It may help if you explain your omega notation. You appear to use $\omega_4^5$ to mean the probability that a select four among five distinct events occur, and such. Is this the case? $\endgroup$ Mar 15, 2022 at 1:39
  • $\begingroup$ I'm sorry guys...@JMoravitz ece.iisc.ac.in/~parimal/2020/spqt/lecture-26.pdf here you can find a definition. $\endgroup$
    – Bmb58
    Mar 15, 2022 at 8:58
  • $\begingroup$ @GrahamKemp yes, it's as you said! and $\omega_5$ is the probability that five events among five distinct events occur. $\endgroup$
    – Bmb58
    Mar 15, 2022 at 8:59
  • $\begingroup$ How did you derive that equation? Using the Law of Total Probability, I obtain $$\color{red}{\omega_3} = \tbinom 20\color{blue}{\omega^5_3}+\tbinom 21\color{darkred}{\omega^5_4}+\tbinom 22\color{blue}{\omega_5}$$ $\endgroup$ Mar 16, 2022 at 1:11

1 Answer 1

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Let $\omega^n_k$ be the probability of an intersection of $k$ specified exchangeable events and complements of $n-k$ other specified events (all distinct).

We have been given 5 known probabilities:$${\def\c{{\small\complement}}{\omega^1_1=\tfrac 12=\mathsf P(E_1)=\mathsf P(E_2)=\ldots=\mathsf P(E_5)\\\omega^2_2=\tfrac 14=\mathsf P(E_1,E_2)=\mathsf P(E_1,E_3)=\ldots=\mathsf P(E_4,E_5)\\\omega^5_1=\tfrac 1{30}=\mathsf P(E_1,E_2^\c,E^\c_3,E^\c_4,E^\c_5)=\ldots=\mathsf P(E_1^\c,E_2^\c,E^\c_3,E^\c_4,E_5)\\\omega^5_3=\tfrac 1{30}=\mathsf P(E_1,E^\c_2,E_3,E^\c_4,E_5)=\ldots\\\omega^5_5=\tfrac 1{30}=\mathsf P(E_1,E_2,E_3,E_4,E_5)}}$$

You seek three unknowns ( $\omega^3_3$, $\omega^4_4$ and $\omega^5_2$) but have derived simultaneous equations with an extra unknown; $\omega^5_4$.

Well, you should just look for more such simultaneous equations...


So, using this notation and the Law of Total Probability, I derive four simultaneous equations with these four unknowns:

$$\begin{align} \color{blue}{\omega^1_1}&=\tbinom 40\color{blue}{\omega^5_1}+\tbinom 41\color{red}{\omega^5_2}+\tbinom 42\color{blue}{\omega^5_3}+\tbinom43\color{darkred}{\omega^5_4}+\tbinom44\color{blue}{\omega^5_5} \\[1ex] \color{blue}{\omega^2_2} &=\tbinom 30\color{red}{\omega^5_2} +\tbinom 31\color{blue}{\omega^5_3}+\tbinom 32\color{darkred}{\omega^5_4}+\tbinom 33\color{blue}{\omega^5_5} \\[1ex] \color{red}{\omega^3_3} &= \tbinom 20\color{blue}{\omega^5_3}+\tbinom 21\color{darkred}{\omega^5_4}+\tbinom 22\color{blue}{\omega^5_5} \\[1ex] \color{red}{\omega^4_4} &=\tbinom 10\color{darkred}{\omega^5_4}+\tbinom 11\color{blue}{\omega^5_5}\end{align}$$

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