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I am trying to find three events $A,B,C$ with the following properties

\begin{gather} P(A|B)>P(A),\\ P(A|C)>P(A),\\ P(A|B\cup C)<P(A). \end{gather}

I have not been able to come up with events satisfying all three, and would appreciate some help.

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  • $\begingroup$ I think it can. Consider flipping two coins. A is the event both coins land on heads, B is the event the first coin lands on head. Then $P(A|B) =1/2 >1/4 = P(A)$ $\endgroup$ – lkjhgfdsa Mar 2 '18 at 12:25
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In fact, I think I've just spotted an answer.

Consider picking a number at random from $\{1,2,\dots,7\}$.

Let $A$ be the event $1$ or $2$ is picked, let $B$ be the event that one of $1$, $3$, or $4$ is picked, and $C$ be the event that one of $1$, $5$ or $6$ is picked.

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    $\begingroup$ Sorry I misread the question and kept giving wrong example, now see.. $\endgroup$ – King Tut Mar 2 '18 at 12:40
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Consider drawing one number from $1$ to $5$ and define following events

  1. $A:$ Number drawn is $1,2$

  2. $B:$ Number drawn is $1,3$

  3. $C:$ Number drawn is $1,4$

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    $\begingroup$ Our answers are basically the same, though I think for yours $P(A|B)=P(A|C)=P(A)$. $\endgroup$ – lkjhgfdsa Mar 2 '18 at 12:43
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    $\begingroup$ Now @lkjhgfdsa $P(A) = 2/5$ so its finally correct $\endgroup$ – King Tut Mar 2 '18 at 12:47
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    $\begingroup$ @lkjhgfdsa Please accept your answer as is very clear and you were able to reach it before! I will like it! $\endgroup$ – King Tut Mar 2 '18 at 12:51
  • $\begingroup$ Ok :) Will do in 2 days when it allows me to! $\endgroup$ – lkjhgfdsa Mar 2 '18 at 12:58
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Here's an example we can reason through intuitively.

Suppose you're buying two lottery tickets with a $\frac1{1000}$ chance of winning. (This chance can be made as small as you like, as long as it's less than $\frac12$.) Let $A$ be the event that neither or both of the tickets win. (They tickets are for different drawings, so they're independent.)

Now,

  • Define $B$ to be the event that the first ticket fails to win. Then $\Pr[A \mid B] > \Pr[A]$ because the likeliest way by far for $A$ to happen is if neither ticket wins, and conditioning on $B$ makes that likelier.
  • Define $C$ to be the event that the second ticket fails to win. We have $\Pr[A \mid C] > \Pr[A]$ for exactly the same reason as above.

However, $B \cup C$ is just the event that at most one ticket wins: the negation of "both tickets win". So $\Pr[A \mid B\cup C] < \Pr[A]$ because conditioning on $B \cup C$ removes one (very unlikely) way for $A$ to happen, and does nothing else.

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  • $\begingroup$ Is $C$ meant to be "the second ticket fails to win"? $\endgroup$ – psmears Mar 2 '18 at 17:14
  • $\begingroup$ It totally is, and that's what I get for copying and pasting. Thanks! Fixed. $\endgroup$ – Misha Lavrov Mar 2 '18 at 17:20

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