MHB How can we find probability space and events

[mathmari]

Hey! :giggle:

Give a probability space and events $A$, $B$ and $C$ such that $P [C \mid A]>P [C \mid A \cup B]$.So we want that $$\frac{P[C\cap A]}{P[A]}>\frac{P[C\cap (A\cup B)]}{P[A\cup B]}=\frac{P[(C\cap A)\cup (C\cap B)]}{P[A\cup B]}$$ But how can we find these events ? :unsure:

 

[I like Serena]

Give a probability space and events $A$, $B$ and $C$ such that $P [C \mid A]>P [C \mid A \cup B]$.

So we want that $$\frac{P[C\cap A]}{P[A]}>\frac{P[C\cap (A\cup B)]}{P[A\cup B]}=\frac{P[(C\cap A)\cup (C\cap B)]}{P[A\cup B]}$$ But how can we find these events ?

Hey mathmari!

It looks as if we're supposed to find an example where the events could be anything.
For instance $A$ and $B$ could be genetic modifications, and $C$ could be whether a certain condition occurs.
Either way, we would need to assign example probabilities to the events and their intersections such that the given expression holds.

 

[mathmari]

It looks as if we're supposed to find an example where the events could be anything.
For instance $A$ and $B$ could be genetic modifications, and $C$ could be whether a certain condition occurs.
Either way, we would need to assign example probabilities to the events and their intersections such that the given expression holds.

So we consider $A$ to be the event that the genetic modification A occurs and $B$ to be the event that the genetic modification B occurs.
Let $P(A)=0.30$, $P(B)=0.40$, $P(A\mid C)=0.25$, $P(B\mid C)=0.35$, $P(C)=0.70$ $P(A\cap B)=0$ and so $P(A\cup B)=P(A)+P(B)=0.30+0.40=0.70$.

Then we have $P(C\cap A)=P(A\mid C)\cdot P(C)=0.25\cdot 0.70=0.175$, $P(C\cap B)=P(B\mid C)\cdot P(C)=0.35\cdot 0.70=0.245$ and $P[(C\cap A)\cup (C\cap B)]=P[C\cap A]+P[C\cap B]=0.175+0.245=0.42$ and so we get at the left side $$\frac{P[C\cap A]}{P[A]}=\frac{0.175}{0.30}\approx 0.583$$ and at the right side $$ \frac{P[C\cap (A\cup B)]}{P[A\cup B]}=\frac{P[(C\cap A)\cup (C\cap B)]}{P[A\cup B]} =\frac{0.42}{0.70} =0.6$$ If all the calculations are correct then the probabilities that I tokk are wrong... :unsure:

 

[I like Serena]

So we consider $A$ to be the event that the genetic modification A occurs and $B$ to be the event that the genetic modification B occurs.
Let $P(A)=0.30$, $P(B)=0.40$, $P(A\mid C)=0.25$, $P(B\mid C)=0.35$, $P(C)=0.70$ $P(A\cap B)=0$ and so $P(A\cup B)=P(A)+P(B)=0.30+0.40=0.70$.

Then we have $P(C\cap A)=P(A\mid C)\cdot P(C)=0.25\cdot 0.70=0.175$, $P(C\cap B)=P(B\mid C)\cdot P(C)=0.35\cdot 0.70=0.245$ and $P[(C\cap A)\cup (C\cap B)]=P[C\cap A]+P[C\cap B]=0.175+0.245=0.42$ and so we get at the left side $$\frac{P[C\cap A]}{P[A]}=\frac{0.175}{0.30}\approx 0.583$$ and at the right side $$ \frac{P[C\cap (A\cup B)]}{P[A\cup B]}=\frac{P[(C\cap A)\cup (C\cap B)]}{P[A\cup B]} =\frac{0.42}{0.70} =0.6$$ If all the calculations are correct then the probabilities that I tokk are wrong...

Yep. It's the wrong way around, so we probably need to increase one probability and decrease another to match.

Btw, if everything would be correct, than the probability of each possible intersection must also be between $0$ and $1$, since otherwise we don't have a valid probability function.

 

[mathmari]

When we consider the example https://mathhelpboards.com/threads/probability-chicken-legs.29067/ with $A=V_1$, $B=V_2$, $C=K$ then we get the following :

$P(V_1)=0.20$, $P(V_2)=0.10$, $P(K)=0.21$
$P(K\mid V_1)=0.80$
$P(K\mid V_1\cup V_2)$
$P(K\mid (V_1\cup V_2))=\frac{P(K\cap (V_1\cup V_2))}{P(V_1\cup V_2)}=\frac{P((K\cap V_1)\cup (K\cap V_2))}{P(V_1\cup V_2)}=\frac{P(K\cap V_1)+P(K\cap V_2)}{0.30}=\frac{0.16+0.04}{0.30}=\frac{0.20}{0.30}\approx 0.67$

So the inequality holds, right? :unsure:

 

[I like Serena]

When we consider the example https://mathhelpboards.com/threads/probability-chicken-legs.29067/ with $A=V_1$, $B=V_2$, $C=K$ then we get the following :

$P(V_1)=0.20$, $P(V_2)=0.10$, $P(K)=0.21$
$P(K\mid V_1)=0.80$
$P(K\mid V_1\cup V_2)$
$P(K\mid (V_1\cup V_2))=\frac{P(K\cap (V_1\cup V_2))}{P(V_1\cup V_2)}=\frac{P((K\cap V_1)\cup (K\cap V_2))}{P(V_1\cup V_2)}=\frac{P(K\cap V_1)+P(K\cap V_2)}{0.30}=\frac{0.16+0.04}{0.30}=\frac{0.20}{0.30}\approx 0.67$

So the inequality holds, right?

Yep. Looks good to me. (Nod)

 

[mathmari]

Yep. Looks good to me. (Nod)

Great! Thank you! (Sun)