# How can we find probability space and events

• MHB
• mathmari
In summary, finding probability space and events involves identifying all possible outcomes of a given situation and assigning a numerical value, known as probability, to each outcome. This can be done through the use of mathematical formulas and techniques, such as the sample space and event diagrams. Probability space and events are important concepts in understanding and predicting the likelihood of certain events occurring, and are commonly used in fields such as statistics, economics, and finance. By understanding how to find probability space and events, we can make more informed decisions and better assess risk in various situations.
mathmari
Gold Member
MHB
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:

mathmari said:
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.

Klaas van Aarsen said:
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:

mathmari said:
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.

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:

mathmari said:
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)

Klaas van Aarsen said:
Yep. Looks good to me. (Nod)

Great! Thank you! (Sun)

## 1. What is a probability space?

A probability space is a mathematical concept used to describe the possible outcomes of an experiment or event. It consists of a sample space (the set of all possible outcomes) and a set of probabilities associated with each outcome.

## 2. How do we determine the sample space?

The sample space is determined by listing all possible outcomes of an experiment or event. This can be done by using a tree diagram, a table, or by using the fundamental counting principle.

## 3. What are events in a probability space?

Events are subsets of the sample space that represent specific outcomes or combinations of outcomes. They can be simple events (one outcome) or compound events (more than one outcome).

## 4. How do we calculate the probability of an event?

The probability of an event is calculated by dividing the number of outcomes in the event by the total number of possible outcomes in the sample space. This can also be written as a fraction, decimal, or percentage.

## 5. Can the probability of an event be greater than 1?

No, the probability of an event cannot be greater than 1. It represents the likelihood of an event occurring and therefore must be between 0 and 1, inclusive.

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