Venn Diagram

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nasi112

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A traveler is looking at upgrades received form airlines and rental car companies and found the following information:

- The probability that he receives either an airline upgrade or a rental car upgrade or both [MATH]= 58[/MATH]%.

- The probability that he receives only an airline upgrade [MATH]= 18[/MATH]%.

- The probability that he receives either an airline upgrade or a rental car upgrade but not both [MATH]= 46[/MATH]%.

Let [MATH]A[/MATH] be the event for airline upgrade and [MATH]B[/MATH] the event for rental car upgrade, and draw a Venn Diagram. Are the events receive a rental car upgrade and receive an airline upgrade independent? Explain/justify your answer.


This is what I did.

[MATH]P(A) + P(B) + P(AB) = 0.58[/MATH]
[MATH]P(A) = 0.18[/MATH]
[MATH]P(A) + P(B) = 0.46[/MATH]
Then,

[MATH]P(B) = 0.28[/MATH]
[MATH]P(AB) = 0.12[/MATH]
I have said that since [MATH]P(BA) \neq P(B) \cdot P(A)[/MATH], the events are not independent. I was shocked that the teacher marked it wrong!


37.png

Which diagram is correct? My diagram or the teacher's diagram. Can anyone help me please?
 
nasi112 said:
A traveler is looking at upgrades received form airlines and rental car companies and found the following information:
- The probability that he receives either an airline upgrade or a rental car upgrade or both [MATH]= 58[/MATH]%.
- The probability that he receives only an airline upgrade [MATH]= 18[/MATH]%.
- The probability that he receives either an airline upgrade or a rental car upgrade but not both [MATH]= 46[/MATH]%.
Click to expand...
You are given: The probability that he receives only an airline upgrade [MATH]= 18[/MATH]%
In your diagram you say that is \(0.06\). So answer your one question.
 
pka said:
You are given: The probability that he receives only an airline upgrade [MATH]= 18[/MATH]%
In your diagram you say that is \(0.06\). So answer your one question.
Click to expand...
Thanks for the reply.
Why is it wrong to say that airline upgrade is event A, and P(A) = the whole circle of A?



lex said:
View attachment 26637
Click to expand...
Thanks for the reply.
The same question I have here. It says airline upgrade is event A, then why you wrote that it is [MATH]P(A \ and \ B')[/MATH]?
 
pka said:
The word only there means A by itself. That is A but not B. A alone.
Click to expand...
If [MATH]A[/MATH] is the event, [MATH]P(A)[/MATH] is its probability. Am I write or wrong?

Is it correct now to say
The probability that he receives only an airline upgrade [MATH]P(A \ and \ B')[/MATH]The probability that he receives an airline upgrade [MATH]P(A)[/MATH]???
 
Your teacher was quite correct, and you were quite wrong.

[MATH]\text {X and Y are mutually exclusive if and only if P(X and Y) = 0.[/MATH]
[MATH]\text {If X and Y are mutually exclusive, P(X or Y) = P(X) + P(Y).}[/MATH]
[MATH]\text {If X and Y are NOT mutually exclusive, P(X or Y) = P(X) + P(Y) - P(X and Y).}[/MATH]
So your adding the joint probability is wrong; it is SUBTRACTED.

Let's go to your problem and define the different events carefully.

A get nothing
B get airline only
C get car only
D get both car and airline.

That is an EXHAUSTIVE set of MUTUALLY EXCLUSIVE events.

[MATH]P(A) + P(B) + P(C) + P(D) = 1.[/MATH]
Let's define E as airline or car, but not both. That is, E is B or C. And B and C are mutually exclusive. So

[MATH]P(E) = P(B) + P(C).[/MATH]
Let's define F as airline or car. That is, F is D or E. And those are mutually exclusive. So

[MATH]P(F) = P(D) + P(E).[/MATH]
So what are we told

[MATH]P(F) = 58\%.[/MATH]
[MATH]P(B) = 18\%.[/MATH]
[MATH]P(E) = 46\%.[/MATH]
[MATH]\therefore 58\% = 46\% + P(D) \implies P(D) = 58\% - 46\% = 12\%.[/MATH]
[MATH]\text {And } 46\% = 18\% + P(C) \implies P(C) = 46\% - 18\% = 28\%.[/MATH]
[MATH]\text {And } P(A) = 1 - 58\% = 42\%.[/MATH]
Independence was not relevant to this exercise, but mutually exclusive was.
 
JeffM said:
Your teacher was quite correct, and you were quite wrong.

[MATH]\text {X and Y are mutually exclusive if and only if P(X and Y) = 0.[/MATH]
[MATH]\text {If X and Y are mutually exclusive, P(X or Y) = P(X) + P(Y).}[/MATH]
[MATH]\text {If X and Y are NOT mutually exclusive, P(X or Y) = P(X) + P(Y) - P(X and Y).}[/MATH]
So your adding the joint probability is wrong; it is SUBTRACTED.

Let's go to your problem and define the different events carefully.

A get nothing
B get airline only
C get car only
D get both car and airline.

That is an EXHAUSTIVE set of MUTUALLY EXCLUSIVE events.

[MATH]P(A) + P(B) + P(C) + P(D) = 1.[/MATH]
Let's define E as airline or car, but not both. That is, E is B or C. And B and C are mutually exclusive. So

[MATH]P(E) = P(B) + P(C).[/MATH]
Let's define F as airline or car. That is, F is D or E. And those are mutually exclusive. So

[MATH]P(F) = P(D) + P(E).[/MATH]
So what are we told

[MATH]P(F) = 58\%.[/MATH]
[MATH]P(B) = 18\%.[/MATH]
[MATH]P(E) = 46\%.[/MATH]
[MATH]\therefore 58\% = 46\% + P(D) \implies P(D) = 58\% - 46\% = 12\%.[/MATH]
[MATH]\text {And } 46\% = 18\% + P(C) \implies P(C) = 46\% - 18\% = 28\%.[/MATH]
[MATH]\text {And } P(A) = 1 - 58\% = 42\%.[/MATH]
Independence was not relevant to this exercise, but mutually exclusive was.
Click to expand...

Thanks for the reply JeffM. I understood what you have explained.

The teacher said
[MATH]P(A|B) = \frac{0.12}{0.4} = 0.3[/MATH]and
[MATH]P(A|B') = \frac{0.18}{0.6} = 0.3[/MATH]and
[MATH]P(A) = 0.3[/MATH]
The conditional probabilities are the same and equal to [MATH]P(A)[/MATH], therefore, the events are independent. (The student can also compare [MATH]P(B|A), P(B|A')[/MATH], and [MATH]P(B)[/MATH])

He also said

event A is for airline upgrade
event B is for car rental upgrade

He also mentioned above [MATH]P(A) = 0.3[/MATH]
When I look at the Venn diagram (teacher version) I see the probability of airline upgrade is 0.3, not 0.18

I am confused. If A is the event for airline upgrade, and P(A) = 0.3, this means the probability of airline upgrade is 0.3, not 0.18
 
The problem you originally quoted need not bother with independence.

It is a quite tricky to communicate here because I used a different notation than you have been using. I do not know why you chose the A and B that you did because they are not mutually exclusive events. Breaking things down into mutually exclusive events frequently makes things a lot clearer. I am going to change my notation to make it compatible with yours.

Your event A is airplane. My event Q is airplane only. My event R is airplane and car. Notice Q and R are mutually exclusive so
P(Q or R) = P(Q) + P(R). Notice as well that Q or R means the same as A.
Thus, P(A) = P(Q) + P(R).

Your event B is car. My event S is car only. My event R is airplane and car. Notice R and S are mutually exclusive so
P(R or S) = P(R) + P(S). Notice as well that R or S means the same as B.
Thus, P(B) = P(R) + P(S).

My event T is neither car nor airplane.

A and B is what? It has to be R, with probability 0.12. R is the only event with both car and airplane. What is probability of B. It is 0.40. Therefore the probability of A given B is just the probability of R given B, 0.12 / 0.4 = 0.3.

My advice to you is to think in terms of mutually exclusive events. You will make fewer errors.
 
JeffM said:
The problem you originally quoted need not bother with independence.

It is a quite tricky to communicate here because I used a different notation than you have been using. I do not know why you chose the A and B that you did because they are not mutually exclusive events. Breaking things down into mutually exclusive events frequently makes things a lot clearer. I am going to change my notation to make it compatible with yours.

Your event A is airplane. My event Q is airplane only. My event R is airplane and car. Notice Q and R are mutually exclusive so
P(Q or R) = P(Q) + P(R). Notice as well that Q or R means the same as A.
Thus, P(A) = P(Q) + P(R).

Your event B is car. My event S is car only. My event R is airplane and car. Notice R and S are mutually exclusive so
P(R or S) = P(R) + P(S). Notice as well that R or S means the same as B.
Thus, P(B) = P(R) + P(S).

My event T is neither car nor airplane.

A and B is what? It has to be R, with probability 0.12. R is the only event with both car and airplane. What is probability of B. It is 0.40. Therefore the probability of A given B is just the probability of R given B, 0.12 / 0.4 = 0.3.

My advice to you is to think in terms of mutually exclusive events. You will make fewer errors.
Click to expand...
Thanks for the clear explanation. Yeah, dividing the events into mutually exclusive events will make the problem simpler.

But the wording is confusing. For example

The probability that he receives either an airline upgrade or a rental car upgrade or both

I cannot know if car upgrade here means (car only = 0.18) or car = 0.3. Maybe, you had figured it out because of the teacher diagram, but what if you have no diagram!
 
The words at the start of the question: 'he receives either an airline upgrade or a rental car upgrade' can be interpreted two ways:
(i) he receives only an airline upgrade or only a rental car upgrade
(ii) he receives only an airline upgrade or only a rental car upgrade or he receives both.
It is clear from the rest of the sentence that the question is about (ii)

Events are sets.
Probabilities are numbers.
The same set can be described in different ways:
[MATH]A \cup B[/MATH] is the same set as [MATH]A \cup B \cup (A \cap B) [/MATH] and the same set as [MATH](A \cap \bar{B}) \cup (B \cap \bar{A}) \cup (A \cap B)[/MATH]
nasi112 said:
I cannot know if car upgrade here means (car only = 0.18) or car = 0.3
Click to expand...
For that reason, it doesn't matter which they mean. It is clear from the full sentence that they mean all of A and all of B.
All we need at the moment is to know what set they are talking about and that is clear.

When it comes to calculating the probability of the event, then you have to take care how the set is described because that will affect how we calculate the probability.
E.g. [MATH]P((A \cap \bar{B}) \cup (B \cap \bar{A}) \cup (A \cap B)) \text{ is simply }P(A \cap \bar{B})+P(B \cap \bar{A})+P(A \cap B) [/MATH]because the sets are disjoint (don't overlap - see Venn Diagram. [MATH] |(A \cap \bar{B}) \cup (B \cap \bar{A}) \cup (A \cap B)| = |(A \cap \bar{B})| + |(B \cap \bar{A})| + |(A \cap B)|[/MATH] ).

But [MATH]P(A \cup B) ≠ P(A)+P(B)[/MATH] (in general), since A and B may overlap (and do here).
The correct rule for this form of the set notation is:
[MATH]\boxed{P(A \cup B) = P(A) +P(B) -P(A \cap B)}[/MATH](You can see why, by looking at the Venn diagram: [MATH]|A\cup B|=|A| + |B| - |A \cap B| \text{ (since } |A \cap B|[/MATH] has been counted twice).
 
lex said:
The words at the start of the question: 'he receives either an airline upgrade or a rental car upgrade' can be interpreted two ways:
(i) he receives only an airline upgrade or only a rental car upgrade
(ii) he receives only an airline upgrade or only a rental car upgrade or he receives both.
It is clear from the rest of the sentence that the question is about (ii)

Events are sets.
Probabilities are numbers.
The same set can be described in different ways:
[MATH]A \cup B[/MATH] is the same set as [MATH]A \cup B \cup (A \cap B) [/MATH] and the same set as [MATH](A \cap \bar{B}) \cup (B \cap \bar{A}) \cup (A \cap B)[/MATH]
For that reason, it doesn't matter which they mean. It is clear from the full sentence that they mean all of A and all of B.
All we need at the moment is to know what set they are talking about and that is clear.

When it comes to calculating the probability of the event, then you have to take care how the set is described because that will affect how we calculate the probability.
E.g. [MATH]P((A \cap \bar{B}) \cup (B \cap \bar{A}) \cup (A \cap B)) \text{ is simply }P(A \cap \bar{B})+P(B \cap \bar{A})+P(A \cap B) [/MATH]because the sets are disjoint (don't overlap - see Venn Diagram. [MATH] |(A \cap \bar{B}) \cup (B \cap \bar{A}) \cup (A \cap B)| = |(A \cap \bar{B})| + |(B \cap \bar{A})| + |(A \cap B)|[/MATH] ).

But [MATH]P(A \cup B) ≠ P(A)+P(B)[/MATH] (in general), since A and B may overlap (and do here).
The correct rule for this form of the set notation is:
[MATH]\boxed{P(A \cup B) = P(A) +P(B) -P(A \cap B)}[/MATH](You can see why, by looking at the Venn diagram: [MATH]|A\cup B|=|A| + |B| - |A \cap B| \text{ (since } |A \cap B|[/MATH] has been counted twice).
Click to expand...
Thank you so much lex for this patient. I started to fill the gaps of my misunderstanding. This problem was a pretty good experience.
 
Yes. You seem to be picking up plenty.
Btw you can show that two events A and B are independent without using conditional probabilities.
[MATH]\boxed{\text{A and B are independent events } \leftrightarrow P(A \cap B)=P(A) \times P(B)}[/MATH]In this case [MATH]P(A) \times P(B) = 0.3 \times 0.4 = 0.12 = P(A \cap B) [/MATH] ✓
[MATH]\therefore[/MATH] A and B are independent events.
 
lex said:
Yes. You seem to be picking up plenty.
Btw you can show that two events A and B are independent without using conditional probabilities.
[MATH]\boxed{\text{A and B are independent events } \leftrightarrow P(A \cap B)=P(A) \times P(B)}[/MATH]In this case [MATH]P(A) \times P(B) = 0.3 \times 0.4 = 0.12 = P(A \cap B) [/MATH] ✓
[MATH]\therefore[/MATH] A and B are independent events.
Click to expand...
Yeah, this is the way I did it but with the wrong probabilities lol:ROFLMAO:

Thanks lex:)
 
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