Probability problem involving 4 Unions (inclusive OR)

[CoinToss]

Homework Statement

Given P(AUBUCUD), expand

The Attempt at a Solution

I approached the solution by procedurally drawing Vann Diagrams, building from AUB to AUBUCUD (Not included); Please do check if my line of reasoning is sound.
(let "n" be the intersection, AND)

1) P(AUB): P(A) + P(B) - P(AnB) [subtract the extra section, this is repeated at the end of subsequent steps]
2) P((AUB)UC): P(AUB) + P(C) - P((AUB)nC)
3) P((AUBUC)UD): P((AUB)UC) + P(D) - P((AUBUC)nD)

Substituting backwards, hence

A(AUBUCUD)= P(AUB) + P(C) - P((AUB)nC) + P(D) - P((AUBUC)nD)
A(AUBUCUD)= P(A) + P(B) - P(AnB) + P(C) - P((AUB)nC) + P(D) - P((AUBUC)nD) (Solved)

Also, do inform me if there is any more efficient way to put this. Thank you for your time.

 

[Ray Vickson]

Homework Statement

Given P(AUBUCUD), expand

The Attempt at a Solution

I approached the solution by procedurally drawing Vann Diagrams, building from AUB to AUBUCUD (Not included); Please do check if my line of reasoning is sound.
(let "n" be the intersection, AND)

1) P(AUB): P(A) + P(B) - P(AnB) [subtract the extra section, this is repeated at the end of subsequent steps]
2) P((AUB)UC): P(AUB) + P(C) - P((AUB)nC)
3) P((AUBUC)UD): P((AUB)UC) + P(D) - P((AUBUC)nD)

Substituting backwards, hence

A(AUBUCUD)= P(AUB) + P(C) - P((AUB)nC) + P(D) - P((AUBUC)nD)
A(AUBUCUD)= P(A) + P(B) - P(AnB) + P(C) - P((AUB)nC) + P(D) - P((AUBUC)nD) (Solved)

Also, do inform me if there is any more efficient way to put this. Thank you for your time.

You have the right idea (and yes, it IS tedious!). However, I believe the question wants you to expand further, so it wants you expand $P((A \cup B)\cap C)$ and $P((A \cup B \cup C)\cap D)$. Can you see how to do that?

 

[CoinToss]

Hello Vickson,

Thank you for your reminder; Using P(AnB)= P(A)*P(B) I believe that they can be solved as below? (some expressions were substituted backwards)

1. P(AnB)= P(A)P(B) [solved]

2. P((AUB)nC) = [P(A)+P(B)-P(AnB)]*P(C)
= [P(A) + P(B) - P(A)P(B)]*P(C)
= P(A)P(C) + P(B)P(C) - P(A)P(B)P(C) [solved]

3. P((AUBUB)nD)= [P(AUB)+P(C)-P((AUB)nC)]*P(D)
= {[P(A) + P(B) - P(AnB)] + P(C) - [P(A)P(C) + P(B)P(C) - P(A)P(B)P(C)]}*P(D)
= [P(A) + P(B) - P(A)P(B) + P(C) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C)]*P(D)
= P(A)P(D) + P(B)P(D) - P(A)P(B)P(D) + P(C)P(D) - P(A)P(C)P(D) - P(B)P(C)P(D) + P(A)P(B)P(C)P(D) [solved]

What a jargon. Thank you for your time.

 

[Ray Vickson]

Hello Vickson,

Thank you for your reminder; Using P(AnB)= P(A)*P(B) I believe that they can be solved as below? (some expressions were substituted backwards)

1. P(AnB)= P(A)P(B) [solved]

2. P((AUB)nC) = [P(A)+P(B)-P(AnB)]*P(C)
= [P(A) + P(B) - P(A)P(B)]*P(C)
= P(A)P(C) + P(B)P(C) - P(A)P(B)P(C) [solved]

3. P((AUBUB)nD)= [P(AUB)+P(C)-P((AUB)nC)]*P(D)
= {[P(A) + P(B) - P(AnB)] + P(C) - [P(A)P(C) + P(B)P(C) - P(A)P(B)P(C)]}*P(D)
= [P(A) + P(B) - P(A)P(B) + P(C) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C)]*P(D)
= P(A)P(D) + P(B)P(D) - P(A)P(B)P(D) + P(C)P(D) - P(A)P(C)P(D) - P(B)P(C)P(D) + P(A)P(B)P(C)P(D) [solved]

What a jargon. Thank you for your time.

No: what you wrote in 1) is true only if the events A and B are independent. Were you told that they are? If not, then what you wrote originally (in your first post) is correct: $P(A \cup B) = P(A) + P(B) - P(AB),$ where I have used the simpler notation $AB$ instead of $A \cap B$.

So, 2) is wrong as well, at least for non-independent events. Start with the set identity $(A \cup B) C = AB \cup AC$. This has nothing to do with probability; it is just "set algebra". OK, so you now have a probability of the form $P(\text{this} \cup \text{that}),$ and you have already figured how to expand such things.

After fixing up 2), you can go on to 3) in a similar way. If you think it was tedious before, just keep going!

 

[CoinToss]

Hello Vickson,

So it would change the scenario if the events were independent. Then if events A, B, C and D could occur without the other(s) happening the 2nd post would be correct?

I think I will get back to this after I familiarize myself with set algebra.
Thanks for your inspection of the equations.

 

[Ray Vickson]

Hello Vickson,

So it would change the scenario if the events were independent. Then if events A, B, C and D could occur without the other(s) happening the 2nd post would be correct?

I think I will get back to this after I familiarize myself with set algebra.
Thanks for your inspection of the equations.

The scenario does not change at all if the events are independent; the only thing that changes is whether or not you have P(AB) = P(A)*P(B) for example; you still have P(AB), etc., present in the final answer.