# Probability problem involving 4 Unions (inclusive OR)

• CoinToss
In summary, the original post discusses a problem of expanding the probability of a union of four events. The user uses set algebra and Vann diagrams to approach the solution, and asks for feedback on their reasoning. In the second post, the user corrects their solution for non-independent events and acknowledges the possibility of tediousness in expanding the probabilities further. They also mention a future plan to study set algebra to better understand the problem.
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.

CoinToss said:

## 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?

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.

Last edited:
CoinToss said:
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!

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.

CoinToss said:
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.

## 1. What is a probability problem involving 4 unions?

A probability problem involving 4 unions is one in which there are 4 different groups or events, and the question asks about the likelihood of at least one of these events occurring. In other words, the problem involves the use of the inclusive OR operator, which means that any one of the events can happen in order for the overall outcome to be considered a success.

## 2. How do you calculate the probability of a problem involving 4 unions?

To calculate the probability of a problem involving 4 unions, you must first determine the individual probabilities of each event occurring. Then, you can use the formula P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) - P(A and B) - P(A and C) - P(A and D) - P(B and C) - P(B and D) - P(C and D) + P(A and B and C) + P(A and B and D) + P(A and C and D) + P(B and C and D) - P(A and B and C and D). This formula takes into account the overlap between events and gives the overall probability of at least one of the events occurring.

## 3. What types of problems commonly involve 4 unions?

Problems involving 4 unions can vary, but some common types include questions about the likelihood of at least one defective item in a batch, the probability of at least one disease being present in a group of individuals, or the chance of winning at least one prize in a lottery with multiple prizes.

## 4. What is the difference between inclusive OR and exclusive OR?

Inclusive OR means that any one of the events can occur in order for the overall outcome to be considered a success. This is denoted by the symbol "or" (A or B or C). Exclusive OR, on the other hand, means that only one of the events can occur. This is denoted by the symbol "xor" (A xor B xor C). In probability problems, exclusive OR is often used when considering mutually exclusive events, while inclusive OR is used when any of the events can happen.

## 5. How can probability problems involving 4 unions be applied in real life?

Probability problems involving 4 unions can be applied in various real-life situations, such as in risk assessment, insurance policies, and medical diagnoses. For example, a doctor may use the probability of at least one disease being present in a group of patients to determine the likelihood of a patient having a particular illness. Similarly, insurance companies may use these types of problems to evaluate the risk of insuring a particular group of individuals.

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