Simple Probability Question(HELP)

[Beowulf2007]

Homework Statement

Given that $$P(T) = P(U) = 0$$ then show that $$P(T \cup U) = 0$$

Given that $$P(T) = P(U) = 1$$ then show that $$P(T \cap U) = 1$$

Homework Equations

I am told that I need to use the following equations.

(1) $$P(T \cup U) = P(T) + P(U)$$ if $$P(T \cap U) = 0$$

(2) $$P(T \cup U) = P(T) + P(U) - P(T \cup U)$$

The Attempt at a Solution

My Solution for question one.

Since We know that $$P(T) = P(U) = 0$$ then by using equation

We get that $$P(T \cup U) = P(T) + P(U) = 0 + 0 = 0$$

My Solution for question two.

Here I have a feeling that I need to use equation two, but how do I deduce
$$P(T \cup U)$$ ??

Best Regards
Beowulf..

 

[e(ho0n3]

(2) $$P(T \cup U) = P(T) + P(U) - P(T \cup U)$$

The last term should be $P(T \cap U)$.

Since We know that $$P(T) = P(U) = 0$$ then by using eqution

We get that $$P(T \cup U) = P(T) + P(U) = 0 + 0 = 0$$

How do you know to use equation (2)? What if $P(T \cap U) \ne 0$?

Here I have a feeling that I need to use equation two, but how do I deduce
$$P(T \cup U)$$ ?? .

Why do you have a feeling? Describe your reasoning.

 

[Beowulf2007]

The last term should be $P(T \cap U)$.

Thanks ;)

How do you know to use equation (2)? What if $P(T \cap U) \ne 0$?
Why do you have a feeling? Describe your reasoning.

From what I see it $P(T \cap U)$ cannot be larger than one for obvious reasons. Because that would give a negative probability.

I am not sure about that, another way of proving question (2)

Could it say If P(C) is the entire probability space where then since P(T) = P(U), then I could say $$P(C) = P(U) + P(T) = 2$$ and then $P(T \cap U) =P(U) \cdot P(T) = 1$
Then $P(T \cup U) = P(c) - P(U) \cdot P(T) = 1$

Is my argument my reasonable? If yes can it be used in (1) too?

Best Regards

Beowulf

 

[e(ho0n3]

From what I see it $P(T \cap U)$ cannot be larger than one for obvious reasons. Because that would give a negative probability.

I will write $T \cap U$ as TU for short. If I understand you correctly, you're saying that P(TU) cannot be greater than 1 because if it is, it's negative?!

Could it say If P(C) is the entire probability space where then since P(T) = P(U), then I could say $$P(C) = P(U) + P(T) = 2$$ and then $P(T \cap U) =P(U) \cdot P(T) = 1$
Then $P(T \cup U) = P(c) - P(U) \cdot P(T) = 1$

Is my argument my reasonable? If yes can it be used in (1) too?

By the axioms of probability, P(C) = 1, not 2. Note that P(U) and P(T) have the same probability. What does that say about C, U and T? Also P(TU) = P(T)P(U) if T and U are independent events. The problem does not mention that T and U are independent.

 

[Beowulf2007]

Hello Again,

The whole problem is as follows: Let (S, E, P) be a probability space and let T and U be events(Nothing is said about the events).

Show, that if

(1) $$P(T) = P(U) = 0$$ then $$P(T \cup U) = 0$$

(2) $$P(T) = P(U) = 1$$ then $$P(T \cap U) = 1$$

My Solultion(1):$$P(T \cup U) = P(T) + P(U) = 0 + 0 = 0$$

My Solultion(2):$$P(T \cap U) = P(T) \cdot P(U) = 1 \cdot 1 = 1$$

Could this be it? What else is there to add?

Sincerely Yours
Beowulf

 

[e(ho0n3]

My Solultion(1):$$P(T \cup U) = P(T) + P(U) = 0 + 0 = 0$$

This will only work if P(TU) = 0. Can you prove that?

My Solultion(2):$$P(T \cap U) = P(T) \cdot P(U) = 1 \cdot 1 = 1$$

This will only work if T and U are independent. Can you prove that?

 

[Beowulf2007]

This will only work if P(TU) = 0. Can you prove that?

This will only work if T and U are independent. Can you prove that?

Question(a)

This will only work if P(TU) = 0? By this you mean proving that P(TU) = emptyset.

The only that I know here is the that is the consequence of the formula P(T U U) = P(T) + P(U).

Is that what you mean??

Question(b)

Since nothing is said about the events being indepedent do I assume they are??

Finally if these two can be proved, does that then complete the solution for (1) and (2)??

Many thanks in advance.

Sincerely Beowulf.

 

[e(ho0n3]

This will only work if P(TU) = 0? By this you mean proving that P(TU) = emptyset.

The only that I know here is the that is the consequence of the formula P(T U U) = P(T) + P(U).

Is that what you mean??

TU is not necessarily empty. Consider the elements in TU individually and their probabilities when considered as events.

Since nothing is said about the events being indepedent do I assume they are??

No. Forget about independence for the moment. If P(X) = 1 where X is some arbitrary subset of the sample space, what can you say about X?

Finally if these two can be proved, does that then complete the solution for (1) and (2)??

All you have to do is show that your results follow from the axioms of probability.

 

[Beowulf2007]

Regarding (1) do I considered the events to the mutually exclusive?? And therefor if P(T) occurs then P(U) also occurs and thus P(T u U ) = P(T) + P(U)?

Or is the explanation more simple??

TU is not necessarily empty. Consider the elements in TU individually and their probabilities when considered as events.



No. Forget about independence for the moment. If P(X) = 1 where X is some arbitrary subset of the sample space, what can you say about X?

Do you mean P's complement here?

All you have to do is show that your results follow from the axioms of probability.

In (2) Is it something to do with the events being mutually inclusive??

SR

Beowolf

 

[e(ho0n3]

You're going about it the wrong way. For problem (1) you're told that P(T) = P(U) = 0. What does this say about T and U? Ditto for problem (2).

 

[Beowulf2007]

You're going about it the wrong way. For problem (1) you're told that P(T) = P(U) = 0. What does this say about T and U? Ditto for problem (2).

Just so we understand each other. Do You mean?

$$T \subseteq U$$ and $$U \subseteq T$$

Therefore their union is defined as

$$T \cup U = T + U$$

Therefore their intersection is defined as

$$T \cap U = T \cdot U$$

And then use this fact to claim that my result in (1) and (2) are true!

How does that sound?

BeoWulf