1. ## 2 Questions about probability+De Morgan laws

about the first Question is known that A and B are Foreign events.
and i asked to find P(A).

about the second Question is known that A and B are event and i asked to decide what value y cannot get.

i tried with the De Morgan laws but i am not sure if i correct in my way.

2. What in the world are Foreign events?

3. Originally Posted by ZOOZ
about the first Question is known that A and B are Foreign events.
and i asked to find P(A).

about the second Question is known that A and B are event and i asked to decide what value y cannot get.

i tried with the De Morgan laws but i am not sure if i correct in my way.
Until we know what you mean by 'foreign' events, there is not enough information to answer these questions.

4. oh well yes i knew is will be wrong to say Foreign events, so what i mean is A and B are mutually exclusive events.

thanks.

5. Originally Posted by ZOOZ
oh well yes i knew is will be wrong to say Foreign events, so what i mean is A and B are mutually exclusive events.
Well then that means $\displaystyle A^c\cap B=B~\&~A\cap B^c=A$.
Use the given.

Moreover, $\displaystyle P(A^c\cap B^c)=1-P(A\cup B)$.

6. thank you for your answer, but i still not sure how can i know what P(A) can be.
i just know that P(A ∩ B) = 0 but i am not sure if is help me here.

can you please give me more tip to solve this question?

thanks again.

7. Originally Posted by ZOOZ
thank you for your answer, but i still not sure how can i know what P(A) can be.
i just know that P(A ∩ B) = 0 but i am not sure if is help me here.
can you please give me more tip to solve this question?
From the given: $\displaystyle 2P(A)=2P(A\cap B^c)=2P(B\cap A^c)=2P(B)$.

And $\displaystyle P(A^c\cap B^c)=1-P(A)-P(B)$

That is all I will do.

8. Originally Posted by ZOOZ
thank you for your answer, but i still not sure how can i know what P(A) can be.
i just know that P(A ∩ B) = 0 but i am not sure if is help me here.

can you please give me more tip to solve this question?

thanks again.
Draw a Karnaugh Table:

$\displaystyle \begin{tabular}{l | c | c | c} & A & A' & \\ \hline B & 0 & x & x\\ B' & x & 2x & 3x \\ \hline & x & 3x & 1 \\ \end{tabular}$

It is simple to solve for x and hence get Pr(A).

Originally Posted by ZOOZ
[snip]
about the second Question is known that A and B are event and i asked to decide what value y cannot get.

i tried with the De Morgan laws but i am not sure if i correct in my way.
As I said earlier, unless you say what sort of events A and B are, this question cannot be answered. But I suggest you draw a Karnaugh Table and make an attempt at it before posting your clarification. Then you can say what you've done and where you get stuck.

9. ok for the 1 question i got P(A)=1/4.
0.5x+1.5x=1

x=0.5

P(A)=x=0.5*0.5=0.25

P(B)=0.5x=0.5*0.5=0.25
do i correct?
thanks to both of you.

10. Originally Posted by ZOOZ
ok for the 1 question i got P(A)=1/4.
0.5x+1.5x=1

x=0.5

P(A)=x=0.5*0.5=0.25

P(B)=0.5x=0.5*0.5=0.25
do i correct?
thanks to both of you.
Yes, Pr(A) = Pr(B) = 1/4.

You still haven't said what sort of events A and B are in question 2.

11. well about question 2 it given that A and B are just events.
is not say's if it an mutually exclusive events.
so i think is just events in Sample space.

thanks.

12. Originally Posted by ZOOZ
well about question 2 it given that A and B are just events.
is not say's if it an mutually exclusive events.
so i think is just events in Sample space.

thanks.
In that case there is not enough information to get a numerical answer.