What is P(B) given conditional probabilities and the complement of A?

[buddingscientist]

Well the first and last I'm having some troubles with, and 2-4 I think the logic I am using is correct but would like his verified since no answers were provided

What is $P(A \cup B)$ if $P(A) = 0.2, P(A \cap B) = 0.1, P(B) = 0.5?$

Would that just be the prob. of being in A or B minus prob of being in both (prob of being in A + prob being in B - A int B). Would it depend on whether they are mutually exclusive or not? (how can we tell if that's all tahts given in the question).
I am kind of half between (A + B) and half between (A + B - AintB). But since A int B was included in the question, would that imply that I should use A + B - A int B = 0.2 + 0.5 - 0.1 = 0.6

---
What is E(X) if Mx(u) = $(1-u)^{-3}, u<1$

To find E(X) find the first derivative:
= -3(1-u)^(-4).-1
= 3(1-u)^(-4)
and then let u -> 0
3(1)^(-4)
=3

Therefor E(X) = 3


---
What is E($X^{3}$) if fx(x) = 2x, 0<x<1

E(X^(3)) = integral (0,1) of 2x.x^3 dx
= int (0,1) 2x^4 dx
= 2/5 x^5 .. (0,1)
= 2/5

Therefor E(X^3)) = 2/5

---
What is c if
g(x) = c|x|, x = -2, -1, 1, 2 is a probability function

For it to be a prob. function, the sum of all the probabilities must equal 1
2c + c + c + 2c = 1
c = 1/6

---
What is $P(\overline{B})$ if $P(B|\overline{A}) = 0.5, P(\overline{A}) = 0.3$ $and P(B|A) = 0.8 ?$

Well I'm a bit stuck on this question;
I used some multiplicative laws to find
$P(A \cap B) = 0.56$
and $P(B \cap \overline{A}) = 0.15$
I'm not sure how to continue from here.

Thanks

 

[xanthym]

(A):
What is $P(A \cup B)$ if $P(A) = 0.2, P(A \cap B) = 0.1, P(B) = 0.5?$

(B):
What is $P(\overline{B})$ if $P(B|\overline{A}) = 0.5, P(\overline{A}) = 0.3$ $and P(B|A) = 0.8 ?$

ITEM (A):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
::: ⇒ P(A ∪ B) = (0.2) + (0.5) - (0.1)
::: ⇒ P(A ∪ B) = (0.6)

ITEM (B):
From problem statement:
P(A^c) = (0.3)
P(A) = 1 - P(A^c) = 1 - (0.3) = (0.7)

P(B | A) = (0.8)
P(B^c | A) = 1 - P(B | A) = 1 - (0.8) = (0.2)

P(B | A^c) = 0.5
P(B^c | A^c) = 1 - P(B | A^c) = 1 - (0.5) = (0.5)

Thus, using above results:
P(B^c) = P(B^c ∩ A) + P(B^c ∩ A^c) =
= P(B^c | A)*P(A) + P(B^c | A^c)*P(A^c) =
= (0.2)*(0.7) + (0.5)*(0.3)
::: ⇒ P(B^c) = (0.29)


~~

 

[HallsofIvy]

Well the first and last I'm having some troubles with, and 2-4 I think the logic I am using is correct but would like his verified since no answers were provided

What is $P(A \cup B)$ if $P(A) = 0.2, P(A \cap B) = 0.1, P(B) = 0.5?$

Would that just be the prob. of being in A or B minus prob of being in both (prob of being in A + prob being in B - A int B). Would it depend on whether they are mutually exclusive or not? (how can we tell if that's all tahts given in the question).
I am kind of half between (A + B) and half between (A + B - AintB). But since A int B was included in the question, would that imply that I should use A + B - A int B = 0.2 + 0.5 - 0.1 = 0.6

$P(A\cup B)= P(A)+ P(B)- P(A \cap B)$ so in this problem, yes, $P(A\cup B)= .2+ .5- .1= 0.6$.
You know that A and B are not mutually exclusive because
$P(A\cap B)$ is not 0!

What is E(X) if Mx(u) = $(1-u)^{-3}, u<1$

To find E(X) find the first derivative:
= -3(1-u)^(-4).-1
= 3(1-u)^(-4)
and then let u -> 0
3(1)^(-4)
=3

Therefor E(X) = 3

Assuming that Mx(u) is the moment generating function, then, yes, E(X) is the coefficient of u in the McLaurin expansion of Mx(u): 3 in this case.

What is E($X^{3}$) if fx(x) = 2x, 0<x<1

E(X^(3)) = integral (0,1) of 2x.x^3 dx
= int (0,1) 2x^4 dx
= 2/5 x^5 .. (0,1)
= 2/5

Therefor E(X^3)) = 2/5

Yes, that's correct.

What is c if
g(x) = c|x|, x = -2, -1, 1, 2 is a probability function

For it to be a prob. function, the sum of all the probabilities must equal 1
2c + c + c + 2c = 1
c = 1/6

Of course.

What is $P(\overline{B})$ if $P(B|\overline{A}) = 0.5, P(\overline{A}) = 0.3$ $and P(B|A) = 0.8 ?$

Well I'm a bit stuck on this question;
I used some multiplicative laws to find
$P(A \cap B) = 0.56$
and $P(B \cap \overline{A}) = 0.15$
I'm not sure how to continue from here.

Thanks

If $P(\overline A)= 0.3$ then P(A)= 1- 0.3= 0.7
$P(B)= P(B|A)P(A)+ P(B|\overline A)P(\overline A)$
= 0.8(0.7)+ 0.5(0.3)= 0.56+ 0.15= 0.71 so
$P(\overline B)= 1- 0.71= 0.29.$

 

[Dr-NiKoN]

Could someone please explain what $A^c$ means? I've never encountered this notation when working with sets? Is it the same as the cartesian product of a set: $A^n$

 

[Data]

$$A^c$$ just means the complement of $$A$$.

 

[buddingscientist]

Ahh thank you all very much, I must have been unaware of the following results:
$P(B)= P(B \cap A) + P(B \cap \overline A)$
$P(B)= P(B|A)P(A)+ P(B|\overline A)P(\overline A)$
and
$P(\overline B)= P(\overline B \cap A) + P(\overline B \cap \overline A)$
$P(\overline B)= P(\overline B|A)P(A)+ P(\overline B|\overline A)P(\overline A)$

The prob. of B is the prob of B if A happens + the prob of B if a doesn't happen. Since either A either happens or it doesn't (duh).

Once again thanks I understand it now.