# Is an answer possible with the given information?

• Sep 15th 2009, 10:59 AM
bartoncl
Is an answer possible with the given information?
P(A U B) = .7 P(A U compliment of B) = .9

Find P(A)

I've racked my brain and cannot come up with an answer.

This is all I've managed to calculate, perhaps I'm just missing a connection.

P(A U compliment of B) = P(compliment of B)
P(B) = 1 - P(compliment of B) = .1
P(compliment of A intersect compliment of B) = 1 - P(A U B) = .3
P(A intersect compliment of B) = P(A U B) - P(B) = .6
P(A|compliment of B) =
P(A intersect compliment of B) / P(compliment of B) = 2/3
"This last one establishes dependence."

That is all I could come up with.
• Sep 15th 2009, 11:42 AM
Opalg
Quote:

Originally Posted by bartoncl
P(A U B) = .7 P(A U compliment of B) = .9

Find P(A)

I've racked my brain and cannot come up with an answer.

This is all I've managed to calculate, perhaps I'm just missing a connection.

P(A U compliment of B) = P(compliment of B)
P(B) = 1 - P(compliment of B) = .1
P(compliment of A intersect compliment of B) = 1 - P(A U B) = .3
P(A intersect compliment of B) = P(A U B) - P(B) = .6
P(A|compliment of B) =
P(A intersect compliment of B) / P(compliment of B) = 2/3
"This last one establishes dependence."

That is all I could come up with.

[Note on spelling. Compliment is when you say something flattering about someone. Complement is "that which completes or makes up the whole".]

I'll write $\displaystyle A^c,\ B^c$ for the complements of A and B.

$\displaystyle P(A^c\cap B^c) = 1 - P(A\cup B) = 1 - .7 = .3$ (you already got that one). In the same way,
$\displaystyle P(A^c\cap B) = 1 - P(A\cup B^c) = 1 - .9 = .1$. Then $\displaystyle P(A^c) = P(A^c\cap B) + P(A^c\cap B^c) = .3 + .1 = .4$. Finally, $\displaystyle P(A) = 1- P(A^c) = .6$.
• Sep 15th 2009, 11:46 AM
Soroban
Hello, bartoncl!

Quote:

$\displaystyle P(A \cup B) \:=\: 0.7 \qquad P(A \cup B') \:=\: 0.9$

Find $\displaystyle P(A).$

Draw a two-circle Venn diagram . . . it will help.

We are given: .$\displaystyle P(A \cup B') \:=\:0.9$

Its complement is: .$\displaystyle P(A \cup B')' \:=\:1 - 0.9 \quad\Rightarrow\quad P(A' \cap B) \:=\:0.1$

$\displaystyle \text{Formula: }\;P(A \cup B) \;=\;P(A) + \underbrace{P(B) - P(A \cap B)}_{\text{This is }P(A'\,\cap\,B)}$

$\displaystyle \text{So we have: }\;\underbrace{P(A \cup B)}_{0.7} \;=\;P(A) + \underbrace{P(A' \cap B)}_{0.1}$

Therefore: .$\displaystyle 0.7 \:=\:P(A) + 0.1 \quad\Rightarrow\quad \boxed{P(A) \:=\:0.6}$