# Thread: Conditional probability and independence

1. ## Conditional probability and independence

A product can have three kinds of defects: A, B, C. It is rejected if$\displaystyle (A \cap B) \cup C$.
It is known that $\displaystyle P(A) = 0.15 , P(B) = 0.25 , P(C) = 0.30$.
It is also known that $\displaystyle P(A|C) = 1.3\cdot P(A)$ and $\displaystyle P(B|C) = 2\cdot P(B)$.
Also A and B are independent.

What is $\displaystyle P[(A \cap B) \cup C]$ ?

Some one can help me with this?

2. Originally Posted by johanS
A product can have three kinds of defects: A, B, C. It is rejected if$\displaystyle (A \cap B) \cup C$.
It is known that $\displaystyle P(A) = 0.15 , P(B) = 0.25 , P(C) = 0.30$.
It is also known that $\displaystyle P(A|C) = 1.3\cdot P(A)$ and $\displaystyle P(B|C) = 2\cdot P(B)$.
Also A and B are independent.

What is $\displaystyle P[(A \cap B) \cup C]$ ?

Some one can help me with this?

I don't see it right now but to get you started, you can find $\displaystyle (A \cap B)$ easily because they are independent. So it is just $\displaystyle P(A)P(B)$. You know $\displaystyle P(C)$ So you are left to find $\displaystyle P(A\cap B\cap C)$

Using the conditional probabilities you can find $\displaystyle P(A\cap C)$ and $\displaystyle P(B\cap C)$

Try to mess with these and see what you can do.

3. Originally Posted by jass10816
I don't see it right now but to get you started, you can find $\displaystyle (A \cap B)$ easily because they are independent. So it is just $\displaystyle P(A)P(B)$. You know $\displaystyle P(C)$ So you are left to find $\displaystyle P(A\cap B\cap C)$

Using the conditional probabilities you can find $\displaystyle P(A\cap C)$ and $\displaystyle P(B\cap C)$

Try to mess with these and see what you can do.
$\displaystyle P(A\cap B\cap C)$ is exactly my problem