# Conditional Probability & Independence

1. Jan 30, 2010

### exitwound

1. The problem statement, all variables and given/known data

One satellite is scheduled to be launched form Cape Canaveral in Florida, and another launching is scheduled for Vandenberg AFB in California. Let A denote the event that the Vandenberg launch goes off on schedule, and let B represent the event that the Cape Canaveral launch goes off on schedule. If A and B are independent events with P(A)>P(B) and P(A u B) = .626, P(A ∩ B) = .144, determine the values of P(A) and P(B).

3. The attempt at a solution

This is a practice problem. I have the answers of .45 & .32. I have no idea how to get them.

I know that if P(A|B) = P(A), then the two probabilities are independent.

I also know that P(A|B)P(B) = P(A ∩ B) .

I don't know how to do this problem.

2. Jan 30, 2010

### rock.freak667

Well then you know that P(A|B)P(B) = P(A ∩ B) simplifies to P(A∩B)=P(A)P(B).

P(A u B) should be the same as P(A)+P(B)

3. Jan 30, 2010

### vela

Staff Emeritus
That's not correct. You'd be double-counting when A and B occur simultaneously. It should be

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$.

4. Jan 30, 2010

### exitwound

I actually got those two equations:

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
and
$P(A\cap B)=P(A)P(B)$

I end up with two equations with two unknowns, but I can't come up with any answers that work.

5. Jan 30, 2010

### vela

Staff Emeritus
Solve for P(A) in the second equation and plug it into the first, then multiply through by P(B). You'll end up with a quadratic equation.

6. Jan 30, 2010

### exitwound

Yup. And it comes out with no solutions. I've tried like 10 times.

7. Jan 30, 2010

### vela

Staff Emeritus
You must be making an algebra error somewhere. I did the same thing and got the expected answer. Why don't you post your work so we can see where you might have gone astray?

8. Jan 30, 2010

### exitwound

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
and
$P(A\cap B)=P(A)P(B)$

$$P(A)=\frac{P(A\cap B)}{P(B)}$$

$$P(A \cup B)=\frac{P(A\cap B)}{P(B)}+P(B)-P(A \cap B)$$

$$P(A \cup B)=P(A\cap B)+P(B)^2-P(A \cap B)P(B)$$

$.626=.144+P(B)^2-.144P(B)$

$P(B)^2-.144P(B)-.770=0$

9. Jan 30, 2010

### vela

Staff Emeritus
You forgot to multiply the lefthand side by P(B).

10. Jan 30, 2010

### exitwound

$$P(B)P(A \cup B)=P(A\cap B)+P(B)^2-P(A \cap B)P(B)$$

$$.626P(B)=.144+P(B)^2-.144P(B)$$
$$0 = .144+P(B)^2-.770P(B)$$

Still no solutions.

11. Jan 31, 2010

### vela

Staff Emeritus
Well, your equation is correct, and it has the aforementioned solutions. All I can suggest is you check your work again carefully.

12. Jan 31, 2010

### exitwound

Oh. I'm a doofus. The coefficient of P(B)^2 is 1, not 0. Thanks for the help.