Help Understanding Conditional Probability

• Oct 26th 2008, 11:33 AM
mlynch18
Help Understanding Conditional Probability
I am in a finite mathematics class and have been given the following Pr[A|B] = 3/4 Pr[B|A] = 3/8 and Pr[A u B] = 1
I have been asked to find Pr[A] Pr[B] and Pr[A n B]
the book has provided the formula:

Pr[A|B] = Pr[A n B] Pr[B|A] = Pr[B n A]
Pr[B] Pr[A]

I am having difficulty understanding where to begin with this. Thanks.
• Oct 26th 2008, 02:13 PM
batman
Note that Pr[A u B] = Pr[A] + Pr[B] - Pr[A n B].
• Oct 26th 2008, 02:23 PM
mlynch18
Quote:

Originally Posted by batman
Note that Pr[A u B] = Pr[A] + Pr[b] - Pr[A n B].

Batman,
Thanks for the reply. Given Pr[A|B] = 3/4 Pr[B|A] = 3/8 and Pr[A u B] = 1. How do I arrive at Pr[A] or Pr[B]?
• Oct 26th 2008, 08:44 PM
mr fantastic
Quote:

Originally Posted by mlynch18
I am in a finite mathematics class and have been given the following Pr[A|B] = 3/4 Pr[B|A] = 3/8 and Pr[A u B] = 1
I have been asked to find Pr[A] Pr[b] and Pr[A n B]
the book has provided the formula:

Pr[A|B] = Pr[A n B] Pr[B|A] = Pr[B n A]
Pr[b] Pr[A]

I am having difficulty understanding where to begin with this. Thanks.

$\displaystyle \Pr(A | B) ~ \Pr(B) = \Pr(A \cap B) \Rightarrow \frac{3}{4} ~ \Pr(B) = \Pr(A \cap B)$ .... (1)

$\displaystyle \Pr(B | A) ~ \Pr(A) = \Pr(A \cap B) \Rightarrow \frac{3}{8} ~ \Pr(A) = \Pr(A \cap B)$ .... (2)

Equate equations (1) and (2): $\displaystyle \frac{3}{4} ~ \Pr(B) = \frac{3}{8} ~ \Pr(A) \Rightarrow \Pr(A) = 2 \Pr(B)$ .... (3)

$\displaystyle \Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$

Substitute from equations (3) and (1):

$\displaystyle \Rightarrow 1 = 2 \Pr(B) + \Pr(B) - \frac{3}{4} \Pr(B)$.

Solve for Pr(B). Hence solve for Pr(A) and $\displaystyle \Pr(A \cap B)$.