# Help with Bayes Thereom

• March 7th 2012, 05:49 PM
kangta27
Help with Bayes Thereom
You are given:

(i.) 0.30 = P (B|A) and ).40 = P(A|B)

(ii.) 0.25 = P((AuB)')

Determine P (AnB)
• March 7th 2012, 07:27 PM
Soroban
Re: Help with Bayes Thereom
Hello, kangta27!

I found a roundabout solution . . .

Quote:

$\text{Given: }\:\begin{Bmatrix} P(B|A) &=& 0.30 & [1] \\ P(A|B) &=& 0.40 & [2] \\ P((A\cup B)') &=& 0.25 & [3]\end{Bmatrix}$

$\text{Determine }P(A\cap B)$

$\text{From [1]: }\:P(B|A) \:=\:\dfrac{P(A\cap B)}{P(A)} \:=\:0.30 \quad\Rightarrow\quad P(A \cap B) \:=\:0.30P(A)\;\;[4]$

$\text{From [2]: }\:P(A|B) \:=\:\dfrac{P(A\cap B)}{P(B)} \:=\:0.40 \quad\Rightarrow\quad P(A\cap B) \:=\:0.40P(B)\;\;[5]$

$\text{Equate [5] and [4]: }\:0.40P(B) \:=\:0.30P(A) \quad\Rightarrow\qiad P(B) \:=\:0.75P(A)\;\;[6]$

$\text{From [3]: }\:P(A \cup B) \:=\:0.75$

$\text{Theorem: }\:P(A \cup B) \;=\; P(A) + P(B) - P(A \cap B)$

$\text{We have: }\qquad0.75 \;=\;P(A) + \underbrace{0.75P(A)}_{[6]} - \underbrace{0.30P(A)}_{[4]}$

. . . . . . . . . . $0.75 \;=\;1.45P(A)$

. . . . . . . . . $P(A) \;=\;\dfrac{0.75}{1.45} \;=\;\dfrac{15}{29}$

$\text{From [4]: }\:P(A \cap B) \:=\:\frac{3}{10}\cdot\frac{15}{29} \:=\:\frac{9}{58}$