# Easy Probability

• Sep 10th 2010, 10:25 AM
Mr Rayon
Easy Probability
If Pr(A) = 4 Pr(B), $Pr(A\cup B) =0.8$ and $Pr(A\cap B) = 0.2$, find:
a) Pr(B)
b) Pr(A)

The bit that confuses me most was probably "Pr(A) = 4 Pr(B)" in the question above. Is this a typo? I've tried figuring this out for few times but I keep getting negative numbers.

Any help would be appreciated.
• Sep 10th 2010, 10:28 AM
Plato
Recall that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
• Sep 10th 2010, 10:32 AM
Mr Rayon
Yeah, I used that formula, just rearranged but kept getting negative numbers, ignoring Pr(B) in the question and assuming it was a typo.
• Sep 10th 2010, 10:42 AM
Plato
$0.8 = 5P(B) - 0.2\, \Rightarrow \,5P(B) = 1$

Where is there a negative?
• Sep 10th 2010, 05:55 PM
Soroban
Hello, Mr Rayon!

Quote:

$\text{If }\text{Pr}(A) = 4\!\cdot\!\text{Pr}(B),\;\;\text{Pr}(A\cup B) = 0.8,\;\;\text{Pr}(A\cap B) = 0.2$

$\text{Find: }\,\text{Pr}(A)\,\text{ and }\,\text{Pr}(B)$

That first equation says: . $\text{Pr}(A)$ is four times $\text{Pr}(B).$

Formula: . $\text{Pr}(A \cup B) \;=\;\text{Pr}(A) + \text{Pr}(B) - \text{Pr}(A \cap B)$

Substitute: . . . $0.8 \quad\;= \;4\!\cdot\!\text{Pr}(B) + \text{Pr}(B) \;\;- \;\; 0.2$

So we have: . $1.0 \;=\;5\!\cdot\!\text{Pr}(B) \quad\Rightarrow\quad \boxed{\text{Pr}(B) \,=\,0.2}$

Then: . $\text{Pr}(A) \:=\:4\!\cdot\!\text{Pr}(B) \;=\;4(0.2) \quad\Rightarrow\quad \boxed{\text{Pr}(A)\;=\;0.8}$