Find P(B): Solving Probability Formula 1

AI Thread Summary
To find P(B) given P(AnB)=0.6 and P(A|B)=0.10, the formula P(A|B)=P(AnB)/P(B) is used. When substituting the values, the calculation yields P(B)=0.6/0.1=6, which is invalid since probabilities cannot exceed 1. The discussion suggests verifying the problem's original parameters to ensure accuracy. The confusion arises from the apparent contradiction between the calculated value and the fundamental properties of probability.
e3027
Messages
4
Reaction score
0
1. Given P(AnB)=0.6 and P(A|B)=0.10, find P(B)



2. P(A|B)=P(AnB)/P(B)




3. I plugged in the numbers in the formula but keep getting that P(B) is 6... which is wrong since P cannot be greater than 0. Am I doing something wrong?
 
Physics news on Phys.org
Instead of plugging in numbers, solve this equation algebraically for P(B):
P(A|B)=P(AnB)/P(B)

When you have P(B) isolated, substitute the values you have for the two other quantities.
 
That's what I did.

P(B)=P(AnB)/P(A|B)=0.6/0.1=6...but this makes no sense to me.
 
I don't see anything wrong in your work, although obviously a probability can't be larger than 1. Are you sure you have the problem written the same as it was given to you?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top