Solve Probability Problem: Find P(B) & P(A/B)

  • Thread starter Thread starter chwala
  • Start date Start date
  • Tags Tags
    Probability
AI Thread Summary
The discussion revolves around solving a probability problem with given values: P(A/B) = 0.8, P(B/A') = 0.6, and P(A) = 0.75. The user is attempting to find P(B) and P(A/B) but is stuck after calculating P(A ∩ B) as 0.6. Participants suggest calculating P(A') and P(A' ∩ B) to combine these probabilities for a solution. The conversation emphasizes the need to clarify the given values and the required outcomes to effectively solve the problem.
chwala
Gold Member
Messages
2,827
Reaction score
415

Homework Statement

## The question is as follows given ##P(A/B)=0.8, P(B/A")=0.6## and P(A)=0.75 then find P(B) and ## P(A/B)##[/B]

Homework Equations

The Attempt at a Solution


##P(A/B)= P(AnB)/P(B) ## where## P(AnB)= P(B/A).P(A)= 0.8*0.75=0.6##
##P(A/B)= 0.6/P(B)## am now stuck here how do i move from here?
 
Physics news on Phys.org
chwala said:

Homework Statement

[/B]
Given ##P(A/B)=0.8, P(B/A')=0.6## and ##P(A)=0.75## then find ##P(B)## and ## P(A/B)##

Homework Equations

The Attempt at a Solution


##P(A/B)= P(A\land B)/P(B) ## where ## P(A\land B)= P(B/A)P(A)= 0.8*0.75=0.6##
##P(A/B)= 0.6/P(B)## am now stuck here how do i move from here?

looks like you're nearly there... calculate ## p(A') ##, then can you find ## p(A' \land B)## ?
 
If you are given P(A \mid B) why do you need to calculate it?

To be clear: are these the items you are given and the items you are to find?
<br /> \begin{align*}<br /> P(A \mid B) &amp; = 0.8 \\<br /> P(B \mid A&#039;) &amp; = 0.6 \\<br /> P(A) &amp; = 0.75 \\<br /> \text{Need} &amp; \\<br /> P(B) &amp; \\<br /> P(A \cap B) \text{ (instead of } &amp; P(A \mid B) \text{?)}<br /> \end{align*}<br />
 
statdad said:
If you are given P(A \mid B) why do you need to calculate it?

To be clear: are these the items you are given and the items you are to find?
<br /> \begin{align*}<br /> P(A \mid B) &amp; = 0.8 \\<br /> P(B \mid A&#039;) &amp; = 0.6 \\<br /> P(A) &amp; = 0.75 \\<br /> \text{Need} &amp; \\<br /> P(B) &amp; \\<br /> P(A \cap B) \text{ (instead of } &amp; P(A \mid B) \text{?)}<br /> \end{align*}<br />
sorry the terms given are P(B/A)= 0.8 , P(B/A")=0.6 and P(A)=0.75 I need to find P(A/B) and P(B)
 
Well, if you have P(B \mid A) and P(A) you can calculate P(A \cap B) . You can also find P(A&#039;) and
then get P(A&#039; \cap B). What will P(A \cap B) and P(A&#039; \cap B) together get you? Once you have that you can finish the questions.
 
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