Need help finding P ( A ∩ B ): Suppose P(A)=P(A)=0.25 and P(B)=P(B)=0.62...

[FONT=&quot]Suppose [/FONT][FONT=&quot][/FONT]P(A)=P(A)=[FONT=&quot]0.25 and [/FONT][FONT=&quot][/FONT]P(B)=P(B)=[FONT=&quot]0.62. If [/FONT][FONT=&quot][/FONT]P(A∣B)=P(A∣B)=[FONT=&quot] 0.323, what is [/FONT][FONT=&quot][/FONT]P(A∩B)?P(A∩B)?[FONT=&quot]

[/FONT]^^ This is the question I'm working on. I must be overlooking something simple. How do I work this problem out?

northwind87 said:

Suppose P(A)=P(A)=0.25 and P(B)=P(B)=0.62. If P(A∣B)=P(A∣B)= 0.323, what is P(A∩B)?P(A∩B)?

^^ This is the question I'm working on. I must be overlooking something simple. How do I work this problem out?

Click to expand...

P(A|B) = P(A AND B)/P(B)?

Is that ALL the information we have? Are A and B mutually exclusive?

@tkhunny: \(\displaystyle P(A \cap B) \) means the same thing as \(\displaystyle P(A\,\mathtt{AND}\,B)\), and in general it is given by:

\(\displaystyle \displaystyle P(A \cap B) = P(A|B)P(B) = P(B|A)P(A)\)

Thus, the OP has all the information needed to solve the problem. A few other points:

• The events cannot be mutually-exclusive, because then \(\displaystyle P(A \cap B) = 0 \), since \(\displaystyle P(A|B) = 0 \) in that instance.
• The events cannot be independent, because then \(\displaystyle P(A|B) = P(A)\)

j-astron said:

@tkhunny: \(\displaystyle P(A \cap B) \) means the same thing as \(\displaystyle P(A\,\mathtt{AND}\,B)\), and in general it is given by:

\(\displaystyle \displaystyle P(A \cap B) = P(A|B)P(B) = P(B|A)P(A)\)

Thus, the OP has all the information needed to solve the problem. A few other points:

• The events cannot be mutually-exclusive, because then \(\displaystyle P(A \cap B) = 0 \), since \(\displaystyle P(A|B) = 0 \) in that instance.
• The events cannot be independent, because then \(\displaystyle P(A|B) = P(A)\)

Click to expand...

Whoops! Right you are. Repaired above.

northwind87 said:

Suppose P(A)=P(A)=0.25 WHY THE REPETITION OF P(A)?

and P(B)=P(B)=0.62. WHY THE REPETITION OF P(B)?

If P(A∣B)=P(A∣B)= 0.323, WHY THE REPETITION OF P( A | B)?

what is P(A∩B)?P(A∩B)? WHY THIS REPETITION?

^^ This is the question I'm working on. I must be overlooking something simple. How do I work this problem out?

Click to expand...

Why all the repititions? They make no sense.

To work with conditional probabilities, you need one BASIC definition and two BASIC theorems.

\(\displaystyle P( \lambda ) > 0 \implies P( \kappa \ | \ \lambda) \equiv \dfrac{P( \lambda \ \bigcap \ \kappa)}{P(\lambda)}.\)

That is the definition. One resulting theorem is:

\(\displaystyle P( \lambda ) > 0 \implies P( \lambda \ \bigcap \ \kappa ) = P( \kappa \ | \ \lambda ) * P ( \lambda ).\)

From that theorem, you get:

\(\displaystyle P( \lambda ) > 0 < P( \kappa ) \implies P( \kappa \ | \ \lambda ) * P( \lambda ) = P( \lambda \ \bigcap \ \kappa ) = P( \lambda \ | \ \kappa ) * P( \kappa ).\)

How does that apply to your problem?

JeffM said:

Why all the repetitions? They make no sense.

Click to expand...

The repetitions are the result of pasting from certain web sites that provide both a formatted and an unformatted version of each piece of math. When one pastes such material, they should delete the first copy of each, as a service to the reader. Perhaps it isn't visible until you do a preview, though I doubt that.

But this is better than pasting from other sites that have their math purely as an image, so all the important stuff just doesn't copy, and the question is useless.

Dr.Peterson said:

The repetitions are the result of pasting from certain web sites that provide both a formatted and an unformatted version of each piece of math. When one pastes such material, they should delete the first copy of each, as a service to the reader. Perhaps it isn't visible until you do a preview, though I doubt that.

But this is better than pasting from other sites that have their math purely as an image, so all the important stuff just doesn't copy, and the question is useless.

Click to expand...

Ahh, thanks. Glad to know it is a program doing the silliness.