Is the Probability Formula P(C∩G) = P(G)P(C|G) Correct?

[princejan7]

Homework Statement

My book says that
P(C∩G) = P(G)P(C|G)

but shouldn't it be

P(C∩G)= P(C)P(C|G)

?

Homework Equations

The Attempt at a Solution

 

[jbunniii]

Homework Statement

My book says that
P(C∩G) = P(G)P(C|G)

but shouldn't it be

P(C∩G)= P(C)P(C|G)

No. To see why, consider the case where $C$ and $G$ are independent. Then we expect $P(C|G) = P(C)$ and $P(C \cap G) = P(C)P(G)$. That is consistent with the formula from the book, not your proposed formula.

 

[Ray Vickson]

Homework Statement

My book says that
P(C∩G) = P(G)P(C|G)

but shouldn't it be

P(C∩G)= P(C)P(C|G)

?

Homework Equations

The Attempt at a Solution

The book is correct. The definitionof conditional probability is
$$P(C|G) \equiv \frac{P(C \cap G)}{P(G)} \text{ if } P(G) \neq 0$$
Sometimes, however, we are given $P(C|G)$ and $P(G)$; in that case we can get $P(C \cap G)$ by 'reversing' the formula above.

 

[haruspex]

My book says that
P(C∩G) = P(G)P(C|G)

but shouldn't it be

P(C∩G)= P(C)P(C|G)

Perhaps you are misreading the notation. P(C|G) means the probability of the event C given that event G occurs. Maybe you read it as the other way around?