MHB How Do I Calculate the Probability of Two Events in Email Marketing?

[tmt1]

Let say I am a company and I email 1 million customers to try to get them to make a purchase on my website, of which 50000 click the link in the email and of those 1000 makes a purchase.

We can say that clicking the link is event A, and making a purchase is event B.

What is $P(B | A)$ or the probably of B if A happens. The formula for this is $$\frac{P(A \cap B)}{P(B)}$$.

How can I figure out $P(A \cap B)$? I know the formula is $P(A) \cdot P(B)$.

I guess $P(A)$ is equal to $\frac{1000}{1000000}$ and $P(B) = \frac{50000}{1000000}$ but these multiplied is $\frac{50,000,000}{1000000000000}$ however I think the answer says that $P(A) \cdot P(B)$ would be $\frac{1000}{1000000}$

 

[HallsofIvy]

P(A) is the probability that

Let say I am a company and I email 1 million customers to try to get them to make a purchase on my website, of which 50000 click the link in the email and of those 1000 makes a purchase.

We can say that clicking the link is event A, and making a purchase is event B.

What is $P(B | A)$ or the probably of B if A happens. The formula for this is $$\frac{P(A \cap B)}{P(B)}$$.

How can I figure out $P(A \cap B)$? I know the formula is $P(A) \cdot P(B)$.

No, it is not! As you said in the immediately previous sentence $P(B|A)= \frac{P(A\cap B)}{P(B)}$ so that $P(A\cap B)= P(A|B)\cdot P(B)$ or, equivalently, $P(A\cap B)= P(B|A)\cdot P(B)$. $P(A\cap B)= P(A)\cdot P(B)$ if and only if A and B are independent- that is, if $P(A|B)= P(A)$ and $P(B|A)= P(B)$.

I guess $P(A)$ is equal to $\frac{1000}{1000000}$

Yes, that is correct.

and $P(B) = \frac{50000}{1000000}$

No, in this case, we are not told what P(B) is. We are told that of the 50000 people who clicked on the e-mail 1000 purchased. That is, $P(B|A)= \frac{1000}{50000}= \frac{1}{50}$.

[quotebut these multiplied is $\frac{50,000,000}{1000000000000}$ however I think the answer says that $P(A) \cdot P(B)$ would be $\frac{1000}{1000000}$[/QUOTE]
I'm not sure were "50,000,000" and "100000000" came from, you had said before 50000 and 1000000, but the fraction is, of course, the same. $P(A)= \frac{5000}{1000000}= \frac{5}{1000}$. And since, as above, $P(B|A)= \frac{1}{50}$, $P(A\cap B)= P(B|A)P(A \frac{5}{1000}\cdot\frac{1}{50}= frac{1}{10000}$.