Conditional Probability (The cookie Poroblem) Gentlemen, 
I have small confusion in finding donditional probability in the "Cookies Problem" describe below: 
Suppose there are two full bowls of cookies. Bowl #1 has 10 chocolate chip and 30 plain cookies, while bowl #2 has 20 of each. Our friend Fred picks a bowl at random, and then picks a cookie at random. 
If I ask, what is the probability that Fred picks a plain cookie given Bowl 1?
People say the right answer is straight forward: 
$
P(\text{Plain|Bowl1}) = \frac{30}{40}= \frac{3}{4}
$
But I'm confused, why we did not invoke the probability of choosing Bowl 1? 
I mean, why we did not use the following: 
$
P(\text{Plain|Bowl1}) = \frac{30}{40}\frac{1}{2}= \frac{3}{8}
$
please clarify, 
Thank you so much, 
 A: When we say something is “given,” we can say that we already observed it. In this case, we already observed you choosing bowl 1, so we don’t need to consider the probability of you choosing it. So if I asked you, what’s the probability that you pull the plain cookie GIVEN that you absolutely must pull from bowl 1, then you basically are acting like bowl 2 doesn’t exist. 
You could prove this with Bayes rule, but that isn’t necessary. 
A: Intuitively: It is the probability of picking the cookie given that it was bowl one.  You do not include the probability for picking that bowl because it was given that it was picked.
A: Compare the questions:
1) What is the probability of a plain cookie given Bowl 1 (is picked already)?
$$P(P|B1)=\frac{30}{40}=\frac34.$$
2) What is the probability of picking Bowl 1 and a plain cookie?
$$P(B1\cap P)=P(B1)\cdot P(P|B1)=\frac12\cdot \frac{30}{40}=\frac38.$$
3) Given a plain cookie is picked, what is the probability that it came out of Bowl 1?
$$P(B1|P)=\frac{P(B1\cap P)}{P(B1\cap P)+P(B2\cap P)}=\frac{3/8}{3/8+2/8}=\frac35.$$
A: We don't care HOW he chose bowl $1$.
We just know that he's already choosing from bowl $1$, now we're just focused on the cookies.
The probability is $\displaystyle \frac{30}{40}$. 
