Conditional probability marble question

[dragon513]

Q. A box contains three blue marbles, five red marbles, and four white marbles. If one marble is drawn at random, find:
a) P(blue|not white)
b) P(not red|not white)

The answer for both a) and b) is 3/8.
However right now I don't even understand the question.

part a) wants possibility of getting a blue ball given that white wasn't drawn.
So... NOT getting a white ball is 8/12.
Given that, getting a blue ball is 3/11.

but 3/11 is not even close to 3/8!

What's wrong with my work?

Thanks in advance!

 

[Corneo]

Why not use the defination of conditional probability.
P(E|F) = \frac {P(E \cap F)} {P(F)}
Since the event of getting a blue and non white is the same as saying getting a blue. P(blue) = 3/12 and P(non white) = 8/12

P (\text{blue} | \text{nonwhite}) = \frac {P(\text{blue})}{P(\text{non-white})} = \frac {\frac {3}{12}}{\frac {8}{12}} = \frac {3}{8}

 

[HallsofIvy]

Q. A box contains three blue marbles, five red marbles, and four white marbles. If one marble is drawn at random, find:
a) P(blue|not white)
b) P(not red|not white)
The answer for both a) and b) is 3/8.
However right now I don't even understand the question.
part a) wants possibility of getting a blue ball given that white wasn't drawn.
So... NOT getting a white ball is 8/12.
Given that, getting a blue ball is 3/11.
but 3/11 is not even close to 3/8!
What's wrong with my work?
Thanks in advance!

No, "given that" getting a blue ball is NOT 3/11- you need to take away more than 1 white ball!

Since you are told that the ball drawn was not white, you can just ignore the white ones- it's as if there were no white balls. There were 3 blue, 5 red, and 4 white marbles. Ignoring the white balls, there are a total of 3+ 5= 8 marbles, 3 of which are red.

Of course, since you are ignoring the white balls, saying the ball drawn is NOT blue is exactly the same as saying it is red- the problems in a) and b) are exactly the same.