So let's suppose we have 2 red balls, 3 blue balls, and 4 green balls in a cup and we take 3 balls without replacing them.
a. What is the probability that 3 balls of the same color are chosen
b. What is the Conditional Probability of choosing 3 green balls from the sample?
So from my understanding, I think solving a is basically ((3/9)(2/8)(1/7)) + ((4/9)(3/8)(2/7))? Because we are unioning the chances of either getting all blue or all green?
b. I have no clue how to tackle this to be honest. I know that conditional is something like (A|B) where
(A|B) = the probability intersection of A and B divided by the probability of B.
That is right. At b. it is asked for probability of choosing 3 green balls given 3 balls of the same color are chosen. Let
$A$:Choosing 3 balls of the same color
$B$: Choosing three green balls.
The asked probability is $P(B|A)=\frac{P(A \cap B)}{P(A)}$
We have $P(A\cap B)=P(B)$, since $B\subset A$. Thus $P(B|A)=\frac{P(B)}{P(A)}$
I leave the rest for you.