- #1
k77i
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Homework Statement
A box contains one yellow, two red, and three green balls. Two balls are randomly chosen without replacement. Define the following events:
A: \{ One of the balls is yellow \}
B: \{ At least one ball is red \}
C: \{ Both balls are green \}
D: \{ Both balls are of the same color \}
Find the following conditional probabilities:
(a) P(A|B) =
(b) P(B|D) =
(c) P(D|C^c) =
Homework Equations
P(A U B) = P(A)P(A|B)
P(A intersect B) = P(A)P(B)
The Attempt at a Solution
Using 6C2, i.e. 6!/[(2!)(4!)], I started by finding the probability of each event:
p(A)= 5/15 = 1/3
p(B)= 9/15 = 3/5
p(C)= 3/15 = 1/5
p(D)= 4/15
I found the answer for a) P(A|B) = P(A U B)/P(B) = 1/3
*I just manipulated the equation P(A U B) = P(A)P(A|B) here
This didn't work for b) P(B|D) I'm not sure why..
As for c) P(D|C^c), I'm not entirely sure how to work compliments into the equations.