SOLUTION: 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:

Question 1146318: 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(D|B^c)=
(c)P(C|D)=

Answer by VFBundy(438) (Show Source): You can put this solution on YOUR website!
A:{One of the balls is yellow}:

Probability the first ball is yellow:
= 1/6

Probability the second ball is yellow:
= 5/6 * 1/5
= 5/30
= 1/6

Probability one of the balls is yellow:
=1/6 + 1/6
= 1/3

B:{At least one ball is red}:

Probability NO balls are red:
= 4/6 * 3/5
= 12/30
= 2/5

Probability at least one ball is red:
= 1 - 2/5
= 3/5

C:{Both balls are green}:

= 3/6 * 2/5
= 6/30
= 1/5

D:{Both balls are of the same color}:

Probability both balls are yellow:
= 0 (because there is only one yellow ball)

Probability both balls are red:
= 2/6 * 1/5
= 2/30
= 1/15

Probability both balls are green:
= 3/6 * 2/5
= 6/30
= 1/5

Probability both balls are the same color:
= 0 + 1/15 + 1/5
= 1/15 + 3/15
= 4/15

Find the following conditional probabilities:

(a) P(A|B)=

Probability at least one ball is yellow (A) given that at least one ball is red (B):

P(A|B) = P(A*B)/P(B)

Probability one ball is yellow AND at least one ball is red (P(A*B)):

Probability first ball is yellow and second ball is red:
= 1/6 * 2/5
= 2/30
= 1/15

Probability first ball is red and second ball is yellow:
= 2/6 * 1/5
= 2/30
= 1/15

Probability one ball is yellow AND at least one ball is red (P(A*B)):
= 1/15 + 1/15
= 2/15

We know P(B) is 3/5.

P(A|B)
= P(A*B)/P(B)
= (2/15)/(3/5)
= 10/45
= 2/9

P(D|B^c)=

Probability both balls are the same color (D) given that NOT at least one ball is red (B^c)...or: Probability both balls are the same color (D) given that zero balls are red (B^c)

P(D|B^c) = P(D*B^c)/P(B^c)

Probability both balls are the same color AND zero balls are red (P(D*B^c)):

Probability both balls are green:
= 3/6 * 2/5
= 6/30
= 1/5

This is the only scenario that works. There is only one yellow ball, so both balls can't be yellow. Also, obviously, both balls can't be red AND zero balls are red.

So...probability both balls are the same color AND zero balls are red (P(D*B^c)):
= 1/5

Since we know P(B) is 3/5, that means P(B^c) is 2/5.

P(D|B^c)
= P(D*B^c)/P(B^c)
= (1/5)/(2/5)
= 5/10
= 1/2

P(C|D)=

Probability both balls are green (C) given that both balls are the same color (D):

P(C|D) = P(C*D)/P(D)

Probability both balls are green AND both balls are the same color (P(C*D)):

This is the same as asking the probability that both balls are green, since if both balls are green, they are obviously the same color.

From earlier, we know the probability that both balls are green is 1/5.

So...probability both balls are green AND both balls are the same color (P(C*D)):
= 1/5

We know P(D) is 4/15.

P(C|D)
= P(C*D)/P(D)
= (1/5)/(4/15)
= 15/20
= 3/4
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