Question

1. If P(A)=0.1, P(B)=0.3, and P(A∪B)=0.37, then
P(A∩B)=

2. If P(A∩B)=0.36, and P(A|B)=0.4, then
  P(B)=

3.

A box contains one yellow, two red, and three green balls. Two balls are randomly chosen without replacement. Define the following events:

A:{A:{ One of the balls is yellow }}
B:{B:{ At least one ball is red }}
C:{C:{ Both balls are green }}
D:{D:{ Both balls are of the same color }}

Find the following conditional probabilities:

(a)  P(A|B)=  
(b)  P(B^c|D)=  
(c)  P(D|C^c)

Answer 1

1) P(A∩B)= P(A)+P(B)-P(A u B)=0.1+0.3-0.37=0.03

2)P(B)=P(A n B)/P(A|B)=0.36/0.4=0.9

3)a)

P(B)=P(at least one is red)=1-P(none is red)=1-(4/6)*(3/5)=3/5

P(A n B)=P(one ball is red and other yellow)=2*(2/6)*(1/5)=4/30

hence P(A|B)=P(A n B)/P(B)=(4/30)/(3/5)=2/9

b)

P(D)=P(both are red)+P(both are green)=(2/6)*(1/5)+(3/6)*(2/5)=8/30

P(Bc n D)=P(both are green)=(3/6)*(2/5)=6/30

hence P(B^c|D)=P(Bc n D)/P(D)=(6/30)/(8/30)=3/4

\c)

P(C^c)=1-P(both are green)=1-6/30=24/30=4/5

P(D n C^c)=P(both are red)=(2/6)*(1/5)=2/30

hence P(D |C^c)=P(D n C^c)/P(C^c)=(2/30)/(4/5)=1/12