Answer:
Explanation:
a.
let A= number 5
B=number >3
P(A)=1/6 {5}
P(B)=3/6 {4,5,6}
p(A∩B)=1/6 {5}
P(AUB)=P(A)+P(B)-P(A∩B)=1/6+3/6-1/6=3/6=1/2
b.
let event C=number<5
D=even number
C=[1,2,3,4}
D={2,4,6}
C∩D={2,4}
P(C)=4/6
P(D)=3/6
P(C∩D)=2/6
P(C∪D)=P(C)+P(D)-P(C∩D)=4/6+3/6-2/6=5/6
c.
let event E=number 2
F= an odd number.
E={2}
F={1,3,5}
E∩F={}
P(E)=1/6
P(F)=3/6
P(E∩F)=0
P(E∪F)=P(E)+P(F)-P(E∩F)=1/6+3/6-0=4/6=2/3