Answer: [tex]\dfrac{11}{36}[/tex]
Step-by-step explanation:
Total outcomes for rolling a pair of six-sided dice n(S)= 6 x 6 =36
Let A be the event that the first die is one and B be the event that the sum of the dice is eight.
A = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)}
P(A)= [tex]=\dfrac{n(A)}{n(S)}=\dfrac{6}{36}[/tex]
B= {(2,6), (6,2), (3,5), (5, 3), (4,4)}
P(B)= [tex]=\dfrac{n(B)}{n(S)}=\dfrac{5}{36}[/tex]
A ∩ B = Ф
P(A ∩ B)=0
P(A or B)= P(A)+P(B)+P(A∩ B)
[tex]=\dfrac{6}{36}+\dfrac{5}{36}+0\\=\dfrac{11}{36}[/tex]
Hence, P(A or B) = [tex]\dfrac{11}{36}[/tex]