Answer:
0.27
Step-by-step explanation:
To solve this problem, we will use Set Theory Approach to Probability.
Let U = the Sample Space which consists of all possible tests.
Of course U = {1,2,3 .... ,30}
So, no elements in that.
U = n = 30
Let Q = the event bearing a multiple of 4 No winning ticket, and,
R = the case that has the No. 19 winning ticket
∴
Q = { 4, 8, 12 , 16 , 20, 24 , 28} , and
R = {19} , so that
Q ∩ R = ϕ
P(Q ∩R ) = 0
As, n(Q) = 7 , n(R) = 1 , we have,
P (Q) = [tex]\frac{n (A)}{n(U)}[/tex] = [tex]\frac{7}{30}[/tex], and P (R) = [tex]\frac{1}{30\\}[/tex]
P (Q∪R) = P(Q ) + P(R) − P(A∩B) = [tex]\frac{7}{30}[/tex] + [tex]\frac{1}{30\\}[/tex] − 0 = [tex]\frac{8\\}{30\\}[/tex]
≅ 0.27 .