Given
[tex]S=\mleft\lbrace1,2,3,4,5,6,7,8,9\mright\rbrace[/tex]
by selecting two numbers from S, the number of possible outcomes is
[tex]n(S)=9\times9=81[/tex]
Let E be the event that selecting two numbers randomly and their sum is 12 with replacement.
[tex]E=\lbrack(3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3)\}[/tex][tex]n(E)=7[/tex]
The probability is
[tex]P(E)=\frac{n(E)}{n(S)}[/tex]
Substitute values, we get
h is
[tex]P(E)=\frac{7}{81}[/tex]
b)without replacement
Areplacementout
[tex]A=\mleft\lbrace(3,9\mright),(4,8),(5,7),(7,5),(8,4),(9,3)\}[/tex][tex]n(A)=6[/tex]
[tex]P(A)=\frac{n(A)}{n(S)}[/tex]
Substitute the values, we get
[tex]P(A)=\frac{6}{81}=\frac{2}{27}[/tex]
The probability that the sum is 12 if selecting two numbers without replacement
[tex]P(A)=\frac{2}{27}[/tex]
