Question

The question is in the picture

Answer 1

Given

`S=\mleft\lbrace1,2,3,4,5,6,7,8,9\mright\rbrace`

by selecting two numbers from S, the number of possible outcomes is

`n(S)=9*9=81`

Let E be the event that selecting two numbers randomly and their sum is 12 with replacement.

`E=\lbrack(3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3)\}`
`n(E)=7`

The probability is

`P(E)=(n(E))/(n(S))`

Substitute values, we get

h is

`P(E)=(7)/(81)`

b)without replacement

Areplacementout

`A=\mleft\lbrace(3,9\mright),(4,8),(5,7),(7,5),(8,4),(9,3)\}`
`n(A)=6`

`P(A)=(n(A))/(n(S))`

Substitute the values, we get

`P(A)=(6)/(81)=(2)/(27)`

The probability that the sum is 12 if selecting two numbers without replacement

`P(A)=(2)/(27)`