Answer:
The change to the face 3 affects the value of P(Odd Number)
Step-by-step explanation:
Analysing the question one statement at a time.
Before the face with 3 is loaded to be twice likely to come up.
The sample space is:
[tex]S = \{1,1,2,2,2,3\}[/tex]
And the probability of each is:
[tex]P(1) = \frac{n(1)}{n(s)}[/tex]
[tex]P(1) = \frac{2}{6}[/tex]
[tex]P(1) = \frac{1}{3}[/tex]
[tex]P(2) = \frac{n(2)}{n(s)}[/tex]
[tex]P(2) = \frac{3}{6}[/tex]
[tex]P(2) = \frac{1}{2}[/tex]
[tex]P(3) = \frac{n(3)}{n(s)}[/tex]
[tex]P(3) = \frac{1}{6}[/tex]
P(Odd Number) is then calculated as:
[tex]P(Odd\ Number) = P(1) + P(3)[/tex]
[tex]P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}[/tex]
Take LCM
[tex]P(Odd\ Number) = \frac{2+1}{6}[/tex]
[tex]P(Odd\ Number) = \frac{3}{6}[/tex]
[tex]P(Odd\ Number) = \frac{1}{2}[/tex]
After the face with 3 is loaded to be twice likely to come up.
The sample space becomes:
[tex]S = \{1,1,2,2,2,3,3\}[/tex]
The probability of each is:
[tex]P(1) = \frac{n(1)}{n(s)}[/tex]
[tex]P(1) = \frac{2}{7}[/tex]
[tex]P(2) = \frac{n(2)}{n(s)}[/tex]
[tex]P(2) = \frac{3}{7}[/tex]
[tex]P(3) = \frac{n(3)}{n(s)}[/tex]
[tex]P(3) = \frac{1}{7}[/tex]
[tex]P(Odd\ Number) = P(1) + P(3)[/tex]
[tex]P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}[/tex]
Take LCM
[tex]P(Odd\ Number) = \frac{2+1}{7}[/tex]
[tex]P(Odd\ Number) = \frac{3}{7}[/tex]
Comparing P(Odd Number) before and after
[tex]P(Odd\ Number) = \frac{1}{2}[/tex] --- Before
[tex]P(Odd\ Number) = \frac{3}{7}[/tex] --- After
We can conclude that the change to the face 3 affects the value of P(Odd Number)
