Answer:
[tex]F\ or\ G= \{5,6,7,8,9,10,11,12\}[/tex]
[tex]P(F\ or\ G) = \frac{2}{3}[/tex]
Step-by-step explanation:
Given
[tex]S= \{7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18\}[/tex]
[tex]E=\{9, 10, 11, 12, 13, 14, 15, 16\}[/tex]
[tex]F=\{5, 6, 7, 8, 9\}[/tex]
[tex]G=\{9, 10, 11, 12\}[/tex]
[tex]H=\{2, 3, 4\}.[/tex]
Solving (a): Outcomes of F or G
F or G is the list of items in F, G or F and G.
So:
[tex]F=\{5, 6, 7, 8, 9\}[/tex]
[tex]G=\{9, 10, 11, 12\}[/tex]
[tex]F\ or\ G= \{5,6,7,8,9,10,11,12\}[/tex]
Solving (b): P(F or G)
The general addition rule is:
[tex]P(F\ or\ G) = P(F) + P(G) - P(F\ and\ G)[/tex]
Where:
[tex]P(F) = \frac{n(F)}{n(S)} = \frac{5}{12}[/tex]
[tex]P(G) = \frac{n(G)}{n(S)} = \frac{4}{12}[/tex]
[tex]P(F\ and\ G) = \frac{n(F\ and\ G)}{n(S)}[/tex]
[tex]F\ and\ G = \{9\}[/tex]
So:
[tex]P(F\ and\ G) = \frac{1}{12}[/tex]
[tex]P(F\ or\ G) = P(F) + P(G) - P(F\ and\ G)[/tex]
[tex]P(F\ or\ G) = \frac{5}{12} + \frac{4}{12} - \frac{1}{12}[/tex]
[tex]P(F\ or\ G) = \frac{5+4-1}{12}[/tex]
[tex]P(F\ or\ G) = \frac{8}{12}[/tex]
[tex]P(F\ or\ G) = \frac{2}{3}[/tex]
