Answer:
[tex]P(G_1) = \frac{5}{13}[/tex]
[tex]P(G_2) = \frac{1}{3}[/tex]
[tex]P(X \ge 1) = \frac{25}{39}[/tex]
Step-by-step explanation:
Given
[tex]G = 5[/tex]
[tex]R = 4[/tex]
[tex]B = 4[/tex]
[tex]n = 13[/tex]
Solving (a): [tex]P(G_1)[/tex]
This is calculated as:
[tex]P(G_1) = \frac{G}{n}[/tex]
[tex]P(G_1) = \frac{5}{13}[/tex]
Solving (b): [tex]P(G_2)[/tex]
This is calculated as:
[tex]P(G_2) = \frac{G - 1}{n - 1}[/tex] -- this is so because the selection is without replacement
[tex]P(G_2) = \frac{5 - 1}{13 - 1}[/tex]
[tex]P(G_2) = \frac{4}{12}[/tex]
[tex]P(G_2) = \frac{1}{3}[/tex]
Solving (c): [tex]P(X \ge 1)[/tex]
Using the complement rule, we have:
[tex]P(X \ge 1) = 1 - P(X = 0)[/tex]
To calculate [tex]P(X = 0)[/tex], we have:
[tex]G = 5[/tex] --- Green
[tex]G' = 8[/tex] ---- Not green
The probability that both selections are not green is:
[tex]P(X = 0) = P(G'_1) * P(G'_2)[/tex]
So, we have:
[tex]P(X = 0) = \frac{G'}{n} * \frac{G'-1}{n-1}[/tex]
[tex]P(X = 0) = \frac{8}{13} * \frac{8-1}{13-1}[/tex]
[tex]P(X = 0) = \frac{8}{13} * \frac{7}{12}[/tex]
Simplify
[tex]P(X = 0) = \frac{2}{13} * \frac{7}{3}[/tex]
[tex]P(X = 0) = \frac{14}{39}[/tex]
Recall that:
[tex]P(X \ge 1) = 1 - P(X = 0)[/tex]
[tex]P(X \ge 1) = 1 - \frac{14}{39}[/tex]
Take LCM
[tex]P(X \ge 1) = \frac{39 -14}{39}[/tex]
[tex]P(X \ge 1) = \frac{25}{39}[/tex]
