Answer:
[tex]P(X = x) = \frac{2^{3-x}}{27}[/tex]
Step-by-step explanation:
Given
[tex]Black = 4[/tex]
[tex]Green = 2[/tex]
[tex]r = 3[/tex] --- selected balls (with replacement)
Required
Determine the probability distribution for the green balls.
Let x be a [tex]random\ variable[/tex] such that:
[tex]x = 0, 1, 2, 3[/tex]
Because it is a selection with replacement
The probability of a green ball being drawn is always:
[tex]P(G) = \frac{Green}{Total}[/tex]
[tex]P(G) = \frac{2}{6}[/tex]
[tex]P(G) = \frac{1}{3}[/tex]
The probability of a black ball being drawn is always:
[tex]P(B) = \frac{Black}{Total}[/tex]
[tex]P(B) = \frac{4}{6}[/tex]
[tex]P(B) = \frac{2}{3}[/tex]
So, the probability distribution of green is:
[tex]P(X = x) = (P(B))^{3-x} * (P(G))^x[/tex]
[tex]P(X = x) = (\frac{2}{3})^{3-x} * (\frac{1}{3})^x[/tex]
Apply law of indices
[tex]P(X = x) = \frac{2^{3-x}}{3^{3-x}} * \frac{1^x}{3^x}[/tex]
[tex]P(X = x) = \frac{2^{3-x}*1^x}{3^{3-x}*3^x}[/tex]
[tex]P(X = x) = \frac{2^{3-x}*1}{3^{3-x+x}}[/tex]
[tex]P(X = x) = \frac{2^{3-x}*1}{3^3}[/tex]
[tex]P(X = x) = \frac{2^{3-x}}{27}[/tex]
