Answer:
[tex]P(x > 1\ and\ x < 2) = \frac{4}{25}[/tex]
Step-by-step explanation:
Given
[tex]S = \{1,2,3,4,5\}[/tex]
[tex]n(S) = 5[/tex]
Required
[tex]P(x > 1\ and\ x < 2)[/tex]
[tex]P(x > 1\ and\ x < 2)[/tex] is calculated as:
[tex]P(x > 1\ and\ x < 2) = P(x > 1) * P(x < 2)[/tex]
Since it is a probability with replacement, we have:
[tex]P(x > 1\ and\ x < 2) = \frac{n(x > 1)}{n(S)} * \frac{n(x < 2)}{n(S)}[/tex]
For x > 1, we have:
[tex]x > 1 = \{2,3,4,5\}\\[/tex]
[tex]n(x > 1) = 4[/tex]
For x < 2, we have:
[tex]x < 2 = \{1\}[/tex]
[tex]n(x < 2) = 1[/tex]
[tex]P(x > 1\ and\ x < 2) = \frac{n(x > 1)}{n(S)} * \frac{n(x < 2)}{n(S)}[/tex]
becomes
[tex]P(x > 1\ and\ x < 2) = \frac{4}{5} * \frac{1}{5}[/tex]
[tex]P(x > 1\ and\ x < 2) = \frac{4}{25}[/tex]