Answer:
[tex]Pr = \frac{1}{6}[/tex]
Step-by-step explanation:
Given
[tex]S = \{1,2,3,4,5\}[/tex]
[tex]n = 5[/tex]
Required
Probability the third term is 3
First, we calculate the possible set.
The first must be prime (i.e. 2, 3 and 5) --- 3 numbers
[tex]2nd \to 4\ numbers[/tex]
[tex]3rd \to 3\ numbers[/tex]
[tex]4th \to 2\ numbers[/tex]
[tex]5th \to 1\ number[/tex]
So, the number of set is:
[tex]S = 3 * 4 * 3 * 2 * 1[/tex]
[tex]S = 72[/tex]
Next, the number of sets if the third term must be 2
[tex]1st \to 2[/tex] i.e. 1 or 5
[tex]2nd \to 3\ numbers[/tex] ---- i.e. remove the already selected first term and the 3rd the compulsory third term
[tex]3rd \to 1\ number[/tex] i.e. the digit 2
[tex]4th \to 2\ numbers[/tex]
[tex]5th \to 1\ number[/tex]
So
[tex]r = 2 * 3 * 1 * 2 * 1[/tex]
[tex]r = 12[/tex]
So, the probability is:
[tex]Pr = \frac{r}{S}[/tex]
[tex]Pr = \frac{12}{72}[/tex]
[tex]Pr = \frac{1}{6}[/tex]
