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Let be the set of permutations of whose first term is a prime. If we choose a permutation at random from , what is the probability that the third term is equal to

Sagot :

MrRoyal

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]

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