Question

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

Answer 1

Answer:

$Pr = \frac{1}{6}$

Step-by-step explanation:

Given

$S = \{1,2,3,4,5\}$

$n = 5$

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

$2nd \to 4\ numbers$

$3rd \to 3\ numbers$

$4th \to 2\ numbers$

$5th \to 1\ number$

So, the number of set is:

$S = 3 * 4 * 3 * 2 * 1$

$S = 72$

Next, the number of sets if the third term must be 2

$1st \to 2$ i.e. 1 or 5

$2nd \to 3\ numbers$ ---- i.e. remove the already selected first term and the 3rd the compulsory third term

$3rd \to 1\ number$ i.e. the digit 2

$4th \to 2\ numbers$

$5th \to 1\ number$

So

$r = 2 * 3 * 1 * 2 * 1$

$r = 12$

So, the probability is:

$Pr = \frac{r}{S}$

$Pr = \frac{12}{72}$

$Pr = \frac{1}{6}$