Question

Two people are to be selected from a group of five employees: Jane, Jan, Jack, John and Joe.

Answer 1

Answer:

a) 10

b) $\frac{4}{10}$

c) $\frac{6}{10}$

d) 20

Step-by-step explanation:

Hi,

There are two ways of solving counting problems:

1. Combination: It is used when order does not matter.

2. Permutation: It is used when order does matter.

a)  

We use combination, order doesn't matter.

There are five total members, we need to select two.

n = 5 and r = 2

$5C2 = 10$

b)

We are fixing one seat for Jack. Hence only four people are left, any one of them can take the other seat.

n = 4  and r = 1

$4C1 = 4$

There are 4 possibilities where Jack will be selected from a total of 10 possibilities (as calculated in part a. )

$\frac{4}{10}$ is the total probability of selection.

c)

To select Jack or Jane, means having one seat fixed for either of them.

$2C1 = 2$

To select any one the other three members, the possibilities are:

$3C1 = 3$

In the rules of probability, and = multiplication; or = addition.

Here, we are considering an "AND" situation.

$2$ × $3$ $= 6$ are the total instances of having Jack or Jane on one seat out of a total of 10 possibilities (as calculated in part a)

The probability then changes to $\frac{6}{10}$

d)

Here we use Permutation since order does matter.

There are 5 people and 2 are to be selected.

n = 5 ; r = 2

$5P2 = 20$

Formula:

Combination: $\frac{n!}{(n-k)! k!}$

Permutation: $\frac{n!}{(n-k)!}$