Answer:
a) 10
b) [tex]\frac{4}{10}[/tex]
c) [tex]\frac{6}{10}[/tex]
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
[tex]5C2 = 10[/tex]
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
[tex]4C1 = 4[/tex]
There are 4 possibilities where Jack will be selected from a total of 10 possibilities (as calculated in part a. )
[tex]\frac{4}{10}[/tex] is the total probability of selection.
c)
To select Jack or Jane, means having one seat fixed for either of them.
[tex]2C1 = 2[/tex]
To select any one the other three members, the possibilities are:
[tex]3C1 = 3[/tex]
In the rules of probability, and = multiplication; or = addition.
Here, we are considering an "AND" situation.
[tex]2[/tex] × [tex]3[/tex] [tex]= 6[/tex] 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 [tex]\frac{6}{10}[/tex]
d)
Here we use Permutation since order does matter.
There are 5 people and 2 are to be selected.
n = 5 ; r = 2
[tex]5P2 = 20[/tex]
Formula:
Combination: [tex]\frac{n!}{(n-k)! k!}[/tex]
Permutation: [tex]\frac{n!}{(n-k)!}[/tex]
