91Ó°ÊÓ

Experiment: Seating five people - $A,B,C,D,E$in a line.

Outcome space $S=\left\{\left(x_1,x_2,x_3,x_4,x_5\right):x_i∈\{A,B,C,D,E\},x_i≠x_j\right\}$

Probability is equal for each event $\left\{x\right\}∈S$

$P\left(X\right)=\frac{\#\text{elements in}X}{\#\text{elements in}S};X⊆S$

The probability that one person is between $A$and $B$?

There are 3 different people that can be between $A$and $B$, and that can be in order $AXB$and $BXA$

And if we consider that triplet one element there are $3!$permutations.

The wanted probability is:

localid="1649670543539" $P\left(X\right)=\frac{\#\text{elements in}X}{\#\text{elements in}S}=\frac{3·2·3!}{5!}=0.3$

Experiment: Seating of five people - $A,B,C,D,E$ in a line.

Outcome space $S=\left\{\left(x_1,x_2,x_3,x_4,x_5\right):x_i∈\{A,B,C,D,E\},x_i≠x_j\right\}$

Probability is equal for each event $\left\{x\right\}∈S$

$P\left(X\right)=\frac{\#\text{elements in}X}{\#\text{elements in}S};X⊆S$

Probability that two people are between $A$and $B$?

There are $3·2$different people that can be between $A$and $B$, and $A$and $B$can be in order $A…B$and $B…A$

And if we consider that four letters one element there are $2!$permutations (the block from $A$to $B$and the remaining letter.

The wanted probability is:

$P\left(X\right)=\frac{\#\text{elements in}X}{\#\text{elements in}S}=\frac{3·2·2·2!}{5!}=0.2$

Experiment: Seating of five people - $A,B,C,D,E$in a line.

Outcome space $S=\left\{\left(x_1,x_2,x_3,x_4,x_5\right):x_i∈\{A,B,C,D,E\},x_i≠x_j\right\}$

Probability is equal for each event $\left\{x\right\}∈S$

$P\left(X\right)=\frac{\#\text{elements in}X}{\#\text{elements in}S};X⊆S$

Probability that three people are between $A$and $B$?

There are $3·2·1$different people that can be between $A$and $B$, and $A$and $B$can be in order $A…B$and $B…A$

There are no remaining letters to arrange. The wanted probability is:

$P\left(X\right)=\frac{\#\text{elements in}X}{\#\text{elements in}S}=\frac{3·2·1·2}{5!}=0.1$