Q.44 - A First Course in Probability

Question

Five people, designated as $A, B, C, D, E$, are arranged in linear order. Assuming that each possible order is equally likely, what is the probability that 

(a) there is exactly one person between $A$ and $B$?

(b) there are exactly two people between $A$ and $B$?

(c) there are three people between $A$ and $B$? 

Step-by-Step Solution

Answer

a). The probability that is exactly one person between $A$ and $B$ is $0.3$.

b). The probability that exactly two people between $A$ and $B$ is $0.2$.

c).  The probability that three people between $A$ and $B$ is $0.1$

Part (a) Step 1: Given Information

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

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

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

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

Part (a) Step 2: Explanation

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 $A X B$ and $B X A$

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

The wanted probability is:

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

Part (b) Step 1: Given Information

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

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

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

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

Part (b) Step 2: Explanation

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 $A\dots B$ and $B\dots A$

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:

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

Part (c) Step 1: Given Information

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

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

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

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

Part (c) Step 2: Explanation

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 $A\dots B$ and $B\dots A$

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

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