Chapter 2, Problem 98SE

To determine

Find the probability that at least one of the five gets her own calculator.

Obtain the probability that at least one of the n individuals gets their own calculator.

Answer to Problem 98SE

The probability that at least one of the five gets her own calculator is 0.633.

The probability that at least one of the n individuals gets their own calculator is $\underset{\_}{1-e^{-1}}$

Explanation of Solution

Given info:

The information is based on five friends named Allison, Beth, Carol, Diane and Evelyn who has identical calculator. They placed the calculator together for a study break and after the break each pick one calculator at random.

Calculation:

Define the event as given below:

$\begin{array}{l}A:\text{ The event Allison gets her own calculator}\text{.}\\ B:\text{ The event Beth gets her own calculator}\text{.}\\ C:\text{ The event Carol gets her own calculator}\text{.}\\ D:\text{ The event Diane gets her own calculator}\text{.}\\ E:\text{ The event Evelyn gets her own calculator}\end{array}$

Since, each student has an equal chance to pick a calculator, the events are equally likely.

The probabilities of the corresponding events are given below:

$P\left(A\right)=\frac{1}{5}$, $P\left(B\right)=\frac{1}{5}$,$P\left(C\right)=\frac{1}{5}$, $P\left(D\right)=\frac{1}{5}$, and $P\left(E\right)=\frac{1}{5}$,

The different cases where two students pick their own calculator:

The number of occurrence that A and B get her own calculatoris $\left\{ABCDE,ABCED,ABDCE,ABDEC,ABEDC,ABECD\right\}$.

The number of occurrence that A andC her own calculator is $\left\{ACBDE,ACBED,ACDBE,ACDEB,ACEDB,ACEBD\right\}$.

The number of occurrence that A andD her own calculator is $\left\{ADBCE,ADBEC,ADCBE,ADCEB,ADECB,ADEBC\right\}$.

The number of occurrence that A andE her own calculator is $\left\{AEDBC,AEBDC,AEDCB,AECDB,AECBD,AEBCD\right\}$

Similarly, the other combinations where two students pick their own calculator can be obtained.

The different cases where three students pick their own calculator:

The number of occurrence that A andB and C her own calculator is $\left\{ABCED,ABCDE\right\}$.

Similarly, the other combinations where three students pick their own calculator can be obtained.

The different cases where four students pick their own calculator:

The number of occurrence that A and B and C and D her own calculator is $\left\{ABCDE\right\}$.

Similarly, the other combinations where four students pick their own calculator can be obtained.

The different cases where five students pick their own calculator:

The number of occurrence that A and B and C and D and E her own calculator is $\left\{ABCDE\right\}$.

Factorial of an integer:

The factorial of a non-negative integer n is given by

$n!=\left(n\right)\times \left(n-1\right)\times \mathrm{...}\times 1$

Substitute 5 for ‘n’

$\begin{array}{c}5!=5\times 4\times 3\times 2\times 1\\ =120\end{array}$

Thus, there are 120 ways that the 5 students pick an identical calculator.

The probability that A and B is obtained as:

$\begin{array}{c}P\left(A\cap B\right)=\frac{N\left(A\cap B\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that A and C is obtained as:

$\begin{array}{c}P\left(A\cap C\right)=\frac{N\left(A\cap C\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that A and D is obtained as:

$\begin{array}{c}P\left(A\cap D\right)=\frac{N\left(A\cap D\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that A and E is obtained as:

$\begin{array}{c}P\left(A\cap E\right)=\frac{N\left(A\cap E\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that B and C is obtained as:

$\begin{array}{c}P\left(B\cap C\right)=\frac{N\left(B\cap C\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that B and D is obtained as:

$\begin{array}{c}P\left(B\cap D\right)=\frac{N\left(B\cap D\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that B and E is obtained as:

$\begin{array}{c}P\left(B\cap E\right)=\frac{N\left(B\cap E\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that C and D is obtained as:

$\begin{array}{c}P\left(C\cap D\right)=\frac{N\left(C\cap D\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that C and E is obtained as:

$\begin{array}{c}P\left(C\cap E\right)=\frac{N\left(C\cap E\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that C and D is obtained as:

$\begin{array}{c}P\left(C\cap D\right)=\frac{N\left(C\cap D\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that D and E is obtained as:

$\begin{array}{c}P\left(D\cap E\right)=\frac{N\left(D\cap E\right)}{N}\\ =\frac{6}{120}\\ =\frac{1}{20}\end{array}$

The probability that A and B and C is obtained as:

$\begin{array}{c}P\left(A\cap B\cap C\right)=\frac{N\left(A\cap B\cap C\right)}{N}\\ =\frac{2}{120}\\ =\frac{1}{60}\end{array}$

The probability that A and B and D is obtained as:

$\begin{array}{c}P\left(A\cap B\cap D\right)=\frac{N\left(A\cap B\cap D\right)}{N}\\ =\frac{2}{120}\\ =\frac{1}{60}\end{array}$

The probability that A and B and E is obtained as:

$\begin{array}{c}P\left(A\cap B\cap E\right)=\frac{N\left(A\cap B\cap E\right)}{N}\\ =\frac{2}{120}\\ =\frac{1}{60}\end{array}$

The probability that A and C and D is obtained as:

$\begin{array}{c}P\left(A\cap C\cap D\right)=\frac{N\left(A\cap C\cap D\right)}{N}\\ =\frac{2}{120}\\ =\frac{1}{60}\end{array}$

The probability that A and C and E is obtained as:

$\begin{array}{c}P\left(A\cap C\cap E\right)=\frac{N\left(A\cap C\cap E\right)}{N}\\ =\frac{2}{120}\\ =\frac{1}{60}\end{array}$

The probability that B and C and D is obtained as:

$\begin{array}{c}P\left(B\cap C\cap D\right)=\frac{N\left(B\cap C\cap D\right)}{N}\\ =\frac{2}{120}\\ =\frac{1}{60}\end{array}$

The probability that B and C and E is obtained as:

$\begin{array}{c}P\left(B\cap C\cap E\right)=\frac{N\left(B\cap C\cap E\right)}{N}\\ =\frac{2}{120}\\ =\frac{1}{60}\end{array}$

The probability that C and D and E is obtained as:

$\begin{array}{c}P\left(C\cap D\cap E\right)=\frac{N\left(C\cap D\cap E\right)}{N}\\ =\frac{2}{120}\\ =\frac{1}{60}\end{array}$

The probability thatA andB and C and D is obtained as:

$\begin{array}{c}P\left(A\cap B\cap C\cap D\right)=\frac{N\left(A\cap B\cap C\cap D\right)}{N}\\ =\frac{1}{120}\end{array}$

The probability that A and B and C and D and E is obtained as:

$\begin{array}{c}P\left(A\cap B\cap C\cap D\cap E\right)=\frac{N\left(A\cap B\cap C\cap D\cap E\right)}{N}\\ =\frac{1}{120}\end{array}$

Addition rule:

For any five events A ,B, C,D and E

$P\left(A\cup B\cup C\cup D\cup E\right)=\left\{\begin{array}{l}P\left(A\right)+P\left(B\right)+P\left(C\right)+P\left(D\right)+P\left(E\right)\\ -P\left(A\cap B\right)-P\left(A\cap C\right)-P\left(A\cap D\right)-P\left(A\cap E\right)\\ -P\left(B\cap C\right)-P\left(B\cap D\right)-P\left(B\cap E\right)-P\left(C\cap D\right)\\ -P\left(C\cap E\right)-P\left(D\cap E\right)+P\left(A\cap B\cap C\right)+\\ P\left(A\cap B\cap D\right)+P\left(A\cap B\cap E\right)+P\left(A\cap C\cap D\right)\\ +P\left(A\cap C\cap E\right)+P\left(A\cap D\cap E\right)+P\left(B\cap C\cap D\right)+\\ P\left(B\cap C\cap E\right)+P\left(C\cap D\cap E\right)+P\left(B\cap D\cap E\right)\\ -P\left(A\cap B\cap C\cap D\right)-P\left(A\cap B\cap C\cap E\right)-\\ P\left(B\cap C\cap D\cap E\right)-P\left(A\cap C\cap D\cap E\right)\\ -P\left(A\cap B\cap D\cap E\right)+P\left(A\cap B\cap C\cap D\cap E\right)\end{array}\right\}$

By using addition rule,

The probability that at least one of the five gets her own calculator is obtained as shown below:

$\begin{array}{c}P\left(A\cup B\cup C\cup D\cup E\right)=\left\{\begin{array}{l}\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}-\frac{1}{20}-\frac{1}{20}-\frac{1}{20}-\frac{1}{20}-\frac{1}{20}-\frac{1}{20}-\frac{1}{20}-\frac{1}{20}\\ -\frac{1}{20}-\frac{1}{20}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}\\ -\frac{1}{120}-\frac{1}{120}-\frac{1}{120}-\frac{1}{120}-\frac{1}{120}+\frac{1}{120}\end{array}\right\}\\ =\left\{\frac{5}{5}-\frac{10}{20}+\frac{10}{60}-\frac{5}{120}+\frac{1}{120}\right\}\\ =\left\{1-\frac{1}{2}+\frac{1}{6}-\frac{1}{24}+\frac{1}{120}\right\}\\ =\frac{120-60+20-5+1}{120}\end{array}$

$\begin{array}{l}=\frac{76}{120}\\ =0.633\end{array}$

Thus the probability that at least one of the 5individuals gets their own calculator is 0.633.

The probability that at least one of the n individuals gets their own calculator is obtained as shown below:

The probability that at least one of the 5individuals gets their own calculator can be expressed as $\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{5!}$

$P\left(A\cup B\cup C\cup D\cup E\cup \mathrm{...}\cup n\right)=\left\{\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{5!}+\mathrm{...}+{\left(-1\right)}^{n-1}\frac{1}{n!}\right\}$

The above expression may also be written in the power series of $e^x$ at $x=-1$, when n is large,

$\begin{array}{l}{e}^x=\left\{1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\mathrm{...}+\frac{x^n}{n!}\right\}\\ e^{-1}=\left\{1+\frac{\left(-1\right)}{1!}+\frac{{\left(-1\right)}^2}{2!}+\frac{{\left(-1\right)}^3}{3!}+\mathrm{...}+\frac{{\left(-1\right)}^n}{n!}\right\}\\ e^{-1}=\left\{1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\mathrm{...}+{\left(-1\right)}^{n-1}\frac{1}{n!}\right\}\\ 1-e^{-1}=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\mathrm{...}+{\left(-1\right)}^{n-1}\frac{1}{n!}\end{array}$

$\begin{array}{c}P\left(\text{at least one of the n individuals get their calculator}\right)=1-e^{-1}\\ =1-0.3679\\ =0.632\end{array}$

$\begin{array}{c}P\left(\text{not at least one of the n individuals get their calculator}\right)=1-0.632\\ =0.3679\end{array}$

When n is large, there 63.2% for the at least one of the n individuals get their calculator. Rather in large group, there 36.8% chance that not at least one of the n individual get their calculator back.

The probability that at least one of the n individuals gets their own calculator is $\underset{\_}{1-e^{-1}}$