Chapter 4.CR, Problem 22CR

To determine

To find:

The probability that the number on the winning ticket is either even or less than 300.

Answer to Problem 22CR

Solution:

The probability of being the number on the winning ticket either even or less than 300 is 0.7917.

Explanation of Solution

Given information:

Raffle tickets for a trip to Miami are assigned three-digit numbers using the digits 0-9.

Steps:

For selection of objects from a group regard to their arrangement is the permutation.

Formula for permutation is used for counting number of selection of objects from a group regard to their arrangement.

Formula used:

Formula to calculate the probability of event $A$ is,

$P\left(A\right)=\frac{\text{favorable}\text{outcomes}}{\text{Number}\text{of}\text{total}\text{outcomes}}$

For any two events $A\text{and}B$, the probability of occurring either $A\text{or}B$ is,

$P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)-P\left(A\text{and}B\right)$.

Formula to calculate the possible number of count to select $r$ objects from $n$ objects in permutation is,

${}_n{P}_r=\frac{n!}{\left(n-r\right)!}$

$n!=n\cdot \left(n-1\right)\cdot \left(n-2\right)\mathrm{...3}\cdot 2\cdot 1$

Here $n$ and $r$ both are positive integers.

Total number of digits is 10.

Required numbers are of 3 digits.

Number of random choices is 3.

Substitute 10 for $n$ and 2 for $r$ in the formula for permutation.

Total possible number of 3 digit number is,

$\begin{array}{rll}{}_{10}{P}_3& =& \frac{10!}{\left(10-3\right)!}\\ & =& \frac{10!}{7!}\\ & =& \frac{10\times 9\times 8\times 7!}{7!}\\ & =& 720\end{array}$

Unit digit of an even number should be any one of 0, 2, 4, 6, 8.

Unit digit of an even number is fixed.

Possible choice for other two digits is 2 out of 10.

There are 5 choices for digit for getting an even number.

Substitute 10 for $n$ and 2 for $r$ in the formula for permutation.

Total possible number of arrangement of first two digits is,

$\begin{array}{rll}{}_{10}{P}_2& =& \frac{10!}{\left(10-2\right)!}\\ & =& \frac{10!}{8!}\\ & =& \frac{10\times 9\times 8!}{8!}\\ & =& 90\end{array}$

Total number of 3 digit even numbers is,

$\begin{array}{r}5\times 90\\ =450\end{array}$

Let $A$ be the event of even number.

Calculate the probability of event $A$.

$P\left(A\right)=\frac{450}{720}$

Hundred place of a number less than 300 should be any one of 0, 1, 2.

Hundred place digit of a number less than 300 is fixed.

Possible choice for other two digits is 2 out of 10.

Substitute 10 for $n$ and 2 for $r$ in the formula for permutation.

Total possible number of arrangement of last two digits is,

$\begin{array}{rll}{}_{10}{P}_2& =& \frac{10!}{\left(10-2\right)!}\\ & =& \frac{10!}{8!}\\ & =& \frac{10\times 9\times 8!}{8!}\\ & =& 90\end{array}$

There are 3 choices for hundred place digit for a number less than 300.

Total number of 3 digit numbers less than 300 is,

$\begin{array}{r}3\times 90\\ =270\end{array}$

Let $B$ be the event of number less than 300.

Calculate the probability of event $B$.

$P\left(B\right)=\frac{270}{720}$

An even number less than 300 has 3 fixed choices (0, 1, 2) for hundredth place and 5 fixed choices (0, 2, 4, 6, 8) for unit place.

There is one choice for the tens place digit.

Substitute 10 for $n$ and 1 for $r$ in the formula for permutation.

Total possible number of arrangement of tens digit is,

$\begin{array}{rll}{}_{10}{P}_1& =& \frac{10!}{\left(10-1\right)!}\\ & =& \frac{10!}{9!}\\ & =& \frac{10\times 9!}{9!}\\ & =& 10\end{array}$

Total number of 3 digit even numbers and less than 300 is,

$\begin{array}{l}3\times 5\times 10\\ =150\end{array}$

$A\text{and}B$ is the event even numbers and less than 300.

Calculate the probability of event $A\text{and}B$.

$P\left(A\text{and}B\right)=\frac{150}{720}$

$A\text{or}B$ is the event of either even number or a number less than 300.

Calculate the probability either $A\text{or}B$.

$\begin{array}{rll}P\left(A\text{or}B\right)& =& P\left(A\right)+P\left(B\right)-P\left(A\text{and}B\right)\\ & =& \frac{450}{720}+\frac{270}{720}-\frac{150}{720}\\ & =& \frac{570}{720}\\ & =& 0.7917\end{array}$

Conclusion:

Thus, the probability of being the number on the winning ticket either even or less than 300 is 0.7917.