Archive for November, 2015

Problem 1163

An experiment consists of rolling two fair dice and adding the dots on the two sides facing up.

Find the probability of the sum of the dots indicated.

A sum loss then 7

 

Solution:-

 

The probability of getting a sum less than 7 is \frac{5}{12}

 

Problem 1162

Is the selection a permutation, a combination, or neither?

Sam rents 5 videos from a video store.

 

Solution:-

 

Use the definition of permutation and combination to determine the type of selection made.

A permutation of a set of n distinct object is an arrangement of the objects in a specific order without repetition.

A combination of a set of n distinct objects is an arrangement of the object without repetition where order is irrelevant.

In the selection process of the problem, the order that the video are rented is not important.

There will not be ant renting video form the video store because the student cannot rent the same copy of a movie multiple time at the same moment in time.

Use the fact that the selection process has no repetition and order is irrelevant to determine the type of selection.

The selection is a combination.

 

 

Problem 1161

Is the selection a permutation, a combination, or neither?

Ten woman in a salon are dying their hair blonde, brunette, black, or red.

 

Solution:-

 

Use the definitions of permutation and combination to decide the type of selection mode.

A permutation of a set of n distinct is an arrangement of the objects in a specific order without repetition.

A combination of a set of n distinct object is an arrangement of the object without repetition where order is irrelevant.

In the selection process of the problem, the order of who gets their hair dyed and the order that the colors are chosen is not important.

There is repetition since there are only four hair colors to choose from and 10 women dying their hair.

Use the fact that the selection process has repetition and order is irrelevant to determine the type of selection.

The selection is neither a permutation or a combination.

Problem 1160

Each of 2 countries sends 3 delegates to a negotiating conference. A rectangular table is used with 3 chairs on each long side. If each country is assigned a long side of the table (operation 1), how many seating arrangement are possible?

 

Solution:-

 

Number of Permutations of n Objects

The number of permutations of n distinct object without repetition is given by

Pn,n = n(n-1)(n-2)*……..2*1= n!

Note the operation 1 is the assignment of a country to the side of a table. There are two sides of the table, so two permutations of how the sides may be assigned to the countries .

P 2,2= 2

Once the side are assigned the seating arrangement of each country can be determined. The arrangements for the countries are separate. As a result, the total number of arrangements (once the sides are chosen) is the product of the number of arrangements for each country.

Arranging the seating means putting the delegates in an order, hence the solution requires permutations.

The number of ways one country’s delegates may be seated is

P 3,3 = 3! = 6

The total number of seating arrangements is given by.

= P2.2*P3,3*P3,3 = 2*3!*3! = 2*6*6 = 72.

Problem 1159

For P = {3,5,9,14}, Q = {1,6,12}, and R = {4,6,9,12}, find P U (Q∩R).

 

Solution:-

 

The union of sets A and B, denoted by A U B, is the set formed by combining all elements of A and all elements of B into one set. Symbolically, this is expressed by the equality

A U B = {x | x ϵ A or x ϵ B}.

Remember that the word ‘or’ is used to mean that x may be an element of set A or set B or both.

The intersection of sets A and B, denoted by A ∩ B, is the set of elements in set A that are also in set B. Symbolically, this is expressed by the equality A ∩ B = {x | x ϵ A and xϵ B}.

Determine the intersection of sets Q and R.

Q ∩ R = {6,12}

Determine the union of P and the intersection of Q and R.

(P U (Q ∩ R)) = {3,5,6,9,12,14}.

 

 

Problem 1158

Let U = {16,25,36,49,64,81} and A = {16,25,36} . Find A’.

 

Solution:-

 

The complete of A (relative to the universal set U), denoted by A’, is the set of elements in U that are not in A.

A’ = {x ϵ U | x Ɇ A}

Thus, determine all the elements of U that are not in A.

The element 49 of the set U, for instance, is not in set A. Therefore, it is an element of set A’.

However, the element 16 is in both sets A and U. Therefore, 16 is not an element of set A’.

The elements of set A’ are shown below.

A’ = {49, 64, 81}

 

Problem 1157

Write the resulting set using the listing method.

{x | x is an old number between 11 and 19, inclusive}

 

Solution:-

 

The listing method listed all the elements of a set between braces { }.

The resulting set will contain all odd number between 11 and 19, inclusive. Remember, inclusive means that the endpoints are included.

An odd number is a integer that is not divisible by 2. The first odd number between 11 and 19, inclusive, is 11.

Therefore are four remaining odd numbers between 11 and 19, inclusive. These odd number are 13,15,17,and 19.

Therefore, the resulting set, written in listing notation, is {11, 13, 15, 17,19}

 

Problem 1156

Write the resulting set using the listing method.

{x | x2=9}

 

Solution:-

 

S = {x | P(x)} means “S is the set of all x such that P(x) is true”

The value of x that satisfy x2 = 9 are 3 and -3.

Thus, {x | x2 = 9} = {3, -3}.

 

Problem 1155

Write the resulting set using the listing method.

{5,6,7}∩ {2,3}

 

Solution:-

 

The intersection of sets A and B, denoted by A ∩ B, is the set of elements in set A that are also in set B. This is stared symbolically below.

A∩B = {x | x ϵ A and xϵ B}

Determine which elements, if any, of {5,6,7} are also elements of {2,3}.

The number 5 is an element if {5,6,7}, but not an element of {2,3}.

The number 6 is an element of {5,6,7}, but not an element of {2,3}.

The number 7 is an element of {5,6,7}, but not an element of {2,3}.

Since the two sets have no elements in common, {5,6,7}∩{2,3} is equal to empty set, denoted { } or ø.

 

Problem 1154

Write the resulting set using the listing method.

{3,9,15} U {9,13,15}

 

Solution:-

 

A U B = {x | x ϵ A or x ϵ B}

Remember that the word ‘or’ is used to mean that x may be an element of set A, set B, or both.

Let A represent the first set. Thus, the elements in set A are 3,9, and 15.

Let B represent the second set. Thus, the elements in set B are 9,13,and 15.

The elements that set A and B have in common are 9 and 15.

When listing the elements in a set, do not list an element more than once.

Finally, list all of the elements that appear in either set A or set B. If an element appears in both of the sets, list it only once.

Therefore , the union of set A and set B, denoted A U B, is equal to {3,9,15,13}.