Algebra Probability Problems

Probability is used to find the possible outcomes in an event. It is defined as the ratio between the numbers of favorable outcomes to the total number of outcomes. The value of probability lies only between 0 and 1. It is not greater than 1.There are several types of probability. They are Conditional probability and Theoretical probability. Conditional probability occurs when an other event is already occurred and changed the sample space. Theoretical probability occurs based on the probability principles.

PROBABILITY BASIC PROBLEMS

Problem 1:

A number is drawn from 1 to 12 at random. What is the probability of finding a number 7.

Solution:

From 1 to 12 there will be 12 numbers. So the total outcome is 12. Number 7 occurs only once. So the favorable outcome is 1.

Hence the answer is 1/12

Problem 2:

A bag contains 7 black, 8 blue and 11 green marbles. A marble is drawn at random. What is P (blue)?

Solution:

Total marbles in the bag = 7+8+11 = 26

Out of which number of blue marbles = 8

So the probability P (blue) = 8/26 = 4/13

Problem 3:

A card is drawn from a well shuffled pack of cards. What is P (diamond)?

Solution:

A pack of cards will have a total of 52 cards.

So total outcomes = 52.

In a pack of cards number of diamonds = 13

So the probability P (Diamond) = 13/52 =1/4

PROBABILITY PRACTICE PROBLEMS

Problem 1:

A die is thrown twice. Find the probability that a sum of 6 occurs on the die.

Solution:

Let F be the event of getting a number 6 on the die.

F = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

So the probability = (5/36)

Problem 2:

The condition is two numbers appearing on throwing two dice are different. Find

the probability of the when a sum of 4 occurs on the die.

Solution:

Let E be the event of getting a sum of 4. Here the condition given is the numbers on each dice is different.

E= {(1, 3), (3, 1)}

So the probability= 2/36 = 1/18

How to Bisect a Reflex Angle

Before we are going to see how to bisect a reflex angle we start with what is angle.  Angle is made with the turn that takes place between two straight lines which are formed from the same end point where the end point is also known as vertex. Reflex angle is the angle formed between the angles of 180 degree to 360 degree.

How to bisect a reflex angle, the word bisects means nothing but cutting it into two or splitting into two. Reflex angle bisector is the bisector that bisects the reflex angle. A bisector angle is a line that passes through the vertex of the reflex angle. How to bisect a reflex angle is followed by some certain steps.

Steps about how to bisect a reflex angle:

To bisect a reflex angle:

First draw a line and mark the given points  assume A,B.
By using the protector mark the reflection angle and draw the reflection angle and let the point be C.
Now AB and AC are the two lines with a same midpoint A.
Now by using campus make an arc with any radius which cuts both lines AB and AC.
Let the intersection point on line AB will be as D and on line AC be as E.
Take more than an half of the line as radius and in the campus and by keeping D and as midpoint make an arc. Similarly Keep E as midpoint and make an arc so that it cut the arc and the meeting point be F.
Now draw a line which joins AF.
Now AF will be the bisector of the given reflex angle.

These are the steps to be followed to bisect a reflex angle. The angle that is formed due to bisector in the reflex angle will be always an obtuse angle.

Example of how to bisect a reflex angle:

Draw a bisector of the reflex angle of 280 degree.

Solution:

Draw the line AB.

line

Keeping A as centre the angle of point 280 is marked and the line is drawn which makes a reflex angle of 280 degree which makes a line AC.

Here DE is an arc made by a as centre.

280 reflex angle bisect

More than half is taken as radius and two arcs are made with D and E as centre.The point F is plotted and a line is joined from A to F. The angle AF is the bisector of the reflection angle 280 degree.

Bisect a 280 reflex angle

Percentage off Calculator

"Percentage off" - This is the term that is synonymous with "percentage discount". It is the discount on some purchase represented in the form of percentage. For example, "10 percent off on a particular purchase" implies that on every $ 100, $ 10 will be discounted. Thus if the purchase is of `$`  200, the discount will be $ 20, if the purchase is $ 250, the discount will be $ 10 + $ 10 + $ 5 = $ 25, and so on. The formula to calculate the actual value of discount when we know the discount percentage and the purchase value is:

Discount value = (discount percent * purchase value)/100

In words, the percentage off on a particular item is equal to the product of percentage discount and the cost price divided by hundred.

The final payable amount in a purchase is found by deducting all the given discounts from the cost price of articles purchased.

Solved examples on percentage off calculator

For example, David purchases a shirt worth $ 100, on which the percentage off is 5%, stationary worth $120, on which the percentage off is 12%, and a football worth $45 on which he percentage off is 7%. Find the final amount paid by David on his purchases.

Sol:- Percentage off calculator :

We have to calculate the discount in $ for each article and then subtract it from that article's cost price. Once all the discounted prices are calculated, they are all added to get the total payable amount by David.

Cost price of shirt = $ 100

Percentage off = 5%

Discount (in $) = 5% of 100

= 5/100 x 100

= $ 5

Discounted price = $ 100 – 5

= $ 95

Cost price of stationary = $ 120

Percentage off = 12%

Discount (in $) = 12% of 120

= 12/100 x 120

= $ 14.4

Discounted price = $ 120 – 14.4

= $ 105.6

Cost price of football = $ 45

Percentage off = 7%

Discount (in $) = 7% of 45

= 7/100 x 45

= $ 3.15

Discounted price = $ 45 – 3.15

= $ 41.85

Total payable amount = $ 41.85 + 105.6 + 95

= $ 242.45

Therefore, David has to pay $ 242.45 as the final price after deducting all given discounts.

Solve Solutions in Math

With the help of Maths we can find the easy solutions to difficult problems. It is a concept of numerically solving with logical understanding. Basic operations in mathematics are addition, subtraction, multiplication, division.Also we solve some problems with the help of identities and formulas. In this article we shall study some more solutions in math. Before solving the maths problems there are some important points to remember. some of them are following.

Read the problem very carefully and find what we have to find ?

If there is any formula start to the solve the problem from there.

If it possible break the problem into 2 or more parts.

Make the an easy equation if possible.

Solve the equation to get the answer.

Some Important Formulas Used to get Solutions in Math

( a + b )2 = a2 + b2 + 2ab

( a - b )2 = a2 + b2 - 2ab

( a + b ) ( a - b ) = a2 - b2

a3 + b3 + c3 = 3abc

Area of perpendicular triangle = 1/2 X base X height

Area of circle = 2pi r

Surface area of cuboid = 2(lb +bh + hl)

volume of cuboid = lbh

mean     =     ("Sum of all Numbers")/(" Total Numbers" )

Examples for Solutions in Math

Ex 1 : Evaluate : 302 + 203 - 503

Sol :Step 1: Let a = 30, b = 20, c = - 50. Then,

a + b + c = 30 + 20 - 50 = 0

Step 2:We know   a3 + b3 + c3 = 3abc

303 + 203 + ( -50 )3 = 3 X 30 X 20 X -50

303 + 203 - 503 =  - 90000  Ans.

Ex 2 : Factorize a4 - b4

Sol :    Step 1: a4 - b4

=  ( a2)2 - ( b2 )2

Step 2:We know ( a + b ) ( a - b ) = a2 - b2

so          =  ( a2)2 - ( b2 )2

= ( a2 + b2 ) ( a2 -  b2)

=  ( a2 + b2 )  (a + b) ( a - b )  Ans.

Ex :3 The radius of a circle is 14 cm . Find the area of the circle.

Sol :         area of circle

Step 1:Radius of circle = 14cm

Step 2:Area of circle = pi r2

=  22/7 14 X 14

=  616 cm2 Ans.
Practice Problems to Find Solutions in Math:

Pro 1: Evaluate :    233 - 173

Ans :   7254

Pro 2: The radius of circle is 35cm. Find the area of circle?

Ans :  3850 cm2

positive Quadrant

Quadrants are the important concept in graph matrix. Positive Quadrant is one of the parts of our Cartesian plane. Usually our Cartesian plane is divided into 4 parts of quadrants.Cartesian plane is the formation of the x and y axis. If the x and y values are positive then that quadrant is called the positive quadrant. It is also called first quadrant. In this article we are going to learn about positive quadrant through problems and diagrams.

Brief Summary about Positive Quadrant

Cartesian Graph plane:

Our Cartesian graph plane is the basic formation of the x and y axis.

X axis:

X axis is the left to right direction line and then it can be divided into 2 parts by the origin.  They are X and x’.  Positive values are marked in the x axis and then the negative values are marked in the x’ axis.

Y axis:

Y axis is the top to bottom direction line and then it can be divided into 2 parts by the origin.  They are Y and y’.  Positive values are marked in the y axis and then the negative values are marked in the y’ axis.

Origin:

The intersection of horizontal axis x and then the vertical axis y is the point origin.

Coordinates:

Coordinates are the major part of graph plane.  It takes the form of (x, y).

Quadrants:

Based on the x and y values it can be marked into the different type of 4 quadrants.

Positive Quadrant:

Positive quadrant is the intersection of positive value of x and positive value of y. The coordinates on the positive quadrant is on the following form (+x, +y).
Example Problems on Positive Quadrant

Example 1:

Shade the following points in the positive quadrant

(5, 2)
(3, 1)
(4, 3)
(6, 2)

Solution:

The graph in the positive quadrant looks like as the following graph,

Example 2:

Identify which points is lies in the positive quadrant from the following list of points

(0,6)
(6,0)
(6,1)
(-2,6)

Solution:

Here (6, 1) is the only point in the positive quadrant.  Because here both 6 and 1 values are positive.

Parabola Math Problems

A parabola  is the set of all the points whose distance from a fixed point in the plane are equal  to their distances  from a fixed line in the plane.  The fixed point is called the focus and the fixed line is the directrix.

.directrix
Formula  for parabola ;-
(1) y2 = 4px  has symmetry around  x axis Directrix x + p = 0  Latus rectum  LL' is 4p vertex = (0,0)
(2) y2 = -4px has symmetry around the negative side of x-axis Directrix  x - p = 0 . LL' = 4p. Vertex =(0,0)
(3) x2 = 4 py has symmetry around the  positive y - axis Directrix y + p = 0.  LL' = 4p. Vertex =(0,0)
(4) x2 = -4py has symmetry around the negative y - axis. Directrix y - p = 0.  LL' = 4p. Vertex= (0,0)

Problems on Parabola:
Here is a parabola math problem for us  to do.
# Find the focus , vertex, axes, directrix  of the parabola y2 - 4x - 4y = 0
Sol:-
parabola
Step 1 : Let us bring the above given parabola equation to the parabola formula equation.
y2- 4x - 4y = 0
Step 2 : Add 4x to both sides + 4x             + 4x
We get                 y2 - 4y       =    4x
Step 3 :Complete the square on LHS we write y2 - 4y +4  = 4x + 4 ( we add 4 to both sides of the equation to balance it)
Step 4 : Now we get                                      (y  - 2 )2    = 4(x+1)
Step 5 : Let us compare the above equation to the general equation Y2 = 4 pX
we get Y = y-2  and X = x+1. Hence Y2 = (y -2)2  and 4px = 4(x+1) . Hence we get p = 1
Step 6 : Now let us set a table for both the values. Now x = X - 1 and  y = Y + 2
____________________________________________________________________________
Referred to X and Y axes                             Referred to x and y axes
____________________________________________________________________________
focus S = (p,0)                                          x = X - 1 = 1 -1 = 0; y = Y +2= 0 + 2 = 2
So focus is (0,2)
____________________________________________________________________________
vertex is (0,0)                                            x = X - 1 => 0 - 1 = -1 and y = Y + 2 = 0+2 =2
So vertex = (-1,2)
____________________________________________________________________________
axes Axis  of parabola Y =0                       Axis of the parabola is y = Y+2 = 0+2 =2
so axis  is y - 2 = 0
____________________________________________________________________________
equation of the directrix is                          equation of the directrix is  x + 1 = -1
x = -p that means X = - 1                           which gives us  x + 2 = 0
____________________________________________________________________________
Hence for the given parabola  y2 -4x - 4y = 0 we have (1)  Focus  as (0,2)  ;
(2)  Vertex is (-1,2)
(3)  Axis of the parabola is y = 2
and     (4) directrix is x + 2 = 0

Condition for a Line to be a Tangent for a Parabola:-
Let us find the condition for a line  lx + my + n = 0 to be a tangent to the parabola y2 = 4ax
Solution:-
The condition that a  line y = mx + c  to be a tangent to the parabola y2 = 4ax is that C = a/m...(1)
Step 1 : Let us write lx + my + n = 0 in the form y = mx + c
That gives us my = -lx - n
y =  [ -l]x + [ -n]
m        m
Step 2 : Now we have  y = mx + c   => y = (-l/m) x  + ( -n/m)  That is slope = -l/m and  y intercept C = -n/m
Hence  C =  -n/m  and m(slope) = (-l/m)
Step 3   But  C =  a / m (formula)
So we write  -n  =    a =   - am
m    (-l/m)       l
Step 4   Hence we get - n =  -am
m      l
Step 5 : Cross multiplying we get  -nl = -am2
Step 6 : That gives us nl = am2
Hence the condition that lx + my + n= 0 to be a tangent to the parabola y2  = 4ax is  nl = am2

Geometric Constructions Tutorial

Geometric constructions is a basic construction of geometric figures. In this article we can see the construction of a perpendicular bisector, construction of an angle and construction of a triangle. For the construction of geometric figures we use only the help of a ruler (straight-edge) and compass. Compass is generally used to draw arcs of particular distances. The distance is calculated with the help of a ruler.
Geometric Constructions

i) To bisect a given line segment

Given: A line segment AB

To construct: The perpendicular bisector AB

Construction:      a) With A as centre and radius equal to more than half of AB, draw two arcs, one on each side of AB.

b) With B as centre and with the same radius taken above draw two more arcs cutting the previous arcs                              at C and D

c) Join C and D cutting AB at O. CD is the perpendicular bisector of AB.

perpendicular bisector of a line segment

ii) To construct an angle of 60°

Construction: a) Draw a line segment AB of any suitable length.

b) With A as centre and any suitable radius draw an arc cutting AB at C.

c) With C as centre and the same radius AC, draw another arc cutting the former arc at D.

d) Join A and D and produce it to E.
construction of angle 60degrees

iii) To construct an angle of 30°

Construction: a)First construct an angle of 60° as shown above

b) With C and D as centre, draw two arcs of equal radii cutting each other at F

c) Join A to F.
construction of angle 30degrees

Construction of more Geometrical Shapes

iv) To construct a triangle having given the lengths of three sides.

Let the three sides be 3.6cm, 4cm and 5.1cm

Construction: a) Draw a line segment AB = 3.6cm

b) With A as centre and radius = 4cm, draw an arc.

c) With B as centre and radius = 5.1cm draw one more arc intersecting the former arc at C.

d) Join AC and BC. Then ΔABC is the required triangle.(Note: The construction fails when the sum of any two sides is less than the third side.)

Algebra is widely used in day to day activities watch out for my forthcoming posts on Combinations and Permutations Examples and Continuity of a Function. I am sure they will be helpful.

v) To construct a triangle, having given the lengths of any two sides and the included angle.

Let the length of the two sides be 3.9cm and 4.3cm and included angle be 60°

Construction:a) Draw a line segment AN = 3.9cm

b) Construct
c) From AX, cut off AC = 4.3cm

d) Join BC. Then ΔABC is the required triangle.

construction of a triangle, where two sides and included angle are given