Multiplication Rule

The multiplication rule is a product used to determine the probability that two events, A and B, both occur.

The multiplication rule follows from the description of conditional probability.

The result is often written as follow, using set notation:

P( A ∩ B ) = P(A | B ) . P(B)

Or

P(A ∩ B) = P(B | A) . P(A)

Where:

P(A) = probability that event A occurs

P(B) = probability that event B occurs

P(A ∩ B) = probability that event A and event B occur

P(A | B) = the conditional probability that event A occur given that event B has occurred already

P(B | A) = the conditional probability that event B occur given that event A has occurred already

For independent an event, that is events which have no influence on one another, the rule simplifies to:

P(A nn B) = P(A). P(B)

That is, the probability of the joint events A and B is equivalent to the product of the individual probabilities for the two events.

Multiplication Rule of Probability

The addition rule helped us resolve problems when we performed one task and wanted to know the probability of two things happening during that task. This topic deals with the multiplication rule. The multiplication rule also deal with two events, but in these problems the events occur as a result of more than one task (rolling one die then another, drawing two cards, spinning a spinner twice, pulling two marbles out of a bag, etc).

When ask to find the probability of A and B, we would like to find out the probability of events A and B happening.

The Multiplication Rule:

Consider events A and B. P(AB)= P(A) P(B).

What The Rule Means:
Expect we roll one die followed by another and want to find the probability of rolling a 4 on the first die and rolling an even number on the second die. Notice in this problem we are not trade with the sum of both dice. We are only commerce with the probability of 4 on one die only and then, as a separate event, the probability of an even number on one dies only.
P(4) = (1)/(6)
P(even) = (3)/(6)
So P(4even) = ((1)/(6) )((3)/(6) ) = (3)/(36)  = (1)/(12)
While the rule can be applied in any case of dependence or independence of events, we should note here that rolling a 4 on one die followed by rolling an even number on the second die are independent events. Each die is treated as a split thing and what happens on the first die does not influence or effect what happens on the second die. This is our basic description of independent events: the outcome of one event does not influence or affect the outcome of another event.

Let's Practice the Multiplication Rule

Assume you have a box with 3 blue marbles, 2 red marbles, and 4 yellow marbles. You are going to draw out one marble, record its color, put it back in the box and draw another marble. What is the probability of pull out a red marble followed by a blue marble?


The multiplication rule says we need to find P(red) P(blue).

P(red) = (2)/(9)
P(blue) =(3)/(9)

P(redblue) = ((2)/(9) )((3)/(9) ) = (6)/(81)  = (2)/(27)
The events in this example were independent. Once the first marble was pull out and its color recorded, it was returned to the box. Therefore, the probability for the second marble was not affected by what happened on the first marble.

Some students find it supportive to simplify before multiplying, but the final answer must always be simplified.


Think the same box of marbles as in the previous example. However in this case, we are going to draw out the first marble, leave it out, and then pull out another marble. What is the probability of pull out a red marble followed by a blue marble?


We can motionless use the multiplication rule which says we need to find P(red) P(blue). But be alert that in this case when we go to pull out the second marble, there will only be 8 marbles left in the bag.

P(red) = (2)/(9)
P(blue) = (3)/(8)

P(redblue) = ((2)/(9) )((3)/(8) ) = (6)/(72)  = (1)/(12)

The events in this example were dependent. When the first marble was pull out and kept out, it affected the probability of the second event. This is what is meant by dependent events.


Assume you are going to draw two cards from a standard deck. What is the probability that the first card is a champion and the second card is a jack (just one of several ways to get “blackjack” or 21).

Using the multiplication rule we get

P(ace) P(jack) = ((4)/(52) )((4)/(51))  = (16)/(2652)  =  (4)/(663)

Notice that this will be the similar probability even if the question had asked for the probability of a jack followed by an ace.

Preparation for Sample Size

Our critical characteristic of preparation of sample size plan is deciding how big your sample size should be. If you amplify your preparation about sample size you add to the exactitude of your estimates, which means that, for any given approximation / size of result, the better preparation about the sample size is  the more “ statistically significant” the outcome will be. In other words, if an examination is too small then it will not sense fallout that is in actuality significant.

Defintion of preparation for sample size

The preparation about sampling is that a division of statistical performs troubled with the assortment of an impartial or casual division of personality explanation within a inhabitants of persons intended to give up some information regarding the population of distress, mainly for the purposes of construction predictions based on arithmetic supposition. Sampling is a central part of data collection.

Sample size process:

Step 1: Essential the population of worry.

Step 2: Indicate a sampling outside, a spot of material or trial likely to estimate.

Step 3: Specifying a sampling system for selecting items or actions from the enclose.

Step 4: Decisive the sample size.

Step 5: Implementing the sampling map.

Step 6: Gather the study sample data and information.

Step 7: Reviewing the example process.

Formula of preparation for sample size:

The formula for sample size = n = t2 * p (1-p) / m2.

Description:

N represents the required sample size.

T represents the confidence level. (Regular value 1.96).

P represents the estimated prevalence of malnutrition in the area

M represents margin of error. (Steady value 0.05).

Example for the preparation of sample size:

Ex:1 Find  the sample size when the population is 15.

Sol:

The sample size = t2 * p(1-p) / m2.

Where, the t is constant value of 1.96,

the m is constant value of 0.05.

Size = (1.96)2 * (15)*(1-15) / (0.05)2.

= 3.8416 *15 *14 / 0.0025.

= 322694.4

So, the sample size is 322694.4

Ex:2 Find  the sample size when the population is 30.

Sol:

The sample size = t2 * p(1-p) / m2.

Where, the t is a constant value of 1.96,

The m is a constant value of 0.05.

Size = (1.96)2*(30)*(1-30) / (0.05)2.

= 3.8416*30*29 / 0.0025.

= 1336876.8

So, the sample size is 1336876.8

Learn Common Ratio

In mathematics, a geometric series is a series with a permanent ratio between successive terms. For example, the series 1/2+1/4+1/8+1/16.....is geometric, so each term except the first can be obtained by multiplying the previous term by 1/2.Geometric series are one of the easiest examples of infinite series with finite sums. Basically, geometric series played an important role in the early development of calculus, and they continue to be central in the study of convergence of series. Geometric series(GS) are used throughout mathematics, and they have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance.

Common ratio:

The terms of a geometric series form a geometric progression(GP), meaning that the ratio of successive terms in the series is constant. Below table shows several geometric series with different common ratios:

Common ratio


Example

10


4 + 40 + 400 + 4000 + 40,000 + ···

1/3


9 + 3 + 1 + 1/3 + 1/9 + ···

1/10


7 + 0.7 + 0.07 + 0.007 + 0.0007 + ···

1


3 + 3 + 3 + 3 + 3 + ···

−1/2


1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/32 + ···

–1


3 − 3 + 3 − 3 + 3 − ···

The behavior of the terms depends on the common ratio r:

If r is between -1 and +1 the terms of the series become smaller and smaller, approaching zero in the limit. The series converges to a sum, as in the case, where r is a half, and the series has the sum one.

If r is greater than 1 or less than -1 the terms of the series become larger and larger. The total sum of the terms also gets larger and larger, and the series has no sum. (The series diverges.)

If r is equal to 1, all of the terms of the series are the same. The series diverges.
If r is minus one the terms take two values alternately (e.g. 2, −2, 2, −2, 2,... ). The sum of the terms oscillates between two values (e.g. 2, 0, 2, 0, 2,... ). This is a different types of divergence and again the series has no sum.

Examples:

Consider the sum of the following geometric series:

s = 1+2/3+4/9+8/27+......

This series has common ratio 2/3. If we multiply through by this common ratio, then the initial one becomes a 2/3, the 2/3 becomes a 4/9, and so on:

2/3 s = 2/3 + 4/9 + 8/27 + 16/81 + .....

Mean Value Theorem

The Mean Value Theorem has very important consequences in differential calculus.

THEOREM: Let the function f such that

i.           continuous in the closed interval [a,b]
ii.            derivable in open interval (a,b)

Then there exists at least one c with  a < c < b such that

calculus formula

The result in the theorem can be expressed as a statement about graph of f: if A(a , f(a)) and B(b , f(b)), are the end points on the graph, then there is at least one point C between A and B,such that the tangent is drawn from C is parallel to the chord AB.

Mean value theorem graph

Mean value theorem is also known as Lagrange’s Mean Value Theorem or First Mean Value Theorem or Law of Mean.

Applications of mean value theorem

1. Let the function be f such that

(i)                  Continuous in interval [a,b]

(ii)                Derivable in interval (a,b)

(iii)              f'(x) = 0 `AA` x  `epsi` (a,b) , then f(x) is constant in [a , b].

2.Let f and g be a functions such that

(i)             f and g are continuous in interval [a,b]

(ii)            f and g are derivable in interval(a,b)

(iii)          f'(x) = g'(x) `AA` x  `epsi`  (a,b) , then f(x) - g(x) is constant in [a,b]

3.Let the function be f such that

(i)             Continuous in interval[a,b]

(ii)            Derivable in interval(a,b)

(iii)          f'(x) > 0 `AA` x  `epsi`  (a,b), then f(x) is strictly increasing function in [a,b]

4.Let the function be f such that

(i)             Continuous in interval[a,b]

(ii)            Derivable in interval (a,b)

(iii)          f'(x) < 0 `AA` x  `epsi`    (a,b), thenf(x)  is strictly decreasing function in[a,b]

Special case of Mean Value Theorem is when f(a) = f(b).Then there exists at least one c with  a < c < b such that f'(c)= 0 . This case is known as Rolle’s Theorem.

Cauchy’s mean value theorem in calculus

Let f and g be functions such that

i.            both are continuous in closed interval [a,b]
ii.            both are derivable in open interval (a,b)
iii.             g'(x) `!=` 0 for any x `epsi` (a,b)  then there exists at least one number c `epsi` (a,b) such that

`(f'(c))/(g'(c))`  =  `(f(b) - f(a))/(g(b) - g(a))`

Mean value theorem example



Verify Rolle's theorem for the function

f (x) = x2 - 8x + 12 on (2, 6)

Since a polynomial function is continuous and differentiable everywhere f (x) is differentiable and continuous (i) and (ii) conditions of Rolle's theorem is satisfied.

f (2) = 22 - 8 (2) + 12 = 0

f (6) = 36 - 48 + 12 = 0

Therefore (iii) condition is satisfied.

Rolle's theorem is applicable for the given function f (x).

\ There must exist c  (2, 6) such that f '(c) = 0

f '(x) = 2x - 8

`=>`  c = 4 `in`(2,6)

Rolle's theorem is verified.

How to Find Inverse Variation

The inverse variation is the product of two variables equals to a constant and the product is not equal to zero. Inverse variation is in the form of y =k/x. xy = k. Inverse variation in which value of one variable increases while the value of the other  variable decreases in value is known as an inverse variation. For example think a trip of 240 miles. Rate(mph)=>20,30,40,60,80,120 and time(h)=>12,8,6,4,3,2.The numbers can be explained. As the rate of speed increase, the numbers of hour require decrease. As the rate of speed decrease, the number of hour requires increase. contrasting in a direct variation, the ratio in each data is not equivalent. The product of the value in each is equal.

Problem on how to find inverse variation:-

Problem 1:-

If y is inversely proportional to find inverse variation x and y = 10 when x = 2. Find the value of y, when x = 15?

Let, k = x/y

Plug x = 2 and y = 10 in the above equation

k = 10/2

k = 5

Now the equation becomes 3 = x/y

Now, plug x = 15

3 = y/15

3 * 15 = y

So, y = 45 when x = 15

Problem 2:-

find inverse variation F vary inversely with the square of m. if F=15 when m=3,find F when m=5?

Solution:-

F=K/m2

15=K/32

15=K/9

K=135

F=135/m2

F=135/52

=135/25

=5.4..

Problem in Equation on how to find inverse variation:-

Find inverse variation problem:-

Complete the table in support of the positive values of x so that yoo (1)/(x^2)

Table

Find the x value using inverse variation method

Solution:-

If yoo (1)/(x^2)then y =(k)/(x^2)

But y =100 when x = 3

Therefore 100 =(k)/(9)

That is k = 900 so y =(900)/(x^2)

When x = 5, y =(900)/(25) = 36

If x = 10, y =(900)/(100) = 9

And if x = 15, y = (900)/(25) = 4

If y = 25, 25 =(900)/(x^2)

That is 25x2 =900

x2 =(900)/(25) = 36

Table

So that answer x = 6.

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