Bernoulli Trial Probability

Introduction to Bernoulli trial probability

Bernoulli Trails probability: Trails of a random experiment are called Bernoulli trails, if they satisfy the following conditions:

(i)   There should be a finite number of trails.

(ii)  The trails should be independent.

(iii) Each trail has exactly two outcomes: success or failure.

(iv) The probability of success remains the same in each trail.

Examples on Bernoulli Trial Probability

For example :  Throwing a die 50 times is a case of 50 Bernoulli trails, in which each trail results in the success (say an even number ) or  the failure ( an odd number ) and the probability of success (p) is same for all 50 throws. Obviously, the successive throws of the die is independent experiments. If the die is fair and have six numbers 1 to 6 written on six faces, then p =`(1)/(2)`  and q=1-p = 1-`(1)/(2)` =`(1)/(2)`  = probability of failure.

Solved Problems on Bernoulli Trial Probability

Question based on Bernoulli trail probability:

Qu:  Six balls are drawn successively from an urn containing 7 red and 9 black balls. Tell whether or not the trails of drawing balls are Bernoulli trails when after each draw the ball is draw the ball drawn is

(i)  Replaced into the urn.

(ii) Not replaced in the urn.

Solution:(i) The number of trails is finite. When the drawing is done with replacement, the probability of success (say, red ball) is p=`(7)/(16)`  which is same for all six trails (draws). Hence, the drawings of balls with replacement are Bernoulli trails.

(ii) When the drawing is done without replacement, the probability of success (i.e. red ball) in first trail is `(7)/(16)` ; in 2nd trail is `(6)/(15)` if the first ball drawn is red or `(7)/(15)` if the first ball drawn is black and so on. Clearly, the probability of success is not same for all trails; hence the trails are not Bernoulli trails.

Number of Possible Combinations

Introduction for number of possible combinations:

In combinatorial mathematics, a k-combination of a finite set S is a subset of k distinct elements of S. Specifying a subset does not arrange them in a particular order; by contrast, producing the k distinct elements in a specific order defines a sequence without repetition, also called k-permutation. As an example, a poker hand can be described as a 5-combination of cards from a 52-card deck: the 5 cards of the hand are all distinct, and the order of the cards in the hand does not matter.

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Formula for Number of Possible Combinations:

Number of possible Combinations formula is commonly used in probability and statistics. Number of possible combination formula solving is the number of possible combinations of several items from a set of items. We can write in a number of ways similar to:

nCr , Where,

r => number of possibilities in the thing,

n => complete possibilities in the thing.

Also it can be understand as “n select r”.

The formula for the number of possible combinations are given as follow:

n!          n(n-1) (n-2) (n-r+1)
nCr = --------- = ----------------------------
r!(n-r!)                  r!

Example Problems for Possible Number Combinations:

1) From the group of 15 employees how many combinations are formed as an union of 3 employees?

Solution:

Solving Here n= 15 employees,

r = 3 teams

n!
nCr = ---------
r!(n-r!)


15!
15C3 = --------------
3! (15 – 3!)


15!
= -----------
3! (12)!


15*14*13
= --------------
3*2*1

= 5 * 7 * 13

15C3 = 455 possible number of combinations

2) From a group of 8 students the teacher chooses 4 students to give a demonstration. How many combinations are possible?

Solution:

n = 8 students

r = 4 students

Formula:

n!
nCr = ---------
r!(n-r!)

8!
8C4 = -----------
4! (8-4)!

8!
= -----------
4! (4)!


8*7*6*5
= -------------
4*3*2*1

15C4 = 70 possible number of combinations

3) In how many techniques can 5 girls be chosen this leaves four more girls to be chosen from the remaining 17 girls.

Solution:

The number of ways of arranging four boys from seventeen is 17x 16x 15x 14x 13

The number of methods of arranging the four among themselves is 5x4x3x2x1.

Therefore the number of ways of choosing the five girls,

17 x 16 x 15 x 14 x 13
= --------------------------
5 x 4 x 3 x 2 x 1

= 6188 combinations.

Quotient Identities

Introduction to quotient identities

In trigonometry, the following are called the quotient ratios. They are:

1. tanx = `sinx/cosx`

2. cotx = `cosx/sinx`

3. secx = `1/cosx`

4. cscx = `1/sinx`

The quotient identities are formed by the two trigonometric ratios sine and cosine functions. These quotient identities are very helpful in proving certain identities in trigonometry. These quotient identities can be manipulated in different form so that based on the need we can make use of them to prove the identities. They also can be used in different application problems to find the heights, distances and angle of elevations and angle depressions etc. Though the above things can be achieved by the two basic trigonometrical ratios sine and cosine functions, but it may take too long time to evaluate the problems involved in it. Hence it is better to know and understand the usage of the quotient trigonometric identities. We can also see some relations among them which will be of very helpful not only in trigonometry but it helps in evaluation of certain problems in calculus as well.

We have the following important identities:

1.(i) 1 + tan^2x = sec^2x

(ii) sec^2x - tan^2x = 1

(iii) tan^2x = sec^2x - 1

2. (i) 1 + cot^2x = csc^2x

(ii) csc^2x - cot^2x = 1

(iii) cot^2x = csc^2x - 1

The above identities are used to check whether the given trigonometric expressions or trigonometric equations are true or not.

Now let us solve few problems on quotient identities.

Example Problems on Quotient Identities

Ex 1: Prove the following identity:

`tanx/(secx ** 1)` +` tanx/(secx+1)` = 2cscx.

Sol: LHS =` tanx/(secx **1) + tanx/(secx+1)`

= tanx `[1/(secx ** 1) + 1/(secx +1)]`

= tanx`[( secx + 1 + secx ** 1)]/[(secx ** 1)( secx + 1)]`

= tanx `[(2 secx)/( sec^2x ** 1)]`

= `(2 tanx secx)/(tan^2x)`

= `[2secx]/tanx`

= 2 cscx = RHS.

2. Prove that (cosecx - sinx) ( secx - cosx) = `1/[tanx+cotx]`

Proof: LHS = (cosecx - sinx) ( secx - cosx)

= `(1/sinx - sinx)(1/cosx - cosx)`

= `(1- sin^2x)/sinx xx (1-cos^2x)/cosx`

= `[cos^2x]/sinx xx [sin^2x]/cosx`

= sinx cosx = `[sinx cosx]/[sin^2x + cos^2x]`

= `1/[[ sin^2x]/[sinx cosx] + [cos^2x]/ [sinx cosx]]`

= `1/[tanx + cotx] ` = RHS.

Hence the proof.

3. Show that tan^2x + cot^2x + 2 = cose2x sec^2x.

Proof: LHS = tan^2x + cot^2x + 2

= tan^2x + cot^2x + 2tanx cotx

= (tanx + cotx)2

= `[sinx/cosx + cosx/sinx]^2`

= `[sin^2x + cos^2 ]/ [ sin^2x cos^2x]`

= `1/[sin^2x] xx 1/[cos^2x] ` = cosec^2x sec^2x = RHS.

Hence the proof.

Practice Problems on Quotient Identities

Prove the following identities:

1. (cosec x - cotx )2 = `[1-cosx]/[1+cosx]`.

2. sec4x - sec^2x = tan4x + tan^2x.

Optimal Probability Equation

Introduction to Optimal Probability Equation:

Science which goes into details, at times sounds like a jargon and resembles a game. A kind of theory of word puzzles. This introduction includes a lot of jargon for little practical purposes. Theory of probability without a doubt  has its jargon. It also contains a big number of equations and formulas.

Brief Explanation of Optimal Probability Equation

The principal investigator will study several problems in optimal deterministic or stochastic control and probability, using the theory of nonlinear partial differential equations. The project consists of three parts. The optimal control of a singularly controlled Brownian motion will be studied using the results for free boundary problems. This approach has yielded regular results when the dimension is equal to two and it is the first multi-dimensional result which yields the construction of the optimal process. It now appears that it can be generalized to higher dimensions. The second part of this project is devoted to the perturbation theory of infinite dimensional Hamilton-Jacobi equations. As a case study, the principal investigator will study an asymptotic problem related to simple exclusion processes. Finally, an application of the finite dimensional perturbation theory to a problem of production planning is described. The model to be studied has failure-prone machines, and the goal is to construct approximate optimal policies by exploiting the hierarchical structure of the model.

Example of Optimal Probability Equation

Let us take example of Optimal probability, suppose, the maximum possible sanction is 500, so optimal deterrence could be achieved with a probability of 2 percent. Alternatively, we could employ a sanction of 100 and a probability of 10 percent. However, half the individuals overestimate the probability by one percentage point and the other half underestimate it by the same amount.

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Summary : Optimal Probability Equation

Sampling methodology in probability equation can be profitably used to estimate characteristics of a population of interest with less cost, faster speed, greater scope and even greater accuracy compared to a complete enumeration. We discuss basic ideas, underlying assumptions and general considerations of sampling by surveying sampling methods.

Greatest Prime Factor

Introduction to greatest prime factor:

In this section we will study about greatest prime factor. Greatest prime factor is also called as greatest common factor. It is used us to calculate the greatest prime factors of two or more numbers. In this section we will see some solved problems for greatest common factor. Let us study greatest prime factor.

Example Problems for Greatest Prime Factor:

Example problem 1: What is the greatest common factor of 4 and 2?

Solution:

First we have to write the prime factorization for 4 and 2. 2 is a prime number.

4 = 2 × 2

Next, find the common factors shared by both of the numbers.

4 = 2 × 2

2 = 2

The only common factor of 4 and 2 is 2, so the greatest common factor is 2.

Answer: The greatest common factor is 2.

Example problem 2: Dante has 12 math books and 18 science books. If he wants to distribute them evenly among some bookshelves so that each bookshelf has the same combination of math and science books, with no books left over, what is the greatest number of bookshelves Dante can use?

Solution:

Write the prime factorization for 12 and 18.

12 = 2 × 2 × 3

18 = 2 × 3 × 3

Next, find the common factors shared by both of the numbers.

12 = 2 × 2 × 3

18 = 2 × 3 × 3

Finally, multiply the common factors to find the greatest common factor.

2 × 3 = 6

The greatest common factor of 12 and 18 is 6. That means that the greatest possible number of bookshelves is 6, because 12 math books could be put onto 6 bookshelves with 2 math books each and 18 science books could be put onto 6 bookshelves with 3 science books each.

The greatest number of bookshelves Dante can use is 6.

Answer: The greatest number of bookshelves Dante can use is 6.

Practice Problems for Greatest Prime Factor:

Practice problem 1: Zahra and Raul are training for a marathon. Zahra runs 9 miles at a time while Raul prefers to run in blocks of 8 miles. At the end of a month, they realize that they have run same total number of miles. What is the smallest number of miles that each must have run?

Practice problem 2: The city of Hillsdale is honoring 15 mothers and 20 fathers as winners of its "Best Parent" contest. The plan is to take several group photographs, each with the same combination of mothers and fathers and no parents left out. What is the greatest number of photos that can be taken?

Solutions for greatest prime factor:

Solution 1: Each of them must have run 72 miles.

Solution 2: The greatest number of photos that can be taken is 5.

Binomial Theorem Probability

Introduction to Binomial theorem probability:-

In a binomial theorem probability, we deal with two outcomes.  They are called 'success' and  'falure'.  These are two mutually
disjoint outcomes.
We denote success by the symbol p and the failure by the symbol q.  Obviously p+q=1
Formula  for binomial distribution is P[X=x] = {ncx.pxqn-x}  x= 0,1,2 ....... n
The two independent constants n and p are called the parameters of the distribution
Mean = np
Variance = npq
Standard deviation =vnpq

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Problems on Binomial Theorem Probability.

Problem 1:- Eight coins are tosses simultaneously.  Find the probability of getting atleast 6 heads.
Solution:-
Number of trials n = 8
Probability of getting a head is 1/2   That is p= 1/2  that mean q = 1/2
Formula  p(x) = P(X-x) = ncxpx. qn-x and x= 0,1,2 ..... n
= 8cx (0.5)x.(0.5)n-x
= 8cx(0.5)8
Probability of getting at-least 6 heads is P(X=6) = P(X=6) + P(X=7)+P(X=8)
= p(6) +p(7)+p(8)
= (0.5)8[ 8c6 + 8c7 + 8c8}
= (0.5)8[ 8c2 + 8c1  + 1]   since 8c6 = 8c2 and  8c7=8c1
= (0.5)8[ 28 +8+1]
=  1(37)           37
----           =  -----    or   0.14 
256              256
The answer is 0.14

Problem 2 :  Given n = 6 and 9P(X=4)= P(X = 2)  ,  find p

Formula  for probability of randon variable is  P(X=x) = ncx.px.qn-x   x = 0,1,2,3........n
Here n= 6  q = p-1   and P(X=x) = 6cx pxq6-x; x=0,1,2 .....6
x=4 and x=2   hence P(X=4) = 6c4p4q2= 6c2p4q2  (since 6c4 = 6c2)
P(X=2) = 6c2p2q4
It is given 9P(X=4) = P(X=2)
hence      9.6c2p4q2 = 6c2p2q4 =   9 p2 =q2  

Let us take positive square root of both sides.
Then we get 3p= q  which can be written as 3p = 1-p
Transposing p to the other side we get        4p = 1    or  p = 1/4 = 0.25
Hence p =¼  or 0.25 

Algebra is widely used in day to day activities watch out for my forthcoming posts on how do you write an algebraic expression and algebra 2 homework solver. I am sure they will be helpful.           
                                              
Practice Problems on Binomial Theorem Probability:-


1.In a Binomial disribution the mean is 12 and standard deviation is 2,  Find n and p ( Answer  n=18,  p=2/3)

2.Ten coins are tossed simultaneously.Find the probability of getting
a) atleast 7 heads                                                                                                                     
b) exactly 7 heads                                                                                                                    
c) atmost 7 heads     
Answer  (a) 11/64     (b) 15/128   (c)=121/28     

3.A pair of dice is thrown 4 times.  Getting a doublet is considered a success, find the
probability of 2 success.   (answr 25/216)   

4) For a binomial distribution, mean = 7 and variance is 16.  Is this possible?  ( Answer  impossible)

5) Find the mean of a binomial distribution where  n = 10 and p = 3/5  (Answer   6)

Arc Length Chord

Introduction to arc length chord:

The arc length and the chord are the terms found in the circle parts where the arc is the curved and a part of the circle. The chord is nothing but the line containing the two end points which lies on the circle. The arc in the circle has two parts namely the major arc and the minor arc. The arc length can be determined with the help of the radius and the central angles. Let us see about the arc length and chord.

About Arc Length and the Chord:

The arc length of the circle are calculated with the help of the central angle values and the radius value of the circle.  The arc length of the circle are found with the help of formula as follows,

Arc length of circle = `(theta/360)`  × (2pr)

Where `theta` = central angle of the circle

r = radius of the circle.

The length of the chord can be determined by using the formula `2(r^(2)-d^(2))^((1)/(2))`

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Problems on Arc Length Chord:

Example 1:

Calculate the chord length of the circle that has the radius measure is about 8 cm that is 5 units from the middle.

Solution:

Now we have to find the length by using the formula as follows,

`2(r^(2)-d^(2))^((1)/(2))`

Length of the chord = `2(8^(2)-5^(2))^((1)/(2))`

= `2(64 - 25)^((1)/(2))`

= `2(39)^((1)/(2))`

= `2 xx 6.24`

= 5.28

Example 2:

Find the arc length of the curve in the circle, where the radius measurement is about 11 cm and the central angle measurement is about 80 degrees?

Solution:

The arc length are determined as follows,

Arc length = `(theta/360)` × (2pr)

Arc length =  `(80/360)` × (2 × 3.14 × 11)

We have to find the fraction of an angle by using the formula as `theta/360`

Let us substitute the value in the formula, then we get the value as

Arc length = 0.222 × 69.08

Arc length = 15.33 cm

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Direct Variation in Math

Introduction to direct variation in math
This topic has more influence in our day today life. It represents the change of a variable with respect to the other or others. If there is an increase in two related variables, it is called the direct variation.

Example:

A person wages can be decided based on the number of hours they work. If they work for more number of hours, the wages will be more.  If w represents the total wages, k the wage for an hour and h representing the number of hours, then it can be related as follows: `w/h` = k

Therefore the total wages is given by: w = kh.

Now let see few problems on this topic direct variation in math.

Examples on Direct Variation in Math

Ex 1: If w varies directly as h and given w = 125 and h = 25, find:

(i) the relation connecting w and h,

(ii) the value of w when h = 35,

(iii) the value of h when w = 250.

Sol: (i) Given: w ∞ h.

This implies, w = kh, where k is the constant of proportionality.

Therefore, when w = 125 and h = 25, we get:

w = kh

125 = k(25)

Therefore, k = `125/25` = 5.

Hence the equation is written as w = 5h.

(ii) When h = 35, w = 5(35)

= 175

Therefore, w = 175.

(iii) When w = 250, 250 = 5h

h = 250/5

h = 50.

Therefore, h = 50.

Ex 2: If s, the speed, varies directly as d, the distance and given s = 60 mph and d = 120 miles, find:

(i) the relation connecting s and d,

(ii) the value of s when d = 150,

(iii) the value of d when s = 45.

Sol: (i) Given: s ∞ d.

This implies, s = kd, where k is the constant of proportionality.

Therefore, when s = 60 and d = 120, we get:

s = kd

60 = k(120)

Therefore, k = 60/120 = 1/2.

Hence the equation is written as s = d/2.

(ii) When d = 150, s = ½(150)

= 75

Therefore, d = 75 mph.

(iii) When s = 45, 45 = ½ (d)

d = 2 * 45

d = 90.

Therefore, d= 90 miles.

More Example on Direct Variation in Math
Ex 3: The fare (F) of a taxi varies directly as the distance (D) travelled. If the distance is 50 km, the cost is $35. What will be the cost if the distance traveled is 125km?

Sol: Given: F ∞ D.

This implies that, F = kD

Therefore, when F = 35, D = 50, we have 35 = k(50)

K = 0.7.

Therefore, the equation can be written as F = 0.7D

Now, when D = 125km, F = 0.7× 125

= 87.5 dollars.

I hope the discussion on the direct variation would have been clear by now on going through those above examples.

Quadrilateral Pentagon Hexagon

Introduction to quadrilateral pentagon hexagon

A quadrilateral has four sides, four angles and two diagonals. The sum of the four angles of a quadrilateral must be 360. . The word quadrilateral is made of the words quad (denote four) and laterals (denote "of sides").There are different types of quadrilaterals are there, they are following,

Types of different quadrilaterals:

Square
Rhombus
Rectangle
Parallelogram
Trapezoid
Kite

Example on Quadrilateral Pentagon Hexagon

Problem: If the four angles of a quadrilateral are equal in measure. Find the measure of each angle.

Solution:

Let assume each angle is x ,

x + x + x + x = 360 degree

4x = 360 degree

Divide both sides by 4,

x=360/4

x=90 degree

Thus, each angle of this quadrilateral is a right angle.

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Example on Quadrilateral Pentagon Hexagon

Introduction:

Pentagon is 5 side polygons. The internal angle of pentagon is 540 degree .it is classified as regular pentagon and irregular pentagon based on its sides length and angle.

Regular pentagon:

All the sides and all the angles are equal then the pentagon is called Regular pentagon.

Irregular pentagon:

If any one side not equal and any one of the angle not equal then the pentagon is called irregular pentagon.

Perimeter of regular pentagon is =5 * side measurement

Example: Find the perimeter of regular pentagon with each side having 7 inches

Solution:

Perimeter =5*7

=35 inches

Example 2: Find the measure of each angle of a regular pentagon

Solution:

We know that regular pentagon having 5 sides

Total internal angle of pentagon is 540 degree

Therefore each angle = 540/5

=108 degree  

Example on Quadrilateral Pentagon Hexagon

Introduction:

A hexagon is a six side polygon. The total of the internal angles of any hexagon is 720 degrees. it is having six rotational symmetry  and reflection symmetry .it is classified as regular hexagon and irregular hexagon based on sides length and angle . All the sides and all the angles are equal then the hexagon is called Regular hexagon. if any one not equal then the hexagon is called irregular polygon.

Perimeter of Hexagon:

Perimeter for hexagon =6 *side length

Example: Find the perimeter of hexagon having each side length is 8 inches.

Solution:

Perimeter of hexagon = 6*side length.

= 6 *8

= 48 inches