Compound Dependent Events

This page is based on compound dependent events which is a study under dependent events in statistics. In probability, an event is a one or more possible outcomes from an experiment. An event consisting of one or more simple events is called compound event. An event is called dependent event, if an event does affect the other event.

Compound Events Dependent - Examples

Ex 1: A bag contains 20 bulbs, in which 5 bulbs are defective bulbs. If 2 bulbs are taken at one by one without replacement, what is the probability of getting good and defective bulbs?

Sol:

Let S be the sample space, n(S) = 20

Number of good bulbs = 20 - 5 = 15

A be the event of getting a good bulb, n(A) = 15

B be the event of getting a defective bulb, n(B) = 5

P(A) = `(n(A))/(n(S))` = `15/20`= `3/4`

Now 19 balls will be remaining in the box, so

P(B) = `(n(B))/(n(S))` = `5/19`

P(A and B) = `3/4` · `5/19` = `15/76`.

P(Good and Defective bulb) = `15/76` .

Ex 2: A jar contains 7 yellow candies, 5 green candies, and 8 blue candies. If 2 candies are taken one by one without replacement, what is the probability of getting yellow and green candy?

Sol:

Let S be the sample space, n(S) = 7 + 5 + 8 = 20

A be the event of getting a yellow candy, n(A) = 7

B be the event of getting a green candy, n(B) = 5

P(A) = `(n(A))/(n(S))` = `7/20`

Now 19 candies will be remaining in a jar.

P(B) = `(n(B))/(n(S))` = `5/19`

P(A and B) =P(A) · P(B) = `7/20` · `5/19` = `7/76`

P(Yellow and Green) = `7/76`

Practice Problems

Pr 1: A bag contains 15 bulbs, in which 7 bulbs are defective bulbs. If 2 bulbs are taken at one by one without replacement, what is the probability of getting good and defective bulbs?

Pr 2: A jar contains 10 yellow candies, 6 green candies, and 4 blue candies. If 2 candies are taken one by one without replacement, what is the probability of getting yellow and green candy?

Ans: 1) ` 4/15` 2) `3/19`

Poisson Normal Distribution

Poisson normal distribution an Introduction:

Normal distributon:

In probability theory and statistics, the normal distribution or Gaussian distribution is an absolutely continuous probability distribution with zero cumulants of all orders higher than two.

Poisson distributon:

In probability theory and statistics, the Poisson distribution  (or Poisson law of large numbers) is a discrete probability distribution that expresses the probability of a number of events occurring in a fixed period of time if these events occur with a known average rate and independently of the time since the last event.


Poisson Distributon in Poisson and Normal Distribution:

The three basic assumptions on poisson distribution are as follows

The probability of one photon arriving in ?t is proportional to ?t when ?t is very small.

P(1;?t) = a?t for small ?t

where a is a constant whose value is not yet determined.

The probability that more than one photon arrives in ?t is negligible when ?t is very small.

P(0;?t)+P(1;?t) = 1 for small ?t

The number of photons that arrive in one interval is independent of the number of photons that arrive in any other non-overlapping interval.

Variance of the Poisson Distribution:

var(k) = E[k2] - E2[k] = at

Note that for a poisson distribution mean and the variance are equal .

Algorithm for poisson distribution:

Algorithm :Poisson random number:

init:

Let L ? e-?, k ? 0 and p ? 1.

do:

Generate uniform random number u in [0,1] and let p ? p × u.

while p > L.

return k - 1.



Normal Distribution in Poisson and Normal Distribution:

Description of normal distribution:

For any of situation in which the absolute value of a continuous variable is changed randomly from trial to trial.

The random improbability or random error.

Some properties of normal distribution Bell curve:

The curve will be Symmetric,

It will be unimodal,

The value will be extends to +/- infinity,

And the area under the curve is always one (it will be the fixed value).

The normal distribution bell curve is fully based on the mean and standard deviation functions.

These are some of the important points of  poisson and normal distribution.

Numerator of Fractions

Introduction to fractions

A fraction is a number that can represent part of a whole. A fraction consists two parts, a numerator and a denominator, the numerator represents the number of equal parts and the denominator represents number of equal parts make up a whole.

For example, `2/3` is a fraction. Here, the number above the fraction bar is called as the numerator (2) and number below the fraction bar is called as the denominator (3) of the fraction.

Here we are going to study about fractions

Sample Problems on Fractions

Here we are going to study about adding and subtracting fractions with same denominators and different numerators.

Example 1

`2/11` + `5/11`

Solution

Here the given problem is to add the fractions `2/11` and `5/11`

Notice that the denominators of both fractions are same, so we can just add the numerators.

So,

`2/11 + 5/11` = `(2 + 5)/11`

= `7/11`

So, the sum of `2/11` and `5/11` is `7/11`

Example 2

`6/14 - 3/14`

Solution

Here the given problem is to subtract `3/14` from `6/14`

Here the denominators are same, so we can just subtract the numerators.

`6/14 - 3/14` = `(6 - 3)/14`

= `3/14`

Few more Sample Problems on Fractions

Here we are going to study about adding and subtracting fractions with different numerators and denominators.

Example 1

`2/5 + 1/4`

Solution

Here the given problem is to add `2/5` and `1/4`

Notice that the denominators are different in the given fractions and we cannot add the fractions directly. The given fractions can be added with the help of least common denominator.

The least common denominator of `2/5` and `1/4` is 20. Because, 20 is the least common multiple of the denominators 5 and 4.

So, the fractions can be re-written as,

`2/5` = `2/5` * `4/4` = `8/20`

`1/4` = `1/4` * `5/5` = `5/20`

Now the problem becomes,

`8/20 + 5/20` = `(8 + 5)/20`

= `13/20`

Example 2

`3/7 - 1/3`

Solution

Here the given problem is to subtract `1/3` from `3/7`

Notice that the denominators are different in the given fractions and we cannot subtract the fractions directly. The given fractions can be subtracted with the help of least common denominator.

The least common denominator of `1/3` and `3/7` is 21. Because, 21 is the least common multiple of the denominators 3 and 7.

So, the fractions can be re-written as,

`3/7` = `3/7` * `3/3` = `9/21`

`1/3` = `1/3` * `7/7` = `7/21`

Now the problem becomes,

`9/21 - 7/21` = `(9 - 7)/21`

= `2/21`

Antiderivative of Log X

Introduction to anti-derivative of log x:

The natural logarithm is the logarithm to the base e, where e is an irrational constant approximately equal to 2.718. The natural logarithm is generally written as ln(x), loge(x) or sometimes, if the base of e is implicit, as simply log(x). Formally, ln(a) may be defined as the area under the graph of `1/x ` from 1 to a, that is as the anti-derivatives or integral,

ln a  = `int_1^a(1/x)dx`

This defines a logarithm because it satisfies the fundamental property of a logarithm:

ln (ab) = ln a + ln b

Source Wikipedia.

Anti-derivative Logarithmic Formulas:

1. `int` `(1/x)` dx = log x + c

2.  `int` e x dx = e x + c

3. `int (dx) / (a^2 - x^2) ` = `(1/(2a)) log [(a + x) / (a - x)] + c`

4. `int (dx) / (x^2 - a^2)` = `(1/(2a)) log [(x - a) / (x + a)] + c`

5. `int (dx) / sqrt(a^2 - x^2)` = `sin^-1(x / a) + c`

6. `int (dx) / sqrt(x^2 - a^2) ` = `log [(x + sqrt(x^2 - a^2)] + c`

Anti-derivative Logarithmic Problems:

Anti-derivative logarithmic problem 1:

Find the anti-derivative of given logarithmic function, log x  with respect to x

Solution:

Given logarithmic function, ` int ` log x. dx

Let,    u = log x                            dv = dx.

`(du)/(dx) ` = `1/x`                                    v = x

du = `1/x` dx

We know anti-derivative parts formula,  `int ` u dv = uv - `int ` v du

`int ` log x. dx  = log x . x - `int`` x ((dx) /x) `

= x. log x - `int` dx

= x. log x - x + c

= x( log x - 1) + c

Answer: Anti-derivative of log x is    x( log x - 1) + c

Anti-derivative logarithmic problem 2:

Find the anti-derivative of given logarithmic function, `(1 + 25x)/x^2`   with respect to x

Solution:

Given function, ` int` `(1 + 25x)/x^2` . dx

`int``(1 + 25x)/x^2`. dx  =` int` `dx/x^2` + `int` ` (25x)/x^2 ` dx            

= `int` `x^(-2) ` dx + ` int `` 25x^(-1)` dx               

= `x^(-1)` +  25 log x + c

Answer:  Anti-derivative of  `(1 + 25x)/x^2`  is x-1 + 25 log x + c

Anti-derivative logarithmic problem 3:

Find the anti-derivative of given logarithmic function, `e^(3x)/(1- e^(3x))` with respect to x

Solution:

Let u = 1- e3x           du = - `3` `e^(3x)` dx      

So, substitute the u and du

`int `  `e^(3x)/(1- e^(3x))` dx = `int`` (-1/3)(du)/u`                                 

= `(-1/3)` ` int` `1/u` du

= `(-1)/3` ln u + c                                                      we know u = 1- e3x

=` (- ln (1-e^(2x)))/3` +c

= `((-1)/3)` ln(e3x -1) + c

Answer:  Anti-derivative of `e^(3x)/(1- e^(3x))` is  `((-1)/3)` ln(e3x -1) + c

Subtracting Mixed Fractions

Introduction subtracting mixed numbers:

Natural number and a fraction is called mixed number. If numerator is less than denominator then it is said to be proper fraction. Improper fraction is formed when numerator is greater than denominator. For subtracting mixed numbers we have to first convert it in to proper fractions and then subtract the numbers so obtained.

Problem Subtracting Mixed Numbers with Different Denominators:

Pro 1 : 2 `3/5` - 3` 5/7`

Sol : Here the both are mixed fraction,

Step 1:  convert into the proper fraction and subtracting mixed numbers

2 `3/5` first we have to change into the proper numbers `(5xx2+3)/5` , we get

`13/5` is the proper numbers

3 `5/7=` `[(7xx3)+5]/7` = `26/7 ` is the proper number

Here we round mixed numbers and adding both functions

`13/5` -` 26/7`

In this step subtracting mixed numbers taking the Least Common Multiplier of 5 and 7 is 35. `[(13 xx 7) - (26 xx 5)]/(35)`  = `(91-130)/35 = (-39)/(35) `= -1.1142

Problem Subtracting Mixed Numbers with same Denominator:

Ex 1:      2 `3/4` - 3 `5/4`

Sol:  Here the fraction 4 is the equivalent fraction mixed number, subtracting both mixed numbers

2 `3/4` first we have to change into the proper numbers `(4xx2+3)/4` we get

`11/4 ` is the proper numbers

3 `5/4 = [(4xx3)+5]/(4) = (17)/(4)`

Here we round mixed numbers and adding both functions

`11/4 ` - `17/4 = (11-17)/(4) = (-6)/(4) = (-3/2) = -1.5`

Tangent Function Graph

Introduction to tangent function graph:
In geometry, the tangent line (or simply the tangent) to a curve at a given point is the straight line that "just touches" the curve at that point (in the sense explained more precisely below). As it passes through the point where the tangent line and the curve meet, or the point of tangency, the tangent line is "going in the same direction" as the curve, and in this sense it is the best straight-line approximation to the curve at that point. The same definition applies to space curves and curves in n-dimensional Euclidean space.

Source: Wikipedia.


Definitions of a Trigonometry Tangent Function:

Definition of tangent function is defined by using the right angle triangle

Tangent of an angle is the ratio of length of the adjacent and the opposite side.
tan x = opposite / adjacent

The tangent is the inverse of the cotangent and is given by
tan x = 1 / cot x

Quotient of sine and cosine functions is called as tan or tangent function.
tan x = sin x / cos x

Properties of a Tangent Function Using the Graph



Example Problem for Tangent of a Graph:

Example 1:

Find the function value of tan 35 o.

Solution:       

Use the tangent's co function identity to solve the problem.

tan x = cot (90o – x)
tan 35o = cot (90o – 35o)
= cot (55o)
tan 35o = 0.70021
The tangent of 60o is 0.70021.

Example 2:

Find the hypotenuse of the right angle triangle using the Pythagorean Theorem and find the tangent of an angle value.

Solution:

Use the Pythagorean Theorem, for finding the hypotenuse

In the given right angle triangle

AC2 = AB2 + BC2

Here,

AB = Opposite side

BC = Adjacent side

AC = Hypotenuse

AC2 = AB2 + BC2

= 42 + 92

= 16 + 81

AC2 = 97

AC = 9.84

Hypotenuse for the given right angle triangle is 9.84

Tangent function:

Tan ? = Sin ?/ Cos ?

= adj / opp

= 4 / 9

Tan ? = 4 / 9

? = tan -1(4/9)

? = 23.96 °

Solving Distance Formula

Introduction to solving distance formula:
Distance formula is a formula which is used to find the distance between two points (x1, y1) and (x2, y2). This formula is based on the Pythagorean Theorem.Solving problems using distance formula is one of the easier method to find the distance between two given points.

The distance d between the points (x1, y1) and (x 2, y2) is given as,

d = v ((x2 - x1)2 + (y2 - y1)2)

Example Problems on Solving Distances Using Distance Formula:


1) Find the distance between the points (-2, -3) and (-4, 4).

Solution:

Let (x1, y1) = (-2, -3)

(x2, y2) = (-4, 4)

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

=  v (((-4 - (-2))2 + (4 - (-3))2)

=  v ((-2)2  + (7)2)

=  v (4 + 49)

= v53

Solving for d, we get d  ˜ 7.28

2) Find the distance between the points (2, 4) and (5, -1).

Solution:

Let (x1, y1) = (2, 4)

(x2, y2) = (5, -1)

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

=  v ((5 - 2)2 + (-1-4)2)

=  v (32 + (-5)2)

=  v (9 + 25)

= v 34

Solving for d, we get d ˜ 5.83

3) Find the distance between the points (-2, 1) and (1, 5).

Solution:

Let (x1, y1) = (-2, 1)

(x2, y2) = (1, 5)

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

=  v ((1 - (-2))2 + (5 - 1)2

=  v (32 + 42)

=  v (9 + 16)

d  = v 25

Solving for d, we get   d = 5

4) If the distance from the point is (1, 2) to the point (3,y) is v8.  Find the value of  y.

Solution:

Let (x1, y1) = (1, 2)

(x2, y2) = (3, y)

d = v8

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

v8 = v ((3 - 1)2 + (y - 2)2

8 = 22 + (y - 2)2

8 = 4 + (y - 2)2

8 - 4 = (y 2 - 4y + 4)

4 = y 2 - 4y + 4

0 =  y 2 - 4y

0 = y (y - 4)

Solving for y. we get

y = 0 ;   y = 4
   


Solve the Problems Using Distance Formula:

Find the distance between the following points,

(1, 4) and (4, 0)

(2, 8) and  (16, 4)

(8, 5) and (9, 6)

(5, 6) and (-12, 40)

Solutions:

d = 5
d = 14.56
d = 1.41
d = 38.01

Tangent and Secant of a Circle

Introduction to tangent and secant of a circle:

A tangent of a circle is a line drawn from a point passing through the circle just at one point and secant is a line passing through two points of a circle. In case, a tangent and a secant are drawn from the same point outside the circle, we can find an interesting relation. Similar is the case when two secants are drawn from the same point outside the circle.

Let us study these two situations.

Tangent and Secant of a Circle – a Tangent and a Secant

Look at the above diagram.

A tangent OT is drawn from O touching the circle at P. From the same point O, a secant OS is drawn passing through the points Q and R on the circle. Join PR and PQ.

The angle OPQ and ORP are congruent as both of them are subtended by the intercepted arc PQ.

Angle POR is subtended both by the chords PQ and PR at O. Therefore the triangles OPQ and ORP are similar.

Applying the rule of similarity,

OP/OR = OQ/OP

or,  OP2 = OQ*OR



Tangent and Secant of a Circle – Two Secants

Look at the above diagram.

Two secants are drawn from the same point O to the circle. One secant passes through the points A and B on the circle and the other passes through the points C and D on the circle. Join BC and DA.

The angles BAD and BCD are subtended by the same arc BD on the circumference on the circle. Hence these two angles are congruent. The angle at O is common to the triangle AOD and BOC. Therefore, the triangles AOD and BOC are similar.

Applying the rule of similarity,

OC/OA = OB/OD

or,   OA*OB = OC*OD 

The two properties derived are very useful in solving problems related circles with tangents and secants.

Adding Negative Exponents

Introduction :

The term exponent in math is used to find the exponential value of the particular value it may be integer  or fraction . We can easy to get  the power of a  number using online calculator ,Consider  the unknown number , Here x is base and y is the power  of x

Adding Exponents means how many times to divide the number .

that is , `a ^(-n) = 1/a^n`

Steps for Adding Negative Exponents :

Negative exponents are added in the same way as the exponents are added with just a negative sign.

The given terms exponents are combined in a way such that the negative exponents are added or combined in case of same base.
In case of different bases , we have to simplify by convert it to positive exponent ,
Then we have to simplify the exponent value . Make the negative exponent to the positive one .
Now simplify further to get the results.
Ex  :   `5^-2 + 5^-4 = 5^((-2-4))`

`=5^-6`

=`1/5^6`

=`1/15625`

= 0.000064



Examples to Add Negative Exponents:

Ex 1:  A number  with negative exponents `8^(-3)`

Sol :     `8^(-3)` = `1/ (8^3)`

=`1/ (8 *8*8)`

= `1/ 512`

=0.00195

Ex 2:    Add the neagtive exponents` 7^-3`    + `5^-2`

Sol :    `=1/7^3 + 1/5^2`

`=1/343 +1/25`

=0.0029 +0.04

=0.0429

Ex 3:   Add the negative exponents `6^-2 + 6^-3`

Sol :  Using the rule , we have to add the exponents as 6 is the common base

= `6^(-2 + -3)`

` 6^(-2-3)`

=`6^-5`



= `1/(6^5)`

= 0.00012

Example 4:

Add the negative exponents  of `5^-3 +5^-5 +5^-2`

Solution:

Use the property to add the negative exponents , we have

=`5^-3 +5^-5 +5^-2`

=`5^(-3 + -5 + -2)`

=`5^(-3 -5 -2)`

= `5^-10`

= `1/(5^10)`

=0.04832

Example 5:

Add the negative exponents  of` 3^-3 + 3^-6`

Solution:

Use `a^-n = 1/a^n`   rule

We  get ,

=`1/3^3 +1/3^6`

=0.03703 +0.0013

=0.03833

Example 6:

Add the negative exponents  of `5^-2 + 5^-3`

Solution:

Use a ^-n = 1/a^n

Then we  get ,

=`1/5^2 +1/5^3` 

=0.04+0.008

=0.048

Practice Problem to Add Negative Exponents:

Pro1 : Add the negative exponents  of` 2^-2 +2^-4`

Ans : 0.3125

Pro2 :Add the negative exponents  of` 3^-3 +3^-2`

Ans : 0.1481