Probability of a Intersection B

Probability of A Intersection B

Probability is the possibility of the outcome of an event of a particular experiment. Probabilities are occurs always numbers between 0 (impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example probability of getting a 6 when rolling a dice is 1/6. In this lesson we will discuss about probability problems using intersection rule.

Probability of a Intersection B – Example Problems

Example 1: A jar contains 5 red candies, 4 orange candies. If three candies are drawn at random, find the probability, that 1 is red candy and 2 are orange candies

Solution:

We have to select 3 candies, from 9 (5 + 4) candies.

n(S) = 9C3 = (9!)/(3!xx6!) = (9xx8xx7)/(3xx2xx1) = 84

Let A = Event of getting 1 red candy

B = Event of getting 2 orange candies

n(A) = 5C1 = `(5!)/(1!xx4!) ` = `5/1` = 5

n(B) = 4C2 =` (4!)/(2!xx2!) ` = `(4xx3)/(2xx1)` = 6

P(A) =` (n(A))/(n(S))` = `5/84`

P(B) = `(n(B))/(n(S))` = `6/84`

P(A intersection B) = P(A) ∙ (B) = `5/84` ∙ `6/84` = `30/7056`

P(A intersection B) = `5/1176` .



Example 2: A box contains 6 yellow marbles, 6 orange marbles. If four marbles are drawn at random, find the probability, that 2 are yellow marbles and 2 are orange marbles.

Solution:

We have to select 4 marbles, from 12 (6 + 6) marbles.

n(S) = 12C4 = `(12!)/(4!xx8!)` = `(12xx11xx10xx9)/(4xx3xx2xx1)` = 495

Let A = Event of getting 2 yellow marbles

B = Event of getting 2 orange marbles

n(A) = 6C2 = `(6!)/(2!xx4!)` = `(6xx5)/(2xx1)` = `30/2` = 15

n(B) = 6C2 = `(6!)/(2!xx4!)` = `(6xx5)/(2xx1)` = `30/2` = 15

P(A) = `(n(A))/(n(S)) ` = `15/495` = `1/33`

P(B) = `(n(B))/(n(S)) ` = `15/495` = `1/33`

P(A intersection B) = P(A) ∙ (B) = `1/33` ∙ `1/33` = `1/1089`

P(A intersection B) = `1/1089` .

Probability of a Intersection B – Practice Problems

Problem 1: A jar contains 4 lemon candies, 4 orange candies. If two candies are drawn at random, find the probability, that 1 is lemon and 1 is orange candy.

Problem 2: If P(A) = `1/5` , P(B) = `1/7` , P(A or B) = `1/9` , find (A and B)?

Answer: 1) `1/49 ` 2) `73/315`

Area of a Square Inscribed in a Circle

A square is a four sided figure; all the four sides are equal. If we situate a square inside the circle, all the four edges or vertices may touches the circle. A square is a quadrilateral, if the four sides of a square touch the circle, the four sides act as a four chords of a circle. The area of a square is measured in square units such as feet, inch, meter etc. In this article we shall discuss about the area of a square inscribed in a circle.

Area of a Square Inscribed in a Circle

If a square is inscribed in a circle and four sides of a square touches a circle.



The diagonals of the circle act as a diameter of the circle. We can find the area of a square is by:

Area of square = side times side

Area of the square = side side

= s2

Example:

Find the area of the square, sides of the square is 9ft.

Solution:

Given: Side = 8ft therefore diameter = 9ft

Area of a circle = side × side

= 9 × 9

= 81square ft.

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If the square does not touch the circle but inscribed in a circle, In such case we can also find the area of the square.

Example:

Find the area of a square inscribed in a circle, side of the square is 6ft.

Solution:

Given: side = 6t

Area of a square = side × side

= 6 × 6

= 36square ft.

The one or two sides of a square may touch the circle.

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Example for Area of a Square Inscribed in a Circle:

Example1:

Find the area of the square inscribed in a circle, sides of the square is 12m.

Solution:

Given: Side = 12

Area of a square = side × side

= 12 × 12

= 144m2.

Example2:

Find the area of the square inscribed in a circle, sides of the square is 23ft.

Solution:

Given: Side = 23ft

Area of a square = side × side

= 23 × 23

= 529square ft.

Poisson Probability Table

Introduction:

The allocation was first introduced by Simeon-Denis poisson (1781–1840) and in print, together with his probability theory, in 1838 in his work (Research on the Probability of Judgments in Criminal and Civil Matters). The job focused on sure random variables N that tally, among other things, the number of separate occurrences that take place during a time-interval of given length. In probability hypothesis and statistics, Poisson division is a discrete probability division expresses probability of a number of actions.

Poisson Probability Table Conditions:

Poisson probability distribution states two conditions

1.The amount of successes in two disjoint time intervals is autonomous.

2.The probability of a success during a small time interval is comparative to the entire length of the time interval.

Poisson probability formula:

The Probability distribution of a Poisson random variable, X  instead of the number of successes happening in a given time interval or a specified area of space is given by the formula:

P(X)=e-`nu` `nu`X / X!

where

X=0,1,2,3....

e = expected value

µ = mean figure of successes in the given time interval or area of space.

Mean and Variance for poisson probability:

If µ is the standard number of successes happening in a given time interval or area in the Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to µ.

E(X) = µ

and

V(X) = s2 = µ

Note: In a Poisson distribution, only one factor, µ is needed to determine the probability of an occurrence.
Poission Probability Table:

The below shows the value of poisson probability for the mean and X values of

`lambda` =0.1 to 5.5

X  =0 to 17

poisson probability table

Examples of Poisson Probability Distribution:

Example 1:

A life assurance salesman sells on the usual 3 life insurance policies per week. Use Poisson's law to work out the possibility that in a given week he will sell

(a) a few policies

(b) 2 or supplementary policies but less than 5 policies.

(c) assume that there are 5 working days per week, what is the probability that in a given day he will sell single policy?

Solution:

Here, µ = 3

(a) "Some policies" means "1 or more policies". We can work this out by finding 1 minus the "zero policies" probability:

P(X > 0) = 1 - P(x0)

Now P(X)=e-`nu`    `nu`X / X!    So P(X`@` )= e-3 30 / 3! = 4.9787 `xx` 10-2

So

Probability= P(X > 0)

=1 - P(x0)

=1-4.9787`xx`10-2

=0.95021

(b)     P(2`<=` X `>=` 5) = P(X2)+P(X3)+P(X4)

= (e-3 32 / 2!) + (e-3 33 / 3!) + (e-3 34 / 4!)

=0.61611

(c)    Average number of policies sold per day:  `3/5 ` = 0.6

So on a given day   P(X)=e-0.6 (0.6)1 / 1!

= 0.32929

Example 2:

A business makes electric motors. The probability an electric motor is imperfect is 0.01. What is the probability that a model of 300 electric motors will contain precisely 5 defective motors?

Solution:

The average number of defectives in 300 motors is µ = 0.01 × 300 = 3

The probability of getting 5 defectives is:

P(X ) = e-3 35 / 5!

= 0.10082

Intersection of Two Sets

Set is a fundamental part of the mathematics. This set concept is applied in every branch of mathematics. Sets are used in relations and functions. The application of sets are geometry,  sequences, Probability etc. Sets are used in everyday life such as a volleyball team, vowels in alphabets, various kinds of geometry shapes etc.

There are two main important operations in sets. They are  union and intersection of two sets. Let us learn the concepts and properties of intersection of two sets. We will learn some example problems and give practical problems about intersection of two sets.

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Intersection of Sets:

Let A and B be any two sets.  A intersection B is the set of all elements which are similar to both A and B. The symbol `nn` is used to denote the intersection. A intersection B is the set of all those elements which belong to both A and B. Symbolically, we write A `nn` B = { x : x `in` A and x `in` B }.

Ex:

Consider the two sets X = { 1, 5, 8, 9 } and Y = { 5, 6, 9, 15 } . Find X intersection Y.

Sol:

We see that 5, 9 are the only elements which are similar to both X and Y. Hence  X ∩ Y = { 5, 9 }

Some Properties of Operation of Intersection:

(i) Commutative law:  A ∩ B = B ∩ A

(ii) Associative law: ( A ∩ B ) ∩ C = A ∩ ( B ∩ C )

(iii) Law of identity: φ ∩ A = φ, U ∩ A = A (Law of φ and U).

(iv) Idempotent law: A ∩ A = A

(v) Distributive law: A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) i. e., ∩ distributes over ∪

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Practice Problems on Intersection of Two Sets:

Problem 1:

Find the intersection of each of the following two sets:

1. X = { 1, 3, 5 }   Y = { 1, 2, 3 }

2. A = [ a, e, i, o, u ]   B = { a, b, c }

3. A = { 1 , 2 , 3 }    B = `Phi`

Sol:

1. X `nn` Y = { 1, 3 }

2. A `nn` B = [ a ]

3. A `nn` B = `Phi`

Problem 2:

If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find

(i) A ∩ B  (ii) B ∩ C (iii) A ∩ C ∩ D

(iv) A ∩ C (v) B ∩ D (vi) A ∩ (B ∪ C)

(vii) A ∩ D (viii) A ∩ (B ∪ D)

Sol:

i) A ∩ B = { 7, 9, 11 }

ii) B ∩ C = { 11, 13 }

iii) A ∩ C ∩ D = Nill

iv) A ∩ C = { 11 }

v) B ∩ D = nill

vi) A ∩ (B ∪ C) = { 7, 9, 11, 13 }

vii) A ∩ D = nill

(viii) A ∩ (B ∪ D) = { 7, 9, 11}

Diameter of Intersecting Lines

Diameter of Intersecting Lines means we have to find the diameter of the lines which are intersecting the circle. Here we are going to learn about the diameter of the intersecting lines. Basically diameter refers the length of the line from one side of the circle to the other side of the circle which passes through the center. If a line intersecting the circle through its center means we ca say that line is the diameter.

Examples for Diameters of Intersecting Lines:

From the above diagram the line AB intersect the circle. This is passes through the center of the circle. The radius of the circle is r then the diameter of intersecting lines are d = 2r. We will see some example problems for it.

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Example 1 for diameter of intersecting lines:

A line AB is intersecting the circle through its center from the one side to another side. The circumference of the circle is 14 cm. Find the diameter of the intersecting lines.

Solution:

A line AB is intersecting the circle through its center from its one side to another side. We have to find the diameter of the intersecting lines. The diameter is d = 2r

Circumference of the circle C = 2πr

Here 2πr = 14 cm

So 2r = `14 / 3.14` = 4.45 cm

So the diameter of the intersecting lines is 4.45 cm

More Examples for Diameters of Intersecting Lines:

Example 2 for diameter of intersecting lines:

A line XY is intersecting the circle through its center from the one side to another side. The area of the circle is 22 cm2. Find the diameter of the intersecting lines.

Solution:

A line XY is intersecting the circle through its center from its one side to another side. We have to find the diameter of the intersecting lines. The diameter is d = 2r

Area of the circle A = πr2

Here πr2 = 22 cm

So r2 = `22 / 3.14` = 7 cm2

Radius r = 2.645 cm

So the diameter of the intersecting lines is 2r = 5.29 cm

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Average Deviation Calculator

The average deviation in which it is defined to calculate the average for the deviations for the given data. It is calculated by taking sum for all values of the deviation divided by the total number  of values in the given data set.

For finding the average deviation value, find the mean for the given data is using the formula,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

Calculate the deviation value by using the calculated mean value, by using the formula given below,

`(x - barx)^2`

Average Deviation is calculated by taking average for the founded deviation value of the data set.

Average Deviation = `"(sum_(K=1)^n (x-barx^2)) / N`

` T It is nothing but the above shown formulas the population of the variance formula in which it is also called as the average deviation.`

Steps to Calculate the Average Deviation:

1. Give the average for all the given dimensions of the data set .
2. Give the difference of the initial value of the data and the average value we have found which is called as mean difference.
3. Take the all absolute value from this mean difference of the given data.
4. Repeat the steps 2 and 3 for all the other given values and find the mean difference to the data set.

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Screen Shot for the Calculator to Find the Average Deviation:

For finding the average for the squared mean difference which is known as the average deviation of the given data set.



Select the formula to find the result and put the values in input field  then press calculate button.



Average Deviation Calculator - Example Problems:

Average deviation calculator - Problem 1:

Calculate the average deviation for the given data set. 35, 36, 37, 38.

Solution:

Mean: Formula for finding mean,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

`barx = (35+36+37+38) / 4`

` barx = 146 / 4`

` barx =` 36.5

Calculate the deviation for the given data set from the mean,

Deviation  =  `(x - barx)^2`

= `((35-36.5)^2+(36-36.5)^2+(37-36.5)^2+(38-36.5)^2)`

Deviation = 5

Calculate the average deviation for the given data,

Average Deviation = `(sum (x-barx)^2) / (n)`

= `5 / 4`

Average Deviation   = 1.25

Average deviation calculator - Problem 2:

Calculate the average deviation for the given data set. 33, 35, 36, 38.

Solution:

Mean: Formula for finding mean,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

`barx = (33+35+36+38) / 4`

` barx = 142 / 4`

` barx =` 35.5

Calculate the deviation for the given data set from the mean,

Deviation  =  `(x - barx)^2`

=`((33-35.5)^2+(35-35.5)^2+(36-35.5)^2+(38-35.5)^2)`

Deviation = 13

Calculate the average deviation for the given data,

Average Deviation = `(sum (x-barx)^2) / (n)`

= `13 / 4`

Average Deviation   =  3.25

Average deviation calculator - Problem 3:

Calculate the average deviation for the given data set. 2, 5, 6, 5, 4, 7, 8, 5.

Solution:

Mean: Formula for finding mean,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

`barx = (2+5+6+5+4+7+8+5) / 8`

` barx = 42/ 8`

` barx =` 5.25

Calculate the deviation for the given data set from the mean,

Deviation  =  `(x - barx)^2`

= `((2-5.25)^2+(5-5.25)^2+(6-5.25)^2+(5-5.25)^2+(4-5.25)^2+(7-5.25)^2+(8-7.25)^2+(5-5.25)^2)`

Deviation = 16.5

Calculate the average deviation for the given data,

Average Deviation = `(sum (x-barx)^2) / (n)`

= `16.5 / 8`

Average Deviation   = 2.0625

Average Deviation Calculator - Practice Problems:

1 Calculate the average deviation for the following 55.3, 56.6, 50.9 and 54.0.

Answer: Average Deviation = 4.47500

2. Calculate the average deviation for the following data

Answer: Average Deviation = 2

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Statistical Power Table

Statistical power is important in math. In math statistical power means probability of reject a false null hypothesis. In statistics to test hypotheses and also test the null hypothesis. The power is equal to 1-beta. In odd position we want to reject our null hypothesis in favor of the alternative. It is more helpful for exam preparation.

Basic Statistical Power of Negative Z-score Table:

In the following statistical power of negative z score table to explain the how to calculate the probability value




For example,

P(0 > Z > 2.31) = P( 0> Z > `oo` ) − P(-1.13 > Z >0)

= 0.5 − 0.1292 (Use statistical power of negative z score table to calculate the probability value)

= 0.3708

Select the first column value -1.13 then choose the right side eight column value 0.03 in the same direction we got an answer as 0.1292

Example Problems for Statistical Power Table:-

Problem 1:-

Calculate the standard deviation range of P(0 > Z > 1.9) by using statistical power table

Solution:

P(0 > Z > 1.9) = P(-1.9< Z<0)

= 0.0287 (Use statistical power of negative z score table to calculate the probability value)

Select the first column value -1.9 then choose the right side eleventh column value 0.0 in the same direction we got an answer as 0.0287

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Problem 2:-

Calculate the standard deviation range P(-2.2 > Z >3.1) by using statistical power table

Solution:

P(-1.2 > Z >2.1) = P(− 2.2 > Z > 0) + P(-3.1 > Z > 0)

= 0.0139+ 0.0010 (Use statistical power of negative z score table to calculate the probability value)

= 0.133

Select the first column value -2.2 then choose the right side eleventh column value 0.0 in the same direction we got an answer as 0.0139

Select the first column value -3.1 then choose the right side eleventh column value 0.0 in the same direction we got an answer as 0.0010

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Adding and Subtracting Vectors

The addition may be represented graphically by placing the start of the arrow b at the tip of the arrow a, and then drawing an arrow from the start of a to the tip of b. The new arrow drawn represents the vector a + b.



To subtract b from a, place the end points of a and b at the same point, and then draw an arrow from the tip of b to the tip of a. That arrow represents the vector a − b.
Source: Wikipedia



Adding and subtracting of vectors operation and example problems are given below.

Operations for on Vectors:

1. Addition of vectors:

Let `vec(OA)` = `veca` , `vec(AB)` =` vecb` . Join OB.

Then `vec(OB)` represents the addition (sum) of the vectors veca and vecb.



This is written as `vec(OA) ` +` vec(AB)` = `vec(OB)` Thus `vec(OB)` = `vec(OA) ` + `vec(AB) ` = `veca ` +` vecb`

This is known as the triangle law of addition of vectors which states that, if two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their sum is represented by the third side taken in the reverse order.

Applying the triangle law of addition of vectors in




ΔABC, we have BC + CA = BA ⇒ BC+ CA = − AB

⇒ AB + BC + CA = 0

Thus the sum of the vectors representing the sides of a triangle taken in order is the null vector.

2. Subtraction of vectors:

If `veca` and `vecb` are given two vectors, then the subtraction of `vecb` from `veca` is defined as the sum of `veca` and − `vecb` and  denoted by `veca` − `vecb` .



`veca ` − `vecb` = `veca` + ( − `vecb` )

Let `vec(OA)` = `veca` and `vec(AB)` =` vecb `

Then `vec(OB)` = `vec(OA)` + `vec(AB)` = `veca` + `vecb`

To subtract `vecb` from `veca` , produce BA toAB' such that AB = AB'.

∴ `vec(AB')` = − `vec(AB)` = −` vecb`

Now by the triangle law of addition

` vec(OB')` = `vec(OA)` +` vec(AB')` = `veca` + ( `-vecb` ) =` veca ` − `vecb`

Example Problems for Adding and Subtracting of Vectors:

Example problem 1:

The position vectors of the points A, B, C, D are `veca` , `vecb` , `2veca` + `3vecb` ,`veca` − `2vecb` respectively. Find `vec(DB) ` and `vec(AC)`

Solution:

Given that

` vec(OA)` = `veca` ; `vec(OB)` =` vecb` ; `vecOC` =` vec2 ` + `3vecb` ; `vec(OD)` = `veca ` − `2vecb `

`vec(DB)` = `vec(OB)` − `vec(OD)` = `vecb ` − (`veca` − `2vecb` ) = `vecb ` − `veca` + `2vecb` = `3vecb` − `veca`

`vec(AC)` = `vec(OC)` − `vec(OA)`

= (`2veca` + `3vecb` ) − `veca` = `veca` + `3vecb `

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Example problem 2:

` vec(OA)` = `2veca` + `3vecb` - `vecc` , `vec(OB)` = `4veca` + `2vecb` + `2vecc` , Find adding and subtracting of vectors

Solution:

Subtracting the two vector,

`vec(AB) ` = `vec(OB)` - `vec(OA)`

`vec(AB)` = `4veca` + `2vecb ` +` 2vecc` - `2veca` - `3vecb` +` vecc`

`vec(AB)` = `2veca` - `vecb ` + `3vecc`

Adding the two vector,

` vec(AB)` = `2veca` + `3vecb` - `vecc` + `4veca` + `2vecb` + `2vecc`

` vec(AB)` = `6veca` +` 5vecb` + `vecc`

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Accurate to Two Decimal Places

Decimal numbers contains decimal places. Here it is very important to know about the place value in decimal numbers.

So what is Decimal Place Value?

It can  be explained by taking example:



let us discuss it using whole number which does not have decimal point in it.

As we go from  right to left the value, the position or the place value gets 10 times bigger.

Similarly as we move from left to right the place value gets 10 times smaller.

What do we get as we move beyond unit place?

Here is where the decimal point comes into picture. If we move beyond unit place the value becomes 1/10, 1/100, 1/1000 and so on. These number which contains decimal point is called DECIMAL NUMBER.

Below is the example for decimal number:



here we can see that the place value of number goes on decreasing as we move beyond decimal points. The value of one in the above example is 1/10 , the place of 2 become 1/100

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Conversion of Decimal Number to Whole Number (accurate to Two Decimal Places)

HOW TO CONVERT  DECIMAL NUMBER TO WHOLE NUMBER?

This can be explained using eg given below:

consider 125.43 and also 125.56

First check the number immediately after the decimal  place. If it is greater than or equal to 5 then add 1 to the unit place and write it as whole number otherwise discard the numbers after decimal point and write the whole number as it is.

In the above example 1)  the number 4 appears immediately after decimal point which is less than 5 .So there is no change and we have to discard all the number proceeding decimal point i,e 4 and 3 and write the number as it is.

whereas in case of example 2) the number which appears immediately is 5 so add one to unit place which becomes 126


Accurate Writing of the Decimal Numbers to 2 Decimal Places

How to write numbers accurate to 2 decimal places?

Even here we have to follow the same rules mentioned above.

examples:

123.4612 ----------------------->123.46

123.4689 ------------------------> 123.47


In the above example 1) check the third number after the decimal point. In this case the third number is 1 which is smaller than 5 so don't add 1 to the number which in 1/100's  place and discard the 3rd digit after the point and write as it is. Here  the final number accurate to 2 decimal point becomes  1 2 3 4 . 4 6

example 2) here the 3rd digit after decimal point is 8 which is greater than 5 so add 1 to the second digit after decimal. Here  the final number accurate to 2 decimal point becomes  1 2 3 4 . 4 7

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