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Probability Rules

- Calculus
- Absolute Maxima and Minima
- Absolute and Conditional Convergence
- Accumulation Function
- Accumulation Problems
- Algebraic Functions
- Alternating Series
- Antiderivatives
- Application of Derivatives
- Approximating Areas
- Arc Length of a Curve
- Area Between Two Curves
- Arithmetic Series
- Average Value of a Function
- Calculus of Parametric Curves
- Candidate Test
- Combining Differentiation Rules
- Combining Functions
- Continuity
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- Derivative Functions
- Derivative of Exponential Function
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- Derivatives
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- Derivatives of Inverse Trigonometric Functions
- Derivatives of Polar Functions
- Derivatives of Sec, Csc and Cot
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- Determining Volumes by Slicing
- Direction Fields
- Disk Method
- Divergence Test
- Eliminating the Parameter
- Euler's Method
- Evaluating a Definite Integral
- Evaluation Theorem
- Exponential Functions
- Finding Limits
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- First Derivative Test
- Function Transformations
- General Solution of Differential Equation
- Geometric Series
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- Hydrostatic Pressure
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- Implicit Differentiation Tangent Line
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- Mechanics Maths
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- Power
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- Probability and Statistics
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- Frequency, Frequency Tables and Levels of Measurement
- Independent Events Probability
- Line Graphs
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- Mutually Exclusive Probabilities
- Probability Rules
- Probability of Combined Events
- Quartiles and Interquartile Range
- Systematic Listing
- Pure Maths
- ASA Theorem
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- Addition and Subtraction of Rational Expressions
- Addition, Subtraction, Multiplication and Division
- Algebra
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- Determinants
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- Differentiation
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- Direct and Inverse proportions
- Disjoint and Overlapping Events
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- Double Angle and Half Angle Formulas
- Drawing Conclusions from Examples
- Ellipse
- Equation of Line in 3D
- Equation of a Perpendicular Bisector
- Equation of a circle
- Equations
- Equations and Identities
- Equations and Inequalities
- Estimation in Real Life
- Euclidean Algorithm
- Evaluating and Graphing Polynomials
- Even Functions
- Exponential Form of Complex Numbers
- Exponential Rules
- Exponentials and Logarithms
- Expression Math
- Expressions and Formulas
- Faces Edges and Vertices
- Factorials
- Factoring Polynomials
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- Factorising expressions
- Factors
- Finding Maxima and Minima Using Derivatives
- Finding Rational Zeros
- Finding the Area
- Forms of Quadratic Functions
- Fractional Powers
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- Fractions
- Fractions and Decimals
- Fractions and Factors
- Fractions in Expressions and Equations
- Fractions, Decimals and Percentages
- Function Basics
- Functional Analysis
- Functions
- Fundamental Counting Principle
- Fundamental Theorem of Algebra
- Generating Terms of a Sequence
- Geometric Sequence
- Gradient and Intercept
- Graphical Representation
- Graphing Rational Functions
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- Graphs
- Graphs and Differentiation
- Graphs of Common Functions
- Graphs of Exponents and Logarithms
- Graphs of Trigonometric Functions
- Greatest Common Divisor
- Growth and Decay
- Growth of Functions
- Highest Common Factor
- Hyperbolas
- Imaginary Unit and Polar Bijection
- Implicit differentiation
- Inductive Reasoning
- Inequalities Maths
- Infinite geometric series
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- Instantaneous Rate of Change
- Integers
- Integrating Polynomials
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- Integrating e^x and 1/x
- Integration
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- Interest
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- Iterative Methods
- L'Hopital's Rule
- Law of Cosines in Algebra
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- Laws of Logs
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- Math formula
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- Notation
- Number
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- Product Rule
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- Scale Drawings and Maps
- Scale Factors
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- Sector of a Circle
- Segment of a Circle
- Sequences
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- Series Maths
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- Special Products
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- Surds
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- The Quadratic Formula and the Discriminant
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- Transformations of Graphs
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- Bias in Experiments
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- Categorical Variables
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- Chi Square Test for Goodness of Fit
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- Chi-Square Distribution
- Combining Random Variables
- Comparing Data
- Comparing Two Means Hypothesis Testing
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- Conducting a Study
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- Confidence Interval for Population Mean
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- Data Analysis
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Jetzt kostenlos anmeldenJust as we have different multiplicative and addition properties for real numbers, the same is the case for **probabilities of events. **

While dealing with only one event, we don’t need to look for specific properties to model the property of the event. But if we have two or more events, then it becomes really tricky to model the properties of such events, and especially relating these events. Hence there are **Probability Rules **to relate different events by examining their properties.

In such cases, we need to relate such events as well as their intersections and unions, let us explore more about these properties, starting with **The Addition rule of ****Probability.**

Consider two events **A **and **B, **such that they are a part of the sample space **S.** Let $P\left(A\right)$ and$P\left(B\right)$ be the probabilities of events A and B, respectively.

The **Addition law of probability **also referred to as the addition rule or the sum rule, states that the probability of both events occurring that is the union of A and B is given by

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

where *$P\left(A\cup B\right)$ *denotes the probability of the occurrence of A or B, and $P(A\cap B)$ denotes the probability of the occurrence of A and B.

We suppose that A and B are non-empty events and that their intersection is not the empty set.

To understand how the above formula is derived, let us visualize sets A and B as a part of a Venn diagram.

We recall that **Venn Diagrams **are diagrams where the sets and the sample space are depicted as geometric figures in order to better understand their unions, complements, and intersections.

Consider the Venn diagram below.

In the above diagram, the green rectangle represents the sample space, and the two blue circles represent events A and B respectively.

If we want to find the probability of the event "A or B", it will be their union, and we can use the Venn diagram to see how we can do it.

If we add the probabilities of A and B, their intersection will be counted twice, instead of once. Hence, we need to subtract the intersection of them. This gives us:

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

For any events, the word ‘**and**’ implies their ’**intersection’ **and the word **’or’ **implies their **‘union’**

The Addition rule can also be extended to three events, namely A, B and C,

$P\left(A\cup B\cup C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$

where the same idea of Venn diagrams can be considered to derive the formula.

Given two events X and Y with probabilities of their occurrence are respectively 0.3 and 0.4. The probability of the occurrence of X and Y is 0.1, find the probability that X or Y occurs.

**Solution**

The probabilities of X and Y are given as $P\left(X\right)=0.3$ and $P\left(Y\right)=0.4$.

The probability of both of them occurring is $P\left(X\cap Y\right)=0.1$.

We are asked to find the probability of X or Y which is nothing but the probability of their union$P\left(X\cup Y\right)$.

Hence, we can use the **Addition rule **to find it,

$P\left(X\cup Y\right)=P\left(X\right)+P\left(Y\right)-P\left(X\cap Y\right)$

Substituting the appropriate values, we get

$P\left(X\cup Y\right)=0.3+0.4-0.1=0.6$

Hence the probability of event X or Y happening is 0.6.

It can be the case sometimes that two events have nothing to do with each other when their intersection is a null set.

Two events, A and B are called **Disjoint events **if their intersection is a null set that is

$A\cap B=\overline{)O}$

Now, in order to find the probability of the union of two disjoint events, we use the addition rule

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

Noting that A nd B are disjoint events, we have

$P(A\cap B)=P\left(\overline{)O}\right)=0$

Now plug in the value of their intersection, we have

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)$

This is tossed twice and the outcomes are noted, find the probability that the first outcome is 1 and the second outcome is an even number.

A dice is tossed twice and the outcomes are noted, find the probability that the first outcome is 1 and, the second outcome is an even number.

**Solution**

Note that 1 is not an even number, so the two events are disjoint in this case. The reason being that the outcomes of an even number appearing does not overlap with the outcome of 1 appearing on the first toss.

Let the two events be A and B respectively,

$P\left(A\right)=\frac{1}{6}$

since 1 is an outcome of 6 possibilities, and,

$P\left(B\right)=\frac{3}{6}$since there are 3 even numbers from all the 6 possibilities.

We want to find $P\left(A\cup B\right)$. We use thus the addition formula for disjoint events as the occurrence of one event does not affect the occurrence of the other. Hence we have,

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)$

$P\left(A\cup B\right)=\frac{1}{6}+\frac{3}{6}=\frac{2}{3}$

Thus, the probability of getting 1 on the first toss and an even number on the second toss is $\frac{2}{3}$.

We consider two events, A and B such that the probability of event A is dependent on event B. In other words, the outcome of A will differ depending on whether B occurred or not. We write an event ‘A|B’ which is read as ‘event A given that event B occurred’.

The probability that governs such events is known as **Conditional Probability, **where an event is based on the condition that another event has happened.

Using the concept of conditional probability, we can outline a formula for the **Product rule. **It is given as follows,

$P\left(A\cap B\right)=P\left(B|A\right)P\left(A\right)$

Verbally, ‘the probability of the occurrence of A and B is equal to the product of the probability of B occurred A and the probability of A itself’. We can also extend this law to three events, namely, A, B and C,

$P\left(A\cap B\cap C\right)=P\left(C|A\cap B\right)P\left(B|A\right)P\left(A\right)$

Similar expressions can be derived for as many events as one wants.

Consider the below example to understand how the occurrence of an event depends on the occurrence of a preceding event.

We consider 2 bags, an orange bag and a black bag. There are 4 candies in the orange bag and 5 candies in the black bag. There are also 2 chocolates in the orange bag and 3 chocolates in the black bag. A sweet is randomly chosen from any of the bags, what can be said about the probability that the chosen sweet is a candy?

**Solution**

Let A be the event that the chosen sweet is a candy and let B be the event that the candy was chosen from the orange bag. Let C be the event that the bag chosen was the black one.

We can see here that the probability of getting a candy depends on the bag we have chosen. If the sweet is chosen from the black bag, the probability is different if it was chosen from the orange bag.

Consider the above diagram, here all the possible events are branched in order to better understand the probabilities.

(i) If the sweet was chosen from the black bag, we say that the **‘probability of getting the candy given that it came from the orange box’** .

According to the events we defined, the probability of this event is denoted as $P\left(A|B\right)$ and is read as ‘A given B has occurred’.

(ii) If the sweet was chosen from the orange box, the probability of getting a candy is denoted as $P\left(A|C\right)$ and is read as ‘A given C has occurred’.

Let us revisit the example we saw earlier, and calculate the probability using the Product rule.

There are 2 bags, an orange bag and a black bag. There are 4 candies in the orange bag and 5 candies in the black bag. There are also 2 chocolates in the orange bag and 3 chocolates in the black bag. A sweet is randomly chosen, find the probability that the chosen sweet is a candy and it came from the black bag.

**Solution**

Let A be the event that the chosen sweet is a candy and let B be the event that the sweet was chosen from the orange bag. Let C be the event that the bag chosen was the black one.

We want to find the probability that the sweet chosen is a candy given that the bag is black, hence we want to find $P\left(A\cap C\right)$.

Using the product rule we have,

$P\left(A\cap C\right)=P\left(A|C\right)P\left(C\right)$

The probability that the sweet came from the black bag is

$P\left(A\cap C\right)=\frac{candiesintheblackbag}{totalsweetsinblackbag}=\frac{5}{8}$

and the probability of choosing the black bag is 1/2 as there are only two bags,

$P\left(C\right)=\frac{1}{2}$

Substituting these values, we get,

$P\left(A\cap C\right)=\frac{5}{8}\xb7\frac{1}{2}=\frac{5}{16}$

Hence, the probability that the sweet is a candy and it came from the black bag is $\frac{5}{16}$.

Two events are **Independent **of each other is the occurrence of one does not affect the occurrence of another in any possible way.

This can be extended to any finite number of events as long as they do not affect the probability of each other. An important property of independent events can be expressed as a formula,

$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$

In other words, the probability of the intersection of two independent events is the product of the individual probabilities.

Jason and William are playing cards, Jason asks William to draw a random card. William draws a queen and puts it back in the deck. Jason asks him to draw another card and asks him the probability that this card is a king followed by the previous queen. What should be William’s answer?

**Solution**

Let A be the event that the card drawn is a queen and B be the second card drawn is a king.

One should note that it doesn’t matter what William chooses as his first card, both events are completely independent of each other.

Calculating the individual probabilities, we get

$P\left(A\right)=P\left(B\right)=\frac{4}{52}=\frac{1}{13}$

As there are four queens and four kings in a deck of 52 cards, we want to find the probability of the intersection of the two events, using the fact that the events are independent.

$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)=\frac{1}{13}\frac{1}{13}=\frac{1}{169}$.

Are the following two events independent?

A : The rising of the sun

B : Tossing a coin and getting a head

**Solution**

YES They are!

Events A and B are independent as they are completely unrelated and the occurrence of one does not affect the other. The rising of the sun is without any doubt, independent of the outcome of flipping a coin.

Suppose a fair coin is tossed once, there are two possible outcomes: Heads (H) or Tails (T), with equal probabilities. Both events are exactly the opposite of one another, if one event occurs, then all the other outcomes except it are known as the **Complementary events **of it. If we get H when we toss the coin, every other event (in this case, T) is a complementary event of it.

The complement of an event, lets say A, is the subset of the sample space which is not contained in A itself. It is denoted by A^{c}.

This implies that the probability of the complement will be given by all the elements that are in the sample space excluding set A.

${A}^{C}=S-A$

$P\left({A}^{C}\right)=P\left(S\right)-P\left(A\right)=1-P\left(A\right)$

which is also known as the **Complement Rule **of probability.

A sample space is defined by the set S = {1,2,5,7,8,9} and a subset of S is given by

A = {2,5,8}. Find the complement of set A.

**Solution**

According to the definition of the complement of a set, all the elements that are in S but not in A form the set of A^{c}, which here form the set {1,7,9}.

Thus the complement of A is A^{c} = {1,7,9}.

A coin is tossed twice and the outcome is observed. What is the probability that at least 1 head is observed?

**Solution**

The sample space is given by S = {HH,HT,TH,TT} and let A be the set (or the event) consisting of the elements when we have more than 1 head.

The complement of this event will be ‘the event when we have less than 1 head’. Hence ${A}^{c}$ will be given by the set {TT} as it is the only element in the sample space where we have less than 1 head (no heads).

Hence, we can now use the complement rule to calculate its probability

$P\left(A\right)=1-P\left({A}^{C}\right)$

$P\left(A\right)=1-\frac{1}{4}=\frac{3}{4}$

Hence, the probability of getting at least 1 head is $\frac{3}{4}$.

- The union and intersection of any number of events can be related by
**The Addition rule of probability**and for any 2 events A and B, it is given by$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P(A\cap B)$. For two disjoint events, A and B, the**Addition rule**is given by $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)$. - For any two events, A and B, the
**Product rule of probability**is given by $P\left(A\cap B\right)=P\left(A|B\right)P\left(B\right)$ and this law can be extended to any number of events in a similar way. - Two events are said to be
**independent**of each other if the occurrence of one does not affect the occurrence of another, and the same goes for any number of events. - If two events are independent of each other, their
**Product rule**is given by $P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$. - The
**c****omplement of an event**, lets say A, is the subset of the sample space which is not contained in A itself. It is denoted by A^{c}. An event and its complement are related by the equation, $P\left(A\right)=1-P\left({A}^{C}\right)$.

More about Probability Rules

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