What are my chances of getting into Harvard?‘ or ‘What’s my probability of getting scholarships from Oxford?‘ we get tongue-tied. There are so many variables at play, it’s difficult to give an accurate answer.

But when you get probability questions in your GRE and GMAT exam syllabus, you don’t have to get flummoxed.

Understanding the basic rules and formulas of probability will help you score high in the entrance exams Advanced Probability Theory Stats Homework, assignment and Project Help, Advanced Probability Theory Assignment Math Tools: Online Math tools to help with homework or studying. StudyDaddy is the place where you can get easy online Statistics homework. Spend less money when buying from our drugstore..

Meaning and definition of ProbabilityAs the Oxford dictionary states it, Probability means ‘The extent to which something is probable; the likelihood of something happening or being the case’. In mathematics too, probability indicates the same – the likelihood of the occurrence of an event.

Examples of events can be :Tossing a coin with the head upDrawing a red pen from a pack of different coloured pensDrawing a card from a deck of 52 cards etc. Either an event will occur for sure, or not occur at all.

Or there are possibilities to different degrees the event may occur. An event that occurs for sure is called a Certain event and its probability is 1.

An event that doesn’t occur at all is called an impossible event and its probability is 0. This means that all other possibilities of an event occurrence lie between 0 and 1. This is depicted as follows: 0<= P(A)<= 1where A is an event and P(A) is the probability of the occurrence of the event.

This also means that a probability value can never be negative.

Every event will have a set of possible outcomes 1 Oct 2007 - Probability theory is fundamentally important to inferential statistical analysis. This is often the case, even when the sample means are significantly different. The study comparing two hypotensive agents could be designed differently. Royal College of Anaesthetists College of Anaesthetists of Ireland .

When a coin is tossed, the possible outcomes are Head and Tail. Similarly when two coins are tossed, the sample space is (H,H), (H,T), (T,H), (T,T) . The probability of head each time you toss the coin is 1/2.

Basic formula of probabilityAs you might know from the list of GMAT maths formulas, the Probability of the occurrence of an event A is defined as:P(A) = (No. of possible outcomes) Another example is the rolling of dice.

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What is the probability of rolling a 5 when a die is rolled?No. of possible outcomes = 6So the probability of rolling a particular number when a die is rolled = 1/6. Compound probabilityCompound probability is when the problem statement asks for the likelihood of the occurrence of more than one outcome.

P(A or B) is the probability of the occurrence of atleast one of the events.

P(A and B) is the probability of the occurrence of both A and B at the same time. Mutually exclusive events:ORNote: For a mutually exclusive event, P(A and B) = 0. or a 5 when a die is rolled?Solution:Taking the individual probabilities of each number, getting a 2 is 1/6 and so is getting a 5.

Applying the formula of compound probability,Probability of getting a 2 or a 5,==> 1/6 + 1/6 – 0==> 2/6 = 1/3.

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Solution:Probability of selecting a black card = 26/52Probability of selecting a 6 = 4/52Probability of selecting both a black card and a 6 = 2/52P(B or 6) = P(B) + P(6) – P(B and 6)= 26/52 + 4/52 – 2/52= 28/52DOES NOT affect the outcome of the other events, they are called independent events. The outcome of the first roll doesn’t affect the second outcome. What is the probability of getting two consecutive tails ?Probability of getting a tail in one toss = 1/2The coin is tossed twice.

Here’s the verification of the above answer with the help of sample space. When a coin is tossed twice, the sample space is (H,H), (H,T), (T,H), (T,T) .

Our desired event is (T,T) whose occurrence is only once out of four possible outcomes and hence, our answer is 1/4.

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If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 blue pens and 1 black pen?SolutionProbability of drawing 1 blue pen = 4/9 Probability of drawing another blue pen = 4/9 Probability of drawing 1 black pen = 3/9 Probability of drawing 2 blue pens and 1 black pen = 4/9 * 4/9 * 3/9 = 48/729 = 16/243 Dependent EventsWhen two events occur, if the outcome of one event affects the outcome of the other, they are called dependent events In fact, we will study probability theory based on the theory number of elements in the sample space S. For a discrete case, the probability of an event A can be .

Consider the aforementioned example of drawing a pen from a pack, with a slight difference.

Example 1: A pack contains 4 blue, 2 red and 3 black pens. If 2 pens are drawn at random from the pack, NOT replaced and then another pen is drawn.

What is the probability of drawing 2 blue pens and 1 black pen?Solution: Probability of drawing another blue pen = 3/8 Probability of drawing 1 black pen = 3/7 Probability of drawing 2 blue pens and 1 black pen = 4/9 * 3/8 * 3/7 = 1/14 Let’s consider another example:Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement. After drawing one card, the number of cards are 51.

Now, the probability of drawing a king and queen consecutively is 1/13 * 4/51 = 4/663 Conditional probabilityThe formula for conditional probability P(A|B), read as P(A given B) isP(A|B) = P (A and B) / P(B) Consider the following example:Example: In a class, 40% of the students study math and science. What is the probability of a student studying science given he/she is already studying math?SolutionP(M) = 0.

60 Complement of an eventA complement of an event A can be stated as that which does NOT contain the occurrence of A In fact, we will study probability theory based on the theory number of elements in the sample space S. For a discrete case, the probability of an event A can be .

A complement of an event is denoted as P(Ac) or P(A’). if A is the event of getting an even number in a die roll, Ac is the event of NOT getting an even number i. if A is the event of randomly choosing a number in the range of -3 to 3, Ac is the event of choosing every number that is NOT negative i.

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What is the probability of getting at least one head?Solution:Probability of getting no head = P(all tails) = 1/32P(at least one head) = 1 – P(all tails) = 1 – 1/32 = 31/32.

Probability Example 1What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled College of Engineering The basic concepts will be illustrated by use of case studies. 310, 311. Elements of Probability and Mathematical Statistics I and II. treatment of experimental data: normal sampling theory; confidence intervals and .

SolutionLet the event of the occurrence of a number that is odd be ‘A’ and the event of the occurrence of a number that is less than 5 be ‘B’. P(A) = 3/6 (odd numbers = 1,3 and 5)P(B) = 4/6 (numbers less than 5 = 1,2,3 and 4)P(A and B) = 2/6 (numbers that are both odd and less than 5 = 1 and 3)Now, P(A or B) = P(A) + P(B) – P(A or B)= 3/6 + 4/6 – 2/6P(A or B) = 5/6. Probability Example 2A box contains 4 chocobars and 4 ice creams.

What is the probability of choosing 2 chocobars and 1 icecream?SolutionAfter taking out 1 chocobar, the total number is 7. Probability of choosing 2nd chocobar = 3/7Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3So the final probability of choosing 2 chocobars and 1 icecream = 1/2 * 3/7 * 2/3 = 1/7 Probability Example 3When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8.

SolutionLet the event of getting a greater number on the first die be G.

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And there are two ways where the number on the first die is greater than the one on the second given that the sum should equal 8, G = (5,3), (6,2) .

Therefore, P(Sum equals 8) = 5/36 and P(G) = 2/36 This article presents a case study of a learner engaged with a probability paradox. are conceptual and epistemological; that engagement with paradox can be a Gregory ChaitinInformation, randomness and incompleteness: Papers on algorithmic information theory St. Olaf College, Northfield, MN (1992, June 24–27)..

Now, P(G|sum equals 8) = P(G and sum equals 8)/P(sum equals 8)= (2/36)/(5/36)