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Syllabus for CAT examination

Number Systems and interconversion between binary,decimal & hexadecimal Systems

Some tips for solving Time and Work problems

Techniques and tricks for solving Profit and Loss problems

Some important tips and tricks for solving two body or relative motion problems

Some Tips for CAT Reading Comprehensions

$ n! = n (n-1) (n-2) ......3*2*1. $

Ex. $ 5! = 5*4*3*2*1 = 120.$

We use permutations when we have to calculate total no. of arrangements that can be made by taking some or all things together at a time.

We use combinations when we have to calculate the no. of groups or different selections that can be made from given things.

An operation that can produce some well defined outcomes is called an experiment.

An experiment in which all possible outcomes are known and the exact outcome can not be predicted in advance is called a random experiment.

The set of all possible outcomes is called Sample Space.

Any subset of sample space is called an event.

Let S be the sample space and let E be an event then, E

$ P(E) = \frac{n(E)}{n(S)}. $

P(S) = 1 & 0 $ \leq $ p(E) $ \leq $ 1

For any event A and B, we have :

P (A U B) = P(A) + P(B) - P(A ∩ B)

If A~ denotes not A, then P(A~) = 1 - P(A).

Let S be the sample space and A & B are two events such that A∩B = NULL set and A U B =1, then P(A) = 1 - P(B).

For any event A, P(A) and probability of its complement P(A~) together forms set of mutually exclusive events such that P(A) + P(A~) = 1

A and B are two dependent events such that the event B will occure if and only if event A is already occured,then the probability of B is given by

P(B/A) = P(B∩A) / P(A) for all P(A) >0

P(A∩B) = P(A) * P(B)

Lets solve some example to understand these concepts clearly.

Ex.1 A bag contains 8 white and 5 black balls. Two balls are drawn at random. Find the probability that they are of the same colour.

Let S be the sample space . Then,

n(s) = number of ways of drawing 2 balls out of $ (8+5) = 13 C 2 = \frac{13*14}{2} = 91.$

Let E = Event of getting both balls of the same colour. Then, n(E) = no. of ways of drawing (2 balls out of 8) or (2 balls out of 5)

=$ (8 C 2) + (5 C 2) = 28 + 10 = 38 $

hence, P(E) = n(E)/n(S) = 38/91

EX.2 Rupesh is known to hit a target in 5 out of 9 shots whereas David is known to hit the same target in 6 out of 11 shots. What is the probability that the target would be hit once they both try?

$ P(A) = \frac{5}{8} $, $ P(B) = \frac{6}{11}$

and P(A∩B) = P(A) * P(B) since, P(A∩B) = P(A) * P(B)

= $ \frac{5}{9} * \frac{6}{11} $

= $ \frac{10}{33} $

The probability that the target would be hit is given by

$ P(AUB) = P(A) + P(B) - P(A∩B) = \frac{5}{6} + \frac{6}{11} - \frac{10}{33} = \frac{79}{99} $

Thank you for reading. Solve some more examples by yourself for practice. Keep visiting.