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probability basics, important probability formulas and solved probability problems - CAT blogs


Hello Aspirants, Let us 1st take a look at some important definations.

$ 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.


$ p = \frac{n! }{(n-r)!} $


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


$ c = \frac{n!}{ (n-r)! (r!)} $


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

Random 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.

Sample space:
The set of all possible outcomes is called Sample Space.

Any subset of sample space is called an event.

Probability of occurance of an event:
Let S be the sample space and let E be an event then, E C S.
$ P(E) = \frac{n(E)}{n(S)}. $

Some Important result on probability:
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).

Mutually Exclusive Events:
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

Conditional probability
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

For independent events
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. HTML5 Icon

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