﻿ probability basics, important probability formulas and solved probability problems - freetests4u.com CAT blogs

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probability basics, important probability formulas and solved probability problems - freetests4u.com CAT blogs

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### Experiment: 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.Event: 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) >0For 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.

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