Shortcuts in Quantitative Aptitude for Competitive Exams – Probability
Shortcuts in Quantitative AptitudeReasoningEnglish
INTRODUCTION
Random Experiment:
It is an experiment which if conducted repeatedly under homogeneous condition does
not give the same result. The total number of possible outcomes of an experiment in any trial
is known as the exhaustive number of events.
For example;
- In throwing a die, the exhaustive number of cases is 6 since any one of the six faces marked
with 1, 2, 3, 4, 5, 6 may come uppermost. - In tossing a coin, the exhaustive number of cases is 2, since either head or tail may turn over.
- If a pair of dice is thrown, then the exhaustive number of cases is 6 x 6 = 36
- In drawing four cards from a well-shuffled pack of cards, the exhaustive number of
cases is 52C4.
Events are said to be mutually exclusive if no two or more of them can occur simultaneously
in the same trial.
For example,
- In tossing of a coin the events head (H) and tail (T) are mutually exclusive.
- In throwing of a die all the six feces are mutually exclusive.
- In throwing of two dice, the events of the face marked 5 appearing on one die and face
5 (or other) appearing on the other are not mutually exclusive.
Out comes of a trial are equally likely if there is no reason for an event to occur in preference
to any other event or if the chances of their happening are equal.
For example,
- In throwing of an unbiased die, all the six faces are equally likely to occur.
- In drawing a card from a well-shuffled pack of 52 cards, there are 52 equally likely
possible outcomes.
The favourable cases to an event are the outcomes, which entail the happening of an event.
For example,
- In the tossing of a die, the number of cases which are favourable to the “ appearance
of a multiple of 3” is 2, viz, 3 and 6. - In drawing two cards from a pack, the number of cases favourable to “drawing 2 aces”
is 4C2. - In throwing of two dice, the number of cases favourable to “getting 8 as the sum” is 5,:
(2,6), (6,2), (4,4), (3,5) (5,3).
Events are said to be independent if the happening (or non¬happening) of one event is
not affected by the happening or non – happening of others.
CLASSICAL DEFINITION OF PROBABILITY
If there are n-mutually exclusive, exhaustive and equally likely outcomes to a random experiment
and ‘m’ of them are favourable to an event A, then the probability of happening of A is denoted
by P (A) and is defined by P(A)= m/n
P(A) can never be negative.
Since, the number of cases in which the event Awill not happen is n- m then the probability P(A )
of not happening of A is given by
The ODDS IN FAVOUR of occurrence of A are given by m: (n – m) or P(A): P (A) .
The ODDS AGAINST the occurrence of A are given by (n-m): m or P (A) : P(A)
ALGEBRA OF EVENTS
Let A and B be two events related to a random experiment. We define
- The event “A or B” denoted by “A∪B”, which occurs when A or B or both occur. Thus,
P(A ∪B)= Probability that at least one of the events occur - The event “A and B”, denoted by ” A∩B”, which occurs when A and B both occur. Thus,
P(A ∩B) = Probability of simultaneous occurrence of A and B. - The event “ Not – A” denoted by A , which occurs when and only when A does not occur.
Thus P(A) = Probability of non-occurrence of the event A. - A ∩B denotes the “ non-occurrence of both A and B”.“A ∩ B” denotes the “ occurrence of
A implies the occurrence of B”.
For example:
Consider a single throw of die and following two events
A = the number is even = {2,4,6}
B = the number is a multiple of 3 = {3,6}
ADDITION THEOREM ON PROBABILITY
ADDITION THEOREM :
If A and B are two events associated with a random experiment, then
P(A ∪B) = P(A) + P(B) – P( A∩B)
ADDITION THEOREM FOR THREE EVENTS:
If A, B, C are three events associated with a random experiment, then
P( A∪B∪C) = P( A) + P(B) + P(C) – P( A∩B) -P(B∩C) – P(A∩C) + P(A∩B∩C)
If A and B are two mutually exclusive events and the probability of their occurence
are P(A) and P(B) respectively, then probability of either A or B occuring is given by
P( A or B) = P(A) + P(B)
=> P(A + B) = P(A) + P(B)
CONDITIONAL PROBABILITY
Let A and B be two events associated with a random experiment. Then P(A/B) represents the
conditional probability of occurrence of A relative to B.
For example:
Suppose a bag contains 5 white and 4 red balls. Two balls are drawn one after the other without replacement. If A denotes the event “drawing a white ball in the first draw” and B denotes the
event “drawing a red ball in the second draw”.
P (B/A) = Probability of drawing a red ball in second draw when it is known Obviously,
P (A/B) is meaning less in this problem.
MULTIPLICATION THEOREM
If A and B are two events, then P(A∩B)
= P (A) P (B/A), if P (A) > 0
= P (B) P (A/B) if P (B) > 0
From this theorem we get
For example:
Consider an experiment of throwing a pair of dice. Let A denotes the event
“ the sum of the point is 8” and B event “ there is an even number on first die”
Then A = {(2,6), (6,2), (3,5), (5,3), (4,4)},
B={(2,1), (2,2),…..,(2,6), (4,1), (4,2),……….(4,6), (6,1), (6,2),….(6,6)}
Now, P(A/B) = Prob. of occurrence of A when B has already occurred = prob. of getting 8 as
the sum, when there is an even number on the first die
INDEPENDENCE
An event B is said to be independent of an event A if the probability
that B occurs is not influenced by whether A has or has not occurred.
For two independent events A and B.
P(A∩B) = P(A) P(B)
Event A1,A2, An are independent if
P(Ai ∩ Aj) = P(Ai)P(Aj) for all i, j,i ≠ j, That is, the events are pair wise independent.
The probability of simultaneous occurrence of (any) finite number of them is equal to the
product of their separate probabilities, that is they are mutually independent.
For example:
Let a pair of fair coin be tossed, here S = {HH, HT, TH, TT}
A = heads on the first coin = {HH, HT}
B = heads on the second coin = {TH, HH}
C = heads on exactly one coin = {HT, TH}
Hence the events are pairwise independent.
Also P(A∩B∩C) = P(<φ>) = 0≠P(A)P(B)P(C)
Hence, the events A, B, C are not mutually independent.
Leave a Reply