Probability
In this chapter we are basically concerned
with drawing conclusions or inferences from random experiment. For these
conclusions and inferences to be reasonably accurate, an understanding of
proper definition of probability is essential. Before going to the mathematical
definition of probability, let us consider a few examples;.
What do we mean when we make the statements
“India will probably win the cricket match”, “I have a fiftyfifty chance of
getting an even number when a die is rolled,”, “Khan is not likely to go to
club tonight,” or “most of students of post graduate classes will likely be
married within three year”? In each case we are expressing an outcome of which
we are not certain, but owing to past information’s or from an understanding of
the structure of the experiment we have some degree of confidence in the
validity of the statement.
The mathematical theory of probability for
finite sample space, provide a set of numbers called weights, ranging from 0 to
1, which provide a means of evaluating the likelihood of occurrence of events
resulting from a random experiment. To every point in the sample space we
assign a weight such that the sum of all weights is 1. If we have reasons to
believe that a certain sample point is quite likely to occur when the
experiment is conducted, weight assigned to it should be close to one. On the
other hand, a weight closer to zero is assigned to a sample point that is
unlikely to occur.
If we have reasons
to believe that all sample points are equally likely then weight assigned to
each sample point would be equal.
To find the
probability of any event ‘A’ we sum all weights assigned to sample points in A.
This sum is called measure or probability of A and is denoted by P(A).
The probability of
any event A is the sum of the weights of sample points in A. Therefore, 0 £P(A) £1, P(f)
= 0 and P(S) = 1.
If a random
experiment can result in any one of N different equally likely outcomes, and if
exactly n of these outcomes correspond to A, then the probability of event A,
P(A) = .
If the weights
cannot be assumed equal, they must be assigned on the basis of prior knowledge
or experimental evidence. For example, if a coin is not balanced, we would
estimate the two weights by tossing the coin a large number of times and
recording the outcomes. The true weights would be the fraction of heads and
tails that occur in the long run. This method of arriving at weights is also
known as relative frequency definition of probability.
For instance, say
there are 2 players x_{1} and x_{2} and they have played ten
matches among themselves, and x_{1 }have won all the previous ten
matches. Now they play the eleventh match and we have to find the probability
of x_{1} winning the eleventh match. For the given information we may
conclude that x_{1} is more superior to x_{2} thus probability
of x_{1} winning the eleventh match would be equal to one.
Remarks
q
If
the probability of certain event is one, it doesn’t mean that event is going to
happen with certainty! Infact we would be just predicting that, the event is
most likely to occur in comparison to other events. Predictions depend upon the
past information and of course also on the way of analysing the information at
hand!!.
q
Similarly
if the probability of certain event is zero, it doesn’t mean that, the event
can never occur!
Example.1
Two
fair coins are tossed. What is the probability that atleast one head occurs?.
Solution
The sample space ‘S’ for this experiment is,
S = {HH, HT, TH, TT}
As the coin is fair
all these outcomes are equally likely. Let ‘w’ be the weight assigned to any
sample point then w + w + w + w = 1
&⇒ w = 1/4.
If ‘A’ is the event
representing the event of at least one head occurring,
Then A = {HT,
TH, HH}
&⇒ P(A) =
Example.2
Let
a die is loaded in such a way that even faces are twice as likely to occur as
the odd faces. Find the probability that a prime number will show up when the
die is tossed.
Solution
Clearly, in this case sample space is S =
{1, 2, 3, 4, 5, 6}
Let weight assigned
to an odd face be w, then weight assigned to an even face would be 2w. Thus we
have, w + 2 w + w + 2 w + w+2w = 1
&⇒ w = 1/9.
Let ‘E’ be the event
that a prime face shows up, E = {2, 3, 5}
&⇒ P(E) = 2 w + w + w = 4 w =
Example.3
In a single cast with two fair dice, what is the chance of
throwing?
(i) two 4¢s
(ii) a doublet
(iii) five and six
(iv) a sum of 7.
Solution:
(i)
There
are 6 ´ 6 equally likely cases (as any face of
any die may turn up)
&⇒ 36 possible
outcomes. For this event, only one outcome (4, 4) is favourable .
\ probability = 1/36.
(ii)
A doublet can occur in
six ways {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}.
Therefore probability of doublet = 6/36 = 1/6.
(iii) Two favourable outcomes {(5, 6), (6, 5)}
Therefore probability = 2/36 = 1/18
(iv) A sum of 7 can
occur in the following cases
{(1, 6), (2, 5), (3,
4), (4, 3), (5, 2), (6, 1)} which are 6 in number. .
Therefore probability =
6/36 = 1/6
Example.4
A special die with numbers 1, –1, 2,
–2, 0 and 3 is thrown thrice. What is the probability that the total is (i)
Zero, (ii) 6?.
Solution
Let a, b and c be the numbers on the
upper faces in first, second and third throw respectively.
(i) Now number of favourable cases
such that sum of upper faces is zero equals to number of integral solution of a
+ b + c = 0 subject to the condition 2 £ a, b, c £
3 i.e. equal to coefficient of t^{0} in (t^{}^{2} + t^{}^{1} + L + t^{3})^{3}
= coefficient of t^{6} in (1 + t + L + t^{5})^{3}
.
= ^{8}C_{2}

3 = 25.
Therefore,
required probability =
(ii) In this case number of
favourable cases will be equal to number of integral solution of a + b + c = 6
i.e. equal to coefficient of t^{6} in (t^{–2} + t^{–1}
+.... + t^{3})^{3} = coefficient of t^{12} in (1 + t +
.... + t^{5})^{3}.
= coefficient of t^{12}
in (1 – t^{6})^{3} (1 – t)^{–3 }= ^{14}C_{2}
– 3 ^{8}C_{2} + 3 = 10.
Therefore required
probability = =
.
Example.5
From a bag containing 5 white, 7 red and 4 black balls a man
draws 3 balls at random, find the probability of being all white.
Solution:
Total number of balls in the bag = 5
+ 7 + 4 = 16
The total number of ways in which 3
balls can be drawn is ^{16}C_{3} = =
560
Thus sample space S for this
experiment has 560 outcomes i.e. n(S) = 560.
Let E be the event of all the three
balls being white. Total number of white balls is 5. So the number of ways in
which 3 white balls can be drawn = ^{5}C_{3 }= =10
Thus E has 10 elements of S, \
n(E) = 10
\
Probability of E, P(E) = =
=
.
Probability Laws:
Often it is easier to calculate the
probability of some events from known probabilities of other events. This may
well be true if the event in question can be represented as the union of two or
more events, or may be as the intersection of certain events or may be as the
complement of some events.
Before going to these laws let us discuss
few definitions which would be used a lot in explaining these laws;
Mutually Exclusive Events:
A set of events is said to be mutually exclusive
if the occurrence of one of them precludes the occurrence of any of the other
events. For instance, when a pair of dice is tossed, the events a sum of 4
occurs’, ‘a sum of 10 occurs’ and ‘a sum of 12 occurs’ are mutually exclusive.
Simply speaking, if two events are mutually exclusive they can not occur
simultaneously. Using set theoretic notation, if A_{1}, A_{2}, L A_{n} be the set of mutually
events then A_{i} Ç
A_{j} = f for i > j and 1 £ i, j £ n.
Independent Events:
Events are said to
be independent if the occurrence or nonoccurrence of one does not affect the
occurrence or nonoccurrence of other. For instance, when a pair of dice is
tossed, the events ‘first die shows an even number’ and ‘second die shows an
odd number’ are independent. As the outcome of second die does not affect the
outcome of first die. It should also be noted that these two events are not
mutually exclusive as they can occur together.
Similarly, let us
think about a ordinary pack of cards. Let two cards are drawn from this pack in
succession without replacement. Then the events ‘first card is an ace’ and
‘second card is an ace’ are not independent (in fact these are dependent).
Remarks
q
Distinction
between mutually exclusive and independent events should be clearly made. To be
precise, concept of mutually exclusive events is set theoretic in nature while
the concept of independent events is probabilistic in nature.
q
If
two events A and B are mutually exclusive, they would be strongly dependent as
the occurrence of one precludes the occurrence of the other.
q
Let
E_{1}, E_{2}, ...., E_{n} be n independent events then
P(E_{1} ÇE_{2} Ç ..... Ç E_{n})
= P(E_{1}) P(E_{2}) ....... P(E_{n}).
Exhaustive Event:
A set of events is
said to be exhaustive if the performance of random experiment always result in
the occurrence of at least one of them. For instance, consider a ordinary pack
of cards then the events ‘drawn card is heart’, drawn card is diamond’, ‘drawn card
is club’ and ‘drawn card is spade’ is set of exhaustive event. In other words
all sample points put together (i.e. sample space itself) would give us an
exhaustive event.
If ‘E’ be an
exhaustive event then P(E) = 1.