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Basic Concepts Of Probability Part 2

Posted on - 04-01-2017

Math

IIT JEE

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 fifty-fifty 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 x1 and x2 and they have played ten matches among themselves, and x1 have won all the previous ten matches. Now they play the eleventh match and we have to find the probability of x1 winning the eleventh match. For the given information we may conclude that x1 is more superior to x2 thus probability of x1 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 t0 in (t-2 + t-1 + L + t3)3 = coefficient of t6 in (1 + t + L + t5)3 .

= 8C2 - 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 t6 in (t–2 + t–1 +.... + t3)3 = coefficient of t12 in (1 + t + .... + t5)3.

= coefficient of t12 in (1 – t6)3 (1 – t)–3 = 14C2 – 3 8C2 + 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 16C3 = = 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 = 5C3 = =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 A1, A2, L An be the set of mutually events then Ai Ç Aj = f for i > j and 1 £ i, j £ n.

Independent Events:

Events are said to be independent if the occurrence or non-occurrence of one does not affect the occurrence or non-occurrence 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 E1, E2, ...., En be n independent events then P(E1 ÇE2 Ç ..... Ç En)
= P(E1) P(E2)
....... P(En).

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.

 
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