Introduction To Probability

Random Experiment

An experiment, whose all possible outcomes are known in advance but the outcome of any specific performance can not predicted before the completion of the experiment, is known as random experiment. An example of random experiment might be tossing of a coin.

Sample-space

A set of all possible outcomes associated with same random experiment is called sample-space and is usually denoted by 'S'. Consider the experiment of tossing a die. If we are interested in the number that shows on the top face, then sample space would be

S₁ = {1, 2, 3, 4, 5, 6}

Event

An event is a subset of sample – space.

In any sample space we may be interested in the occurrence of certain events rather than in the occurrence of a specific element in the sample space.

Simple Event

If an event is a set containing only one element of the sample-space, then it is called a simple event.

Compound Event

A compound event is one that can be represented as a union of sample points.

For instance, the event of drawing a heart from a deck of cards is the subset A ={heart} of the sample space S = {heart, spade, club, diamond}. Therefore A is a simple event. None the event B of drawing a red card is a compound event since B = {heart U diamond} = {heart, diamond}.

PROBABILITY

If a random experiment can result in any one of N different equally likely outcomes, and if exactly n of these outcomes favours to A, then the probability of event A, P (A) = $\frac{n}{N}$.

i.e. $\frac{\text{favourable}\text{cases}}{\text{total}\text{number}\text{of}\text{cases}}$

Remarks:

1. 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!!
2. 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?

Sol: 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 $\Rightarrow$ w = ¼

If 'A' is the event representing the event of atleast one head occuring,

Then A = {HT, TH, HH} $\Rightarrow$ P(A) = $\frac{3}{4}$

Example -2: In a single cast with two fair dice, what is the chance of throwing

(i) two 4's (ii) a doublet

(iii) five - six (iv) a sum of 7

Sol: (i) There are 6 $\times$ 6 equally likely cases ( as any face of any die may turn up)

$\Rightarrow$ 36 possible outcomes. For this event, only one outcome (4-4) is favourable

$\therefore$ 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

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,.. A_n be the set of mutually events then A_i $\cap$ A_j =$\varphi$ for i $\ne$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.

Remarks

$\bullet$ 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.

$\bullet$ 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.

Exhaustive Event

A set of events is said to be exhaustive if the performance of random experiment always result in the occurrence of atleast 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.

SET THEORETIC PRINCIPLES

If 'A' and 'B' be any two events of the sample space then

$\bullet$ A$\cup$B would stand for occurrence of atleast one of them.

$\bullet$ A $\cap$ B stands for simultaneous occurrence of A and B.

$\bullet$ $\bar{A}$ (or A') stands for non-occurrence of A

$\bullet$ $\bar{A}\cap \bar{B}$(or A' $\cap$ B') stands for non-occurrence of both A and B.

$\bullet$ A $\subseteq$ B stands for 'the occurrence of A implies the occurrence of B'.

$\bullet$ If A and B are any two events, then P(A$\cup$B) = P(A) + P(B) - P(A$\cap$B)

$\bullet$ If A and B are mutually exclusive, P(A$\cup$B) = P(A) + P(B).

As in this case no sample point would be present in A$\cap$B.

$\bullet$ P(A') = 1 - P(A)

As, A$\cup$A' = S and A and A' are mutually exclusive

$\Rightarrow$ P(A$\cup$A') = P(A) + P(A') = P(S) = 1 $\Rightarrow$ P(A') = 1 - P(A).

$\bullet$ P(A$\cap$B') = P(A) - P(A$\cap$B)

As A$\cap$B' and A$\cap$B are mutually exclusive events and (A$\cap$B') $\cup$ (A$\cap$B) = A

$\Rightarrow$ P(A$\cap$B') + P(A$\cap$B) = P(A) $\Rightarrow$ P(A$\cap$B') = P(A) - P(A$\cap$B)

Similarly, P(A'$\cap$B) = P(B) - P(A$\cap$B)

$\bullet$ P(A'$\cup$B') = 1 - P(A$\cap$B)

As, (A' $\cup$ B') $\cup$ (A $\cap$ B) = S

$\Rightarrow$ P(A' $\cup$ B') + P(A $\cap$ B) = 1 $\Rightarrow$ P(A' $\cup$ B') = 1 - P(A$\cap$B)

Similarly, P(A' $\cap$ B') = 1 - P(A$\cup$B)

$\bullet$ P(exactly one of A, B occurs)

= P(A$\cap$B') + P(A'$\cap$B) = P(A) + P(B) - 2P(A$\cap$B)

= P(A$\cup$B) - P(A$\cap$B) = P(A' $\cup$ B') - P(A' $\cap$ B')

If A, B, C are any three events of the sample space, then

$\bullet$ P(A$\cup$B$\cup$C) = P(A) + P(B) + P(C) - P(A $\cap$ B) - P(A$\cap$C) - P(B$\cap$C) + P(A$\cap$B$\cap$C)

$\bullet$ P (Exactly one of A, B, C occurs)

= P(A) + P(B) + P(C) - 2P(A$\cap$B) - 2A(A$\cap$C) - 2P(B$\cap$C) + 3P(A$\cap$B$\cap$C)

$\bullet$ P(Exactly two of A, B, C occur) = P(A$\cap$B) + P(B$\cap$C) + P(A$\cap$C) - 3P(A$\cap$B$\cap$C)

$\bullet$ P (at least two of A, B, C occur) = P(A$\cap$B) + P(B$\cap$C) + P(A$\cap$C) - 2P(A$\cap$B$\cap$C)

$\bullet$ If A₁, A₂ L , A_n are 'n` events, then P(A₁ $\cup$ A₂ L $\cup$A_n)

= $\sum _{r=1}^{n}P(A_r)-\sum _{1\le r_1<r_2\le n}\left(P(A_{r_1}\cap A_{r_2}\right)$ +$\sum _{1\le r_1<r_2<r_3\le n}P\left(A_{r_1}\cap A_{r_2}\cap A_{r_3}\right)$- L+${\left(-1\right)}^{n-1}P\left(A_1\cap A_2\cdots \cap A_n\right)$

$\bullet$ P (A $\cup$ B)$⩾$max (P(A), P(B), P(A) + P(B) - 1)

as A $\subseteq$ A$\cup$B $\Rightarrow$ P(A) $⩽$P(A$\cup$B)

Similarly, B $\subseteq$ A $\cup$ B $\Rightarrow$ P(B) $⩽$P(A$\cup$B) $\Rightarrow$ P(A$\cup$B)$⩾$Max (P(A), P(B))

Also P(A$\cup$B) = P(A) + P(B) - P(A $\cap$ B)

P(A) + P(B) - 1 $⩽$ P(A$\cup$B) $⩽$P(A) + P(B) (As 0 $⩽$P(A $\cap$ B) $⩽$1)

Max (P(A) + P(B) - 1, P(A), P(B)) $⩽$P(A $\cup$ B) $⩽$P(A) + P(B)

$\bullet$ If out of m + n equally likely, mutually exclusive and exhaustive cases, m cases are favourable to an A event and n are not favourable to the an event A, m : n is called odds in favour of A, n : m is called odds against the event A.

Example -6: Let A, B, C be three events. If the probability of occurring one event out of A and B is 1 - a, out of B and C is 1 - 2a, out of C and A is 1 - a and that of occurring three events simultaneously is a², then prove that the probability that at least one out of A, B, C will occur is greater than or equal to 0.5.

Sol: Probability that exactly one event out of A and B occur is p(A) + p(B) - 2p(A$\cap$B) and probability that exactly one event out of B and C occur is p(B) + p(C) - 2p(B$\cap$C) and so on.

Now, p(A$\cup$B$\cup$C) = p(A) + p(B) + p(C) - p(A$\cap$B) - p(A$\cap$C) - p(B$\cap$C) + p(A$\cap$B$\cap$C)

=$\frac{1}{2}$[p(A)+ p(B) - 2p(A$\cap$B) + p(B) + p(C)-2p(B$\cap$C)+p(C)+p(A)-2p(A$\cap$C)] +p(A$\cap$B$\cap$C)

= $\frac{1-a+1-2a+1-a}{2}+a^2$= a² - 2a + $\frac{3}{2}$

Let a² - 2a +$\frac{3}{2}$ = y $\Rightarrow$ a² - 2a + $\frac{3}{2}$-y = 0

Since a is real, so 4 - 4($\frac{3}{2}$ - y)$⩾$0 $\Rightarrow$ y$⩾$1/2