PROBABILITY

        If n represents the total number of equally mutually exclusive and exhaustive possible outcomes of an experiment and m of them are   favourable to the event A, then the probability of the event A is defined as

        P(A) = n(E)/n(S) = m/n

        This is known as classical definition of probability

Note:

       (i)     for any event A S

       (ii)    0 < P(A) < 1, P(A) ε R

       (iii)    P(A) + P() = 1

      (iv)   If a cases are favourable to an event A and b cases are favourable to an event (i.e. unfavourable to A) then P(A) = a/(a+b) and P() = b/(b+a).

         We say that odds in favour of A are a : b

         and odds against of A are b : a

Rule 1: In the case of mutually exclusive events the chance of happening of one or other of them is the sum of the chances of the separate events. e.g. if A and B are mutually exclusive events, P(AυB) = P(A) + P(B). If E₁, E₂, ......, E_n are mutually exclusive events, then P(E₁υE₂υ ...... UE_n) = .

Therefore, if there are n possible outcomes of an experiment which are mutually exclusive, exhaustive and equally likely then, since the probability associated with each outcome is the same (say x) and since they are mutually exclusive, the probability of occurrence of one of them is the sum nx which must be equal to 1(because they are also exhaustive).

That is, nx = 1 => x = 1/n.

Corollary: Since A and A' are mutually exclusive events, P(AUA') = P(A')=1.

If out of m + n equally likely, mutually exclusive and exhaustive cases, m cases are favourable to an event A and n are not favourable to an event A. m : n is called odds in favour of A, n : m is called odds against A, probability of occurrence of event A is m/(m+n) and that of non-occurrence of it is m/(m+n).

Notation:  Let A and B by two events, then

(i)     A' or or A^c stands for the non-occurrence of negation of A.

(ii)    AUB stands for the occurrence of at least one of A or B.

(iii)    AB stands for the simultaneous occurrence of A and B.

(iv)   A'B' stands for the non-occurrence of both A and B.

(v)    AB stands for "the occurrence of A implies occurrence of B".

Illustration:

A die is rolled. What is the probability that outcome will be an even number?

Solution:

When the die is rolled the possible outcomes are

S = {1, 2, 3, 4, 5, 6,} and A = outcome will be even number i.e.
{2, 4, 6}

So, P(A) = n(E)/n(S) = 3/6 = 1/2

Illustration:

Two dice are thrown simultaneously. What is the probability of obtaining a total score of seven?

Solution:

There are six possible ways as to the number of points on the first die; and to each of these ways, there corresponding 6 possible number of points on second die. Hence total number of ways

S = 6 × 6 = 36

We now find out how many ways are favourable to the total of 7 points. This may happen only in following ways : (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3).

Hence, required Probability = 6/36 =1/6.

Illustration:

 Suppose a die is rolled, can we find the probability that either odd number or a number divisible by 4 comes?

Solution:

When a die is rolled 1, 2, 3, 4, 5 and 6 are possible outcomes. So,
S = {1, 2, 3, 4, 5, 6}.

Let   event A : odd number i.e. A = {1, 3, 5}

        event B : number divisible by 4 i.e. B = {4}

Here, we have represented the sample space and event in Venn Diagram.

We have to find probability of either A or B, i.e. P(A U B)

Clearly, A U B = {1, 3, 4, 5} i.e. A + B

Hence, P(A U B) = P(A) + P(B)

 = 3/6 +1/6 = 2/3.

Now, think about a different situation: A die is rolled, can we calculate the probability that the number, which comes is either odd or divisible by 3?

Sample space is S = {1, 2, 3, 4, 5, 6}.

Define, event A : odd number i.e. A = {1, 3, 5}

event B : divisible by 3 i.e. B = {3, 6}

we can represent sample space and even in Venn diagram.

Now, we have to find probability that either A or B occurs i.e. P(A U B)

Clearly, A U B = P{1, 3, 5, 6}

Can we write A U B = A + B?

Let's do it, A U B = 1, 3, 5, 3, 6; this is incorrect for obvious reasons since we counted 3 twice.

From above illustration, it can be concluded that when we write
A U B = A + B we count the common element twice, once while counting the elements of A and once while counting the elements of B. Therefore in order to get correct result we should subtract it (common) element once i.e.

A U B = A + B - A B {A B element common in both A and B}

                 = 1, 3, 5, 4, 6, - 3

                 = 1, 3, 5, 6

        Hence the probability

P(A U B) = P(A) + P(B) - P(AB) = (3/6) + (2/6) - (1/6) = 2/3

Note:

1.     In previous case, P(AUB) = P(A) + P(B) was correct because A∩B = f. So,        we get a very important result.

P(AUB) = P(A) + P(B) when A and B are mutually exclusive events.

P(AUB) = P(A) + P(B) - P(A∩B), when A and B are not mutually exclusive events.

2.     This particular case of two events can easily be generalized for the three events A, B and C

P(AUBUC) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(C∩A) + P(A∩B∩C).

Illustration:

        The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Out of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two and a 40% chance of passing in exactly two. Prove that
m + p + c = 27/20.

Solution:

         1 - (1 - m) (1 - p) (1 - c) = 0.75                               ............ (i)

        (1 - m) pc + m (1 - p) c + mp (1 - c) + mpc = 0.5              ............ (ii)

        (1 - m) pc + m (1 - p) c + mp (1 - c) = 0.4                  ........... (iii)

        (ii) - (iii) gives : mpc = 0.1 = 1/10 From equation (i), we get

         p + c + m - (mp +mc + pc) + mpc = 0.75                    ............. (iv)

From (ii),

        pc + mc + mp - 3mpc + mpc = 0.5

        => pc + mc + mp = 0.7

Substituting in (iv), we get

        m + p + c = 0.75 - 0.1 + 0.7 = 1.35 = 135/100 = 27/20

INTERSECTION AND UNION OF SETS OF EVENTS

Let E₁, E₂, ......, E_n be n events, then P(E₁UE₂U ...... UE_n), represents the probability of occurrence of at least one of the events from E₁, E₂, ......, E_n and P(E₁∩E₂∩ ...... ∩E_n) represents the occurrence of all the events together.

In general, we have

P(E₁UE₂U ... UE_n)=  P(E_i∩E_j∩E_k)-...+(-1)^n-1P(E₁∩E₂∩...∩E_n).

Corollary: If A and B are any two events,

P(AUB) = P(A) + P(B) - P(A∩B).

Note that P(A∩B') = P(A) - P(A∩B).

Also for any two events A and B,

P(exactly one of them occurs)

= P(E₁) + P(E₂) - P(E₁∩E₂)

= P(E₁UE₂) - P(E₁∩E₂).

If E₁, E₂ and E₃ be three events, then

(i)     P(at least two of E₁, E₂, E₃ occur)

        = P(E₂∩E₃) + P(E₃∩E₁) + P(E₁∩E₂) - 2P(E₁∩E₂∩E₃),

(ii)    P(exactly two of E₁, E₂, E₃ occur)

        = P(E₂∩E₃) + P(E₃∩E₁) + P(E₁∩E₂) - 3P(E₁∩E₂∩E₃),

(ii)    P(exactly one of E₁, E₂, E₃ occur)

        = P(E₁) + P(E₂) + P(E₃) - 2P(E₃∩E₁) + 2P(E₁∩E₂) + 3P(E₁∩E₃).

        If events are mutually exclusive, P(E₁UE₂U ...... UE_n) =.