Unit-I: Sets and Functions
Unit-II: Algebra
Unit-III: Coordinate Geometry
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability

Bayes Theorem

If an event  \text{ A} can occur with one of the  n mutually exclusive and exhaustive events  \operatorname{B_1,B_2,...,B_n} and the probabilities  P\left(\dfrac{A}{B_1}\right),P\left(\dfrac{A}{B_2}\right),...,P\left(\dfrac{A}{B_n}\right) are known, then

 P\left( \dfrac{B_i}{A} \right)=\dfrac{P(B_i)\cdot P\left( \dfrac{A}{B_i} \right)}{\sum_{i=1}^{n}P(B_i)\cdot P\left( \dfrac{A}{B_i} \right)}


 \text{A=} event what we have,

 \text{B=} event what we want.


 \begin{aligned} P(A\cap B_i)&=P(A)\cdot P\left( \dfrac{B_i}{A} \right)\\&=P(B)\cdot P\left( \dfrac{A}{B_i} \right)\\\implies P\left( \dfrac{B_i}{A} \right)&=\dfrac{P(B_i)\cdot P\left( \dfrac{A}{B_i} \right)}{P(A)}\\&=\dfrac{P(B_i)\cdot P\left( \dfrac{A}{B_i} \right)}{\sum_{i=1}^{n}P(A\cap B_i)} \end{aligned}

Value of Testimony:

If  P_1 and  P_2 are the probabilities of speaking the truth of two independent witnesses  A and  B , then,

 P ( their combined statement is true)  =\dfrac{P_1P_2}{P_1P_2+(1-P_1)(1-P_2)}

In this case, it has been assumed that we have no knowledge of the event except the statement made by  \text{A} and  \text {B} .

However, if  \text{P} is the probability of the happening of the event before their statement, then,

 \text{ P } ( their combined statement is true )  =\dfrac{PP_1P_2}{PP_1P_2+(1-P)(1-P_2)(1-P_2)}

Here, it has been assumed that the statement given by all the independent witnesses can be given in two ways only so that if all the witnesses tell falsehoods they agree in telling the same falsehood. If this is not the case and  \text {C} is the chance of their coincidence testimony, then the probability that the statement is true  =PP_1P_2 probability of the statement is false  =(1-P)\cdot C ( 1-P_1 )( 1-P_2 ) However, the chance of coincidence testimony is taken only if the joint statement is not contradicted by any witness.

Some Additional Properties:

If  \text{A} and  \text {B} are two events such that,

(i)  A  B , then (i)  P(A)  P(B) .

(ii)  P(A\cap B)  P(B)  P(A)  P(A\cap B)  P(A)+P(B)

(iii) For any two independent events  A and  B

 P\left\{ (A\cup B)\cap(\overline{A}\cap \overline{B}) \right\}  \dfrac{1}{4}

(iv) If  \text {A} and  \text {B} are two events, then

 \max\left\{ 0,1,...,P(\overline{A})-P(\overline{B}) \right\}  P(A\cap B)  \min\left\{ P(A),P(B) \right\}

(v) The event which is not mutually exclusively are known as compatible events.

Odds in Favor and Against an Event:

If  x is the number of cases favorable to the occurrence of event  \text {A} and  \text{ y } that for event  A^1 , then the odds in favor of  A are  x:y and the odds against  A are  y:x . In this case,  P(A)=\dfrac{x}{x+y} and  P(A^1)=\dfrac{y}{x+y}

Some Standard Results:

(1) When three dice are throwing, let  K(3  K  18) be the sum of numbers on them. Then the number of ways of getting total count  'K'

 = \left\{ \begin{array}{cl}\dfrac{(K-1)()K-2}{2} &,\ \text{if}\  3\le K\le 8\\ \dfrac{(19-K)(20-K)}{2}&,\ \text{if}\ 13\le K\le 18\\25&,\ \text{if}\ K=9\ \text{or}\ 12\\ 27&,\ \text{if}\ K=10\ \text{or}\ 11 \end{array} \right.

(2) Suppose a word is given. Let the number of vowels be  'V' , number of constants be  'C' and the total number of letters be  'T'. If the letters of the word are arranged at random, then the probability that

(i) Relative Positions of vowels and constants do not change  =\dfrac{V!C!}{T!}

(ii) The order of vowels does not change  =\dfrac{1}{\text{No. of ways of arranging the vowels}}

(iii) The order of constants does not change  =\dfrac{1}{\text{No. of ways of arranging the constants}}

(iv) The order of the vowels as well as constants does not change  =\dfrac{1}{\text{No. of ways of arranging the vowels or constants}}

(3) Out of  n pairs of shoes, if  K(<n) shoes are selected at random, the probability that there is no pair is  \dfrac{^nC_k\cdot2^k}{^{2n}C_k}

(4) Let  A be a set containing  'n' elements. Then

(i) The probability that two Subsets of  \text {A} selected at random, have  'r' elements in common  =^C_r\cdot \dfrac{3^{n-r}}{4^n} or  ^nC_r\left( \dfrac{3}{4} \right)^n\dfrac{1}{3^r}

(ii) The probability that  'r' subsets of  \text{ A } selected at random have no element in common is  \left( \dfrac{2^r-1}{2^r} \right)^n

(iii) Let  \text { A } be the set containing in elements. If  \text { B } and  \text{ C } are two subsets of  \text {A} , then the probability that  C\cup B=A is  \left( \dfrac{3}{4} \right)^m

(5) Let  S be a finite set containing  'n' elements. Then

        (i) The total number of binary operations on  'S' is equal to  n^{n^2}.

        (ii) The total number of commutative binary operations on  'S' is equal to  n^{\dfrac{n(n+1)}{2}}

(6) Let  x and  y be random variables on the sample space  S and  'K' is a real number. Then,

        (i)  E(kX)=kE(x)

        (ii)  E(x+y)=E(x)+E(y)


Probability theory is applied in everyday life in risk assessment and modeling. The insurance industry and markets use actuarial science to determine to price and make trading decisions.

Governments apply probabilistic methods in environmental regulation, entitlement analysis, and financial regulation. Probability is used to design games of chance so that sasinos can make a guaranteed profit, yet provide payouts to players that are frequent enough to encourage continued play.

Geometrical Applications:

(i) If a point is taken at random on a given straight line segment  \text{ AB } , the chance that it falls on a particular segment  \text { PQ } of the line segment is  \dfrac{PQ}{AB} .

(ii) If a point is taken at random on the area  \text { S } which includes an area  \sigma , the chance that the point falls on  \sigma is  \dfrac{\sigma}{S} .

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