Course Outlines

         Axioms of probability, some theorems on probability, Baye's theorem.

    Discrete random variables: probability distribution, cumulative distribution, mathematical expectation, moments, mean and variance; uniform distribution and binomial distribution:- mean and variance, binomial probability table, recurrence relation.

    Continuous random variables: probability density, cumulative distributions, mean and variance, Chebyshev's theorem and laws of large numbers; Normal distribution ­– properties, mean and variance, area under standard normal curve, normal approximation to binomial distribution.

 Introduction

ü   Systematic study of probability began in the seventeenth century

ü   Game theory and gambling

ü   involve in uncertainty case

Some terms

ü  Experiment

Process which when perform gives different results (tossing a coin, throwing a dice)

ü  Sample space

Set of all possible outcomes of random experiment is called sample

Example S = {1, 2, 3, 4, 5, 6} throwing a dice

        S = {H, T} tossing a coin

        S = {HH, HT, TH, TT} tossing two coins

        S = {BB, BG, GB, GG} having two child

Throwing two dice simultaneously, the sample space S is

S = {(1, 1), (1, 2), (1, 3), …,(6, 6)}, 36 points

Tossing three coins, the sample space S is given by

S = {HHH, HHT, …, TTT}, 8 points

ü  Events

          Subset of sample space

          For S = {1, 2, 3, 4, 5, 6}, some events are {1, 3, 5}, {2, 4, 6}, {2, 3, 5} etc. How many events can be formed ? 64 events

ü  Simple event

containing only one element

S = {1, 2, 3, 4, 5, 6}, some events are {1, 3, 5}, {2, 4, 6}, {2, 3, 5}

ü  compound event

containing the combination of two or more simple events

 

S = {HH, HT, TH, TT} tossing two coins

{HH, TT}, {HT, TH}

ü  Exhaustive events

Totality contains all possible outcomes

S = {1, 2, 3, 4, 5, 6}

ü  Equally likely events

Each has equal chance to occur

ü  Sure events

          contains all the possible outcomes (sample space)

ü  Impossible events

never occur, no chance to occur, f

ü  Mutually exclusive events

Two events are said to be mutually exclusive if they have no elements in common

A Ç B = f

ü  Complementary events

A Ç B = f and A È B = S

S = {1, 2, 3, 4, 5, 6}, some events are A = {1, 3, 5}, B = {2, 4, 6}

A Ç B = f

A È B = {1, 2, 3, 4, 5, 6} = S

Operation on events

ü Union

ü intersection

ü Difference

ü Complement [ S - E = complement of event S]

Example 1

A bag contains 8 red, 3 white and 9 blue balls. If 3 balls are drawn at a random, determine the probability that

(i) all 3 balls are red

(ii) all 3 balls are white

(iii) 2 balls are red and 1 ball is white C(8, 2) ´ C(3, 1)

(iv) at least 1 ball is white

(v) one ball of each color is drawn.

Solution

(i) No. of favourable cases (m) = C(8, 3)

Total No. possible cases (n) = C(20, 3)

Exercise 5.1

1. Let A and B are mutually exclusive events with P(A) = 0.3, P(B) = 0.5. Find (a) P(AÈB)                  (b) P(A')                (c) P(B').

Solution

Here, P(A) = 0.3, P(B) = 0.5

(a) P(AÈB) = P(A) + P(B)

(b) P(A') = 1 - P(A)

2.     Let A, B  and C are mutually exclusive events with P(A) = 0.2, P(B) = 0.3, P(C) = 0.4. Find (a) P(AÈBÈC)     (b) P(AÈB')'

Solution

Here, P(A) = 0.2, P(B) = 0.3, P(C) = 0.4

(a) P(AÈBÈC) = P(A) + P(B) + P(C)

(b) P(AÈB')' = 1 - P(AÈB')

                        = 1 - [ P(A) + P(B')]

                        = 1 - [ P(A) + 1 - P(B)]

                        = ………

3.     What is the probability of getting a total of 11  or 7 when a pair of dice is rolled ?

Solution

The sample space S when a pair of dice is rolled is given by

S = {(1, 1), (1, 2), (1, 3), …., (6, 6)}, 36 equally likely outcomes.

Let A be the event that the sum of two numbers of a sample points is 7 and B be the event that the sum of two numbers of a sample points is 11. Then

A = {(3, 4), (4, 3), (2, 5), (5, 2), (1, 6), (6, 1)}

B = {(5, 6), (6, 5)}

Here, A and B are  mutually exclusive events of sample space S containing 36 equally likely outcomes.

P(A or B) = P(A È B)

                  = P(A) + P(B)

4.     A dice is rolled once. Find the probability of

(a) an odd number or a 5.

(b) an even number or a 5 ?

Solution

The sample space S when a dice is rolled once is given by

S = {1, 2, 3, 4, 5, 6}, 6 equally likely outcomes.

Let A = {an odd number}

           = {1,3, 5}

B = {5 occur}

        = {5}

AÇB = {5}

Now, P(AÈB) = P(A) + P(B) - P(AÇB)

                        = +  -

                       = …..

5.     a pair of dice is rolled. Find the probability that either first die shows 3 or the sum is 6 or 7 .

The sample space S when a pair of dice is rolled is given by

S = {(1, 1), (1, 2), (1, 3), ….(6, 6)}, 36 equally likely outcomes.

A = {first dice shows 3} = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}, 6 equally likely outcomes.

B = {sum is 6 or 7} = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}, 11 equally likely outcomes 

A Ç B = {(3, 3), (3, 4)}

Now,

P(AÈB) = P(A) + P(B) - P(AÇB)

                = + -

                = ……

 Exercise 5.1

1.     Let A and B are mutually exclusive events with P(A) = 0.3, P(B) = 0.5. Find (a) P(AÈB)                  (b) P(A')                (c) P(B').

Solution

Here, P(A) = 0.3, P(B) = 0.5

(a) P(AÈB) = P(A) + P(B)

(b) P(A') = 1 - P(A)

2.     Let A, B  and C are mutually exclusive events with P(A) = 0.2, P(B) = 0.3, P(C) = 0.4. Find (a) P(AÈBÈC)     (b) P(AÈB')'

Solution

Here, P(A) = 0.2, P(B) = 0.3, P(C) = 0.4

(a) P(AÈBÈC) = P(A) + P(B) + P(C)

(b) P(AÈB')' = 1 - P(AÈB')

                        = 1 - [ P(A) + P(B')]

                        = 1 - [ P(A) + 1 - P(B)]

                        = ………

3.     What is the probability of getting a total of 11  or 7 when a pair of dice is rolled ?

Solution

The sample space S when a pair of dice is rolled is given by

S = {(1, 1), (1, 2), (1, 3), …., (6, 6)}, 36 equally likely outcomes.

Let A be the event that the sum of two numbers of a sample points is 7 and B be the event that the sum of two numbers of a sample points is 11. Then

A = {(3, 4), (4, 3), (2, 5), (5, 2), (1, 6), (6, 1)}

B = {(5, 6), (6, 5)}

Here, A and B are  mutually exclusive events of sample space S containing 36 equally likely outcomes.

P(A or B) = P(A È B)

                  = P(A) + P(B)

4.     A dice is rolled once. Find the probability of

(a) an odd number or a 5.

(b) an even number or a 5 ?

Solution

The sample space S when a dice is rolled once is given by

S = {1, 2, 3, 4, 5, 6}, 6 equally likely outcomes.

Let A = {an odd number}

           = {1,3, 5}

B = {5 occur}

        = {5}

AÇB = {5}

Now, P(AÈB) = P(A) + P(B) - P(AÇB)

                        = +  -

                       = …..

5.     a pair of dice is rolled. Find the probability that either first die shows 3 or the sum is 6 or 7 .

The sample space S when a pair of dice is rolled is given by

S = {(1, 1), (1, 2), (1, 3), ….(6, 6)}, 36 equally likely outcomes.

A = {first dice shows 3} = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}, 6 equally likely outcomes.

B = {sum is 6 or 7} = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}, 11 equally likely outcomes 

A Ç B = {(3, 3), (3, 4)}

Now,

P(AÈB) = P(A) + P(B) - P(AÇB)

                = + -

                = ……

 Exercise 5.1

1.     Let A and B are mutually exclusive events with P(A) = 0.3, P(B) = 0.5. Find (a) P(AÈB)                  (b) P(A')                (c) P(B').

Solution

Here, P(A) = 0.3, P(B) = 0.5

(a) P(AÈB) = P(A) + P(B)

(b) P(A') = 1 - P(A)

2.     Let A, B  and C are mutually exclusive events with P(A) = 0.2, P(B) = 0.3, P(C) = 0.4. Find (a) P(AÈBÈC)     (b) P(AÈB')'

Solution

Here, P(A) = 0.2, P(B) = 0.3, P(C) = 0.4

(a) P(AÈBÈC) = P(A) + P(B) + P(C)

(b) P(AÈB')' = 1 - P(AÈB')

                        = 1 - [ P(A) + P(B')]

                        = 1 - [ P(A) + 1 - P(B)]

                        = ………

3.     What is the probability of getting a total of 11  or 7 when a pair of dice is rolled ?

Solution

The sample space S when a pair of dice is rolled is given by

S = {(1, 1), (1, 2), (1, 3), …., (6, 6)}, 36 equally likely outcomes.

Let A be the event that the sum of two numbers of a sample points is 7 and B be the event that the sum of two numbers of a sample points is 11. Then

A = {(3, 4), (4, 3), (2, 5), (5, 2), (1, 6), (6, 1)}

B = {(5, 6), (6, 5)}

Here, A and B are  mutually exclusive events of sample space S containing 36 equally likely outcomes.

P(A or B) = P(A È B)

                  = P(A) + P(B)

4.     A dice is rolled once. Find the probability of

(a) an odd number or a 5.

(b) an even number or a 5 ?

Solution

The sample space S when a dice is rolled once is given by

S = {1, 2, 3, 4, 5, 6}, 6 equally likely outcomes.

Let A = {an odd number}

           = {1,3, 5}

B = {5 occur}

        = {5}

AÇB = {5}

Now, P(AÈB) = P(A) + P(B) - P(AÇB)

                        = +  -

                       = …..

5.     a pair of dice is rolled. Find the probability that either first die shows 3 or the sum is 6 or 7 .

The sample space S when a pair of dice is rolled is given by

S = {(1, 1), (1, 2), (1, 3), ….(6, 6)}, 36 equally likely outcomes.

A = {first dice shows 3} = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}, 6 equally likely outcomes.

B = {sum is 6 or 7} = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}, 11 equally likely outcomes 

A Ç B = {(3, 3), (3, 4)}

Now,

P(AÈB) = P(A) + P(B) - P(AÇB)

                = + -

                = ……

7. (c) P(AÇB') = P(A) - P(A Ç B)

Solution

We have

P(AÇB') = P(A - B)

                  = P(A) - P(AÇB)

(d) P(A'ÇB) = P(B) - P(A Ç B)

Solution

We have

P(A'ÇB) = P(B - A)

                  = P(B) - P(AÇB)

(e) P(A'ÇB') = 1 - P(A) - P(B) + P(A Ç B

Solution

We have

P(A'ÇB') = P(AÈB)'

                    = 1 - P(AÈB)

                    = 1 - [P(A) + P(B) - P(AÇB)]

                    = 1 - P(A) - P(B) + P(AÇB)

 (f) P(AÇB) < 2P(A Ç B) £ P(A) + P(B)

Solution

We have

AÇB Í A Þ P(AÇB) £ P(A)

AÇB Í B Þ P(AÇB) £ P(B)

Adding, we get

2P(A Ç B) £ P(A) + P(B)

or, P(A Ç B) <  P(A) + P(B)

Alternatively,

We have P(AÈB) = P(A) + P(B) - P(AÇB)

or, P(AÇ B) = P(A) + P(B) - P(AÈB)

Since, P(A È B) > 0, so

P(AÇ B) < P(A) + P(B), proved.

(g) P(AÇB)  + 1 < P(A) + P(B)

Solution

We have

P(AÈB) = P(A) + P(B) - P(AÇB)

or, P(AÇB) + P(AÈB) = P(A) + P(B)

or,  P(AÇB) + 1 ³ P(A) + P(B) proved.

Exercise 5.2

1.     If the probability that a man will alive in 20 years is 0.7 and the probability that his wife will alive in 20 years is 0.9, what is the probability that neither will alive in 20 years?

Solution

Let A denotes the event that the man will alive 20 years and B denotes the event that the woman will alive 20 years, then

P(A) = P(A will alive) = 0.7,

P(B) = P(B will alive) = 0.9

P(A') = P(A will not alive) = 1 - 0.7= 0.3

P(B') = P(B will not alive) = 1 - 0.9= 0.1

P(A'ÇB') = P(both will not alive)

                    = P(A') ´ P(B')

                    = 0.3 ´ 0. 1

                    = 0.03

P(both will alive) = P(AÈ B)

= P(A) ´ P(B)

= 0.7 ´ 0.9

= 0.63

2.     There are 8 mangoes and 4 oranges in one bag and 6 mangoes and 6 oranges in another bag. If one fruit from each bag is drawn what is the probability that

(a) both are mangoes

(b) both are oranges

(c) one is mango and other is an orange.

Solution

Let M1 and R1 denotes the events of drawing a mango and an orange from the first bag and M2 and R2 denote the events of drawing a mango and an orange from the second bag respectively.

P(M1) = , P(M2) = , P(R1) = , P(R2) =

(a) P(M1ÇM2) = P(M1) ´ P(M2) = ….

(b) P(R1ÇR2) = P(R1) ´ P(R2) = ….

(b) P(M1ÇR2) + P(M2ÇR1)  = P(M1) ´ P(R2) + P(M2) ´ P(R1)

4.     One bag contains 4 white balls and 2 black balls; another contains    3 white and 5 black balls. If one ball is drawn randomly from each bag, find the probability that

        (a)   both are white

        (b)   both are black.

        (c) one is white and one is black

Solution

Let W1 and W2 are the events that the white balls are drawn from first and second bag respectively and let B1 and B2 are the events that the black balls are drawn from first and second bag respectively. 

Multiplication law of probability

Theorem



The probability of any two events in a sample space S is given by

        P(A Ç B) = P(A) ´ P(B/A)

If A and B are independent, then

        P(A Ç B) = P(A) ´ P(B)

Proof

By the definition of conditional probability, we have

or, P(AÇB) = P(A) ´ P(B/A)

Similarly, P(AÇB) = P(B) ´ P(A/B)

If the events A and B are independent, then

either P(B) = P(B/A) or P(A) = P(A/B)

Hence, P(AÇB) = P(A)´ P(B)

Theorem

If A, B and C are any three events in a sample space S such that P(AÇB) ¹ 0, then

P(A Ç B Ç C) = P(A). P(B/A).P(C/AÇB)

Proof

We have

P(A Ç B Ç C) = P[(AÇB)ÇC]

                            = P(AÇB).P(C/AÇB)

                            = P(A).P(B/A).P(C/AÇB)                        proved.

If A, B and C are independent if

(a) the events are pair-wise independent, i.e.

        P(AÇB) = P(A) ´ P(B)

        P(AÇC) = P(A) ´ P(C)

        P(BÇC) = P(B) ´ P(C) and

(b)   P(AÇBÇC) = P(A) ´ P(B) ´ P(C)

ü  If an random experiment the events A1, A2, A3,… Ak, can occur, then

        P(A1Ç A2Ç A3Ç… ÇAk) = P(A1) . P(A2 / A1) . P(A3 / A1 Ç A2)….

                                                                P(Ak/A1ÇA2ÇA3Ç....ÇAk-1)

ü  If the events A1, A2, A3,… Ak, are independent, then

        P(A1Ç A2Ç A3Ç… ÇAk) = P(A1) . P(A2) . P(A3)….P(Ak)