Handout 04 Probability Theory

Handout 04 Probability Theory

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4. Elementary Probability We live in a world in which we are unable to forecast the future with complete certainty. Our need to cope with uncertainty leads us to the study and use of probability theory. In many instances, we, as concerned citizen, will have some knowledge about the possible outcomes of a decision. By organizing this information and considering it systematically, we will be able to recognize our assumptions, communicate our reasoning to others, and make a sounder decision than we could by using a shot-in-the dark approach. Basic Terminology in Probability In general, probability is the chance something will happen. Probabilities are expressed as fractions ( 1/6, ½ , 8/9 ) or as decimals (0.167, 0.500, 0.889) between zero and one. Assigning a probability zero means that something can never happen; a probability 1 indicates that something will always happen. (1)

A Random experiment

An experiment which produces different results even though it is repeated a large number of times under essentially similar conditions is called a random experiment (2)

Sample Space and Events

The set S of all possible outcomes of a random experiment is called a Sample space. An element in S, is called a sample point or a simple event. (3)

A Simple Event A simple event is the most basic outcome of a random experiment.

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An Event

An event A is a collection of simple events. or, in other words, a subset of the Sample space S. The empty set � is called the impossible event, and S, the sample space is called the certain or sure event. (5)

The Probability of an Event The probability of event A is defined as n(A) no. of favourable cases to A P(A) = = n(S) Total possible cases This is a classical approach. We observe that (i) 0 ≤ P(A) ≤ 1 (ii) P(�) = 0 (iii) P(S) = 1

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Probability Space

If the probability of each sample point in a sample space is known then it is called the probability space, such that (i) pi � 0 (ii) � pi = 1 (7)

Complementary Events

� The complementary of an event A is the event A that occurs when A does not occur., i.e. the event � consisting of all simple events that are not in event A. We will denote the complement of A by A c or A.

Muhammad Naeem, Assistant Professor Department of Mathematics

Handout 04 Probability Theory

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Complementary Relationships The sum of the probabilities of complementary events equals 1. That is � P(A)+P(A) = 1

Example (1) A fair coin is tossed ten times and the up face is recorded after each toss. What is the probability of event A: {observe at least one head}

Solution Since we know the probability of the complement of A, we use the relation ship for the complementary events: � 1023 1 = = .999 P(A) = 1 – P(A) = 1 1024 1024 That is, we are virtually certain of observing at least one head in 10 tosses of the coin. Properties of Events If A and B are two events then they may be (i) Equally Likely (ii) Exhaustive (iii) Mutually Exclusive (1)

Equally Likely Events

Two events A and B are said to be equally likely, when one event is as likely to occur as the other. In other words, each event should occur in equal number in repeated trials. For example when a fair coin is tossed, the head is as likely to appear as the tail. (2)

Exhaustive Events

Events are said to be collectively exhaustive, when the union of mutually exclusive events is the entire sample space S. A group of mutually exclusive and exhaustive events is called a partition of the sample space. For instance, event A and Ac form a partition as they are mutually exclusive and their union is the entire sample space. (3)

Mutually Exclusive Events

Two events A and B of a single experiment are said to be mutually exclusive or disjoint if and only if they cannot both occur at the same time. Example (2) Give a collectively exhaustive list of the possible outcomes of tossing two dice. Also give the probability for each of the following totals in the rolling of two dice: 1, 2, 5, 6, 7, 10 and 11. Solution An exhaustive list of all possible outcomes of tossing two dice: ( dice 1, dice 2) S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) } P(1) = 0/36 = 0, P(2) = 1/36, P(5) = 4/36, P(6) = 5/36, P(7) = 6/36, P(10) = 3/36, P(11) = 2/36

Muhammad Naeem, Assistant Professor Department of Mathematics

Handout 04 Probability Theory

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Addition Law for Mutually Exclusive Events We know that two events are mutually exclusive when A�B = �. Consequently, for two mutually exclusive events A and B P(A�B) = P(�) = 0 Thus, when events A and B are mutually exclusive (in the sense that they cannot occur at the same time), then P(A�B) = P(A) + P(B) Generalization let the events a, B and C are mutually exclusive, then P(A�B�C) = P(A) + P(B) + P(C) Compound Events An event can often be viewed as a composition of two or more other events. Such events are called Compound Events; they can be formed (composed) in two ways. For example A�B, A�B. Additive Law for not Mutually Exclusive Events The probability of the union of events A and B is the sum of the probabilities of events A and B minus the probability of the intersection of events A and B, i.e. P(A�B) = P(A) + P(B) � P(A�B) Example (3) (i)

Find the probability that on a single draw from a pack of playing cards we draw a spade or a face card or both. A fair dice is rolled once. You win Rs.5/- if the outcome is either even or divisible by 3. What is the probability of winning the game? A customer enters a food store. The probability that the customer buys (a) bread is 0.60 (b) milk is 0.50 and (c) both bread and milk is 0.30. What is the probability that the customer would by either bread or milk or both?

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Conditional Probability Let us consider the sample space given by throwing a die. i.e. S = {1,2,3,4,5,6}. Consider the event A = {4}. i.e. the number 4 appears on the die as a result of the throw. Then the probability of A, denoted by P(A) is 1/6. Further take the event B = {2, 4, 6}. Now if we say that the event B had already A taken place and we consider the probability of A. Then we write P( ) or P(A�B) and is given by B A P(A�B) P( ) = P(B) B i.e. the conditional probability of A given that the event B has already occurred Example (4) A man tosses two fair dice. What is the conditional probability that the sum of two dice will be 7, given that,(i) the sum is odd,(ii) the sum is greater than 6,(iii) the two dice had the same out come? Solution The sample space S for this experiment consists of the following 36 equally likely out comes; S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

Muhammad Naeem, Assistant Professor Department of Mathematics

Handout 04 Probability Theory

Let

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A = {the sum is 7}, B = {the sum is odd} C = {the sum is greater than 6}, and D = {the two dice had the same outcomes}.

Then A = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, B = {(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)}, C = {(1,6), (2,5), (2,6), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}, D = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}, A�B = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, A�C = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, A�D = � 3 18 So, P(A) = , P(B) = , 36 36 6 21 P(D) = P(C) = , 36 36 6 6 P(A�C) = , P(A�B) = , 36 36 P(A�D) = 0 Hence using the definition of conditional probability, we get P(A�B) 3 36 1 P(A/B) = = � = P(B) 36 18 3 P(A�C) 3 36 2 P(A/C) = = � = P(C) 36 21 7 P(A�D) 36 P(A/D) = =0� =0 P(D) 3 Multiplication Law for Probability

(General Rule)

If A and be any two events defined in a sample space S, then P(A�B) = P(A) P(B/A), provided P(A) � 0 = P(B) P(A/B), provided P(B) � 0 The conditional probability of B given A has occurred is P(A�B) , provided P(A) � 0 P(B/A) = P(A) by multiplying both sides with P(A), we get P(A�B) = P(A) P(B/A) This is called the general rule of multiplication for probabilities. Example (5) A box contains 15 items, 4 of which are defective and 11 are good. Two items are selected. What is the probability that the first is good and second is defective? Solution Let A denotes the event that the first item selected is good and B, the event that the second item is defective. Then we need to calculate the probability P(A�B) = P(A).P(B/A)

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Handout 04 Probability Theory

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11 15 Given the event A has occurred, there remains 14 items of which 4 are defective. Therefore the probability of selecting the defective after a good has been selected, 4 i.e. P(A/B) = 14 11 4 44 Hence P(A�B) = P(A).P(B/A) = � = = 0.16 15 14 210 Now

P(A) =

Example (6) Two cards are dealt from a pack of ordinary playing cards. Find the probability that the second card dealt is a heart? Solution Let H1 represents the event that the first card dealt is a heart, and H2, the event that the second card dealt is a heart. Then P(the second card is a heart) = P(the first card is a heart and the second card is a heart) + P(the first card is not a heart and the second card is a heart) _ i.e. P(H2) = P(H1�H2) + P(H1�H2) _ _ = P(H1) P(H2/H1) + P(H1) P(H2/H1) 1 13 17 1 13 12 39 13 = = =� � �+� � �= + �52 51 � �52 51 � 17 68 68 4 Multiplication Law The probability of the events A and B when they both occur is equal to the product of their respective probabilities and denote it by the event A�B P(A�B) = P(A) �� P(B) We will discuss the two cases later, whenever the two events are independent or dependent. Example (7) During winter, Mr. Fazal experiences difficulty in starting his two cars. The probability that the first car starts is 0.80 and the probability that the second car starts is 0.40. There is a probability of 0.30 that both the cars start. (a) define the events involved and use probability notation to show the probability information given above. (b) What is the probability that at least one car starts? (c) What is the probability that Mr. Fazal cannot start either of the two cars. Solution We denote Event A: that first car starts Event B: that second car starts Event A�B: that both cars start Event A�B: that either of the two cars start, so Event A�Bc: that A starts but B cannot start Event Ac� B: that B starts but A cannot start Hence given information is P(A) = 0.80 � P(Ac) = 0.20

Muhammad Naeem, Assistant Professor Department of Mathematics

Handout 04 Probability Theory

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P(B) = 0.40 � P(Bc) = 0.60 P(A�B) = 0.30 Now

i.e.

Ac� Bc = (A�B)c denote the event that no car starts Therefore 1- P( (A�B) c ) denotes that at least one car starts 1- P( (A�B)c ) = 1 – P(Ac� Bc) = 1 – {P(Ac) � P(Bc) } = 1 – (0.20 � 0.60) = 1 – 0.12 = 0.88 alternatively we find the probability: P(A�Bc) + P(Ac� B) + P(A�B) = 0.80�0.60 + 0.20�0.40 + 0.80�0.40 = 0.88

Exercise 23.1 (1) Sol.

(2) Sol.

(3) Sol.

(4) Sol.

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Sol.

What is the probability of obtaining at least one head in tossing six fair coins? total possible cases = 2 6 = 64 Event A that at least one head is obtained So Ac is the event of obtaining no head As P(A) + P(Ac) = 1 � P(A) = 1 – P(Ac) Now Ac = {TTTTTT} that all six outcomes are tails Therefore P(Ac) = 1/64 Hence P(A) = 1 – 1/64 = 63/64 In rolling two fair dice. What is the probability of obtaining a sum greater than 10 or a sum divisible by 6? Two fair dice are rolled, n(S) = 36 Event A that sum is greater than 10 and event B is that sum divisible by 6 n(A) = 3, n(B) = 6 and n(A�B) = 1 so P(A�B) = P(A) + P(B) – P(A�B) = 3/36 + 6/36 – 1/36 = 8/36 = 2/9 Three screws are drawn at random from a lot of 100 screws. 10 of which are defective. Find the probability of the event that all 3 screws drawn are non defective, assuming that we draw (a) with replacement (b) without replacement. Defective = 10, good = 90 and total = 100 Experiment : 3 are drawn (i) with replacement required probability = 90/100 × 90/100 × 90/100 = (0.9)3 = 72.9% (ii) without replacement required probability = 90/100 × 89/99 × 88/98 = 72.65% Three boxes contain five chips each, numbered from 1 to 5 and one chip is drawn from each box. Find the probability of the event E that the sum of the numbers on the drawn chips is greater than 4. Box I = {1,2,3,4,5}, Box II = {1,2,3,4,5} and Box III = {1,2,3,4,5} one number is drawn from each box, so total possible drawl = (5)3 = 125 event E that the sum of the numbers is greater than 4 so Ec = {(1,1,1), (1,1,2), (1,2,1), (2,1,1)} therefore P(Ec) = 4/125 � P(E) = 1 – 4/125 = 121/125 A batch of 100 iron rods consists of 25 oversized rods, 25 undersized rods, and 50 rods of the desired length. If two rods are drawn at random without replacement, what is the probability of obtaining (a) two rods of the desired length (b) one of the desired length (c) none of the desired length (d) two undersized rods. oversized = 25, undersized = 25, desired length = 50, total = 100

Muhammad Naeem, Assistant Professor Department of Mathematics

Handout 04 Probability Theory

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Experiment : two are drawn without replacement 50 49 � = 0.2474 100 99 This is the case of without replacement 50 50 (b) P(1 is of desired length) = 2� � = 0.505 100 99 st Actually these are the two either or cases (i) 1 of desired length and 2 nd is not of desired length (ii) 1st is not of desired length and 2nd of desired length 50 49 (c) P(no of desired length) = 1 � = 1 - 0.2474 = 0.7526 100 99 50 49 (d) P(2 are undersized) = 100 � 99 = 0.2474 (7) If a certain kind of tire has a life exceeding 25000 miles with probability 0.95. What is the probability that a set of these tires on a car will last longer than 25000 miles? Sol. p = 0.95, i.e. the probability of a tire having life exceeding 25000 miles P(a set of 4) = (0.95)( 0.95)( 0.95)( 0.95) = (0.95)4 (8) If a certain kind of tire has a life exceeding 25000 miles with probability 0.95. What is the probability that at least one of the tires will not last for 25000 miles? Sol. Probability that a tire has life exceeding 25000 miles = 0.95 Let A be the event that P(no tire has a life exceeding 25000 miles) = (0.05)4 therefore P(at least one) = 1 � (0.05)4 (11) If we inspect sheets of paper by drawing 3 sheets without replacement from every lot of 100 sheets, what is the probability of getting 3 clean sheets although 8% of the sheets contains impurities? Sol. p = 0.92, the probability of a sheet to contain purity P(3 clean sheets) = (0.92)3 (a)

P(2 are of desired length) =

Independent Events Events A and B are independent if the occurrence of B does not alter the probability that A has occurred, i.e., events A and B are independent if P(A/B) = P(A) When events A and B are independent it will also be true that P(B/A) = P(B) Events that are not independent are said to be dependent. Example (8) Consider an experiment of tossing a fair coin twice and recording the up face on each toss. The following events are defined: A: {first toss is a head} B: {second toss is a head} Does knowing that event A has occurred affect the probability that B will occur? Solution Intuitively the answer should be no, since what occurs on the first toss should in no way affect what occurs on the second toss. The sample space for this experiment is S = {HH, TH, HT, TT} and A = { HH, HT}, B = { HH, TH} Each of these simple events has a probability of ¼. Thus, P(B) = P(HH) + P(TH) = ¼ + ¼ = ½ and P(A) = P(HH) + P(HT) = ¼ + ¼ = ½

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Handout 04 Probability Theory

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Now, what is P(B/A)? P(A�B) P(HH) 1/4 1 = = = P(B) P(A) P(A) 1/2 2 We can now see that P(B) = ½. Knowing that the first resulted in a head does not affect the probability that the second toss will be a head. Hence the two events A and B are independent. P(B/A) =

Example (9) Consider the experiment of tossing a fair die and define the following events A = {observe an even number} B = {observe a number less than or equal to 4} Are events A and B are independent? Solution We first calculate P(A)= P(2) + P(4) + P(6) = ½ P(B)= P(1) + P(2) + P(3) + P(4) = 4/6 = 2/3 P(A�B) = P(2) + P(4) = 2/6 = 1/3 Now assume that B has occurred, the conditional probability of A given B is (A�B) 1/3 P(A/B) = = = ½ = P(A) P(B) 2/3 Therefore the events A and B are independent. Note that if we calculate the conditional probability of B given A, our conclusion is the same: (A�B) 1/3 P(B/A) = = = 2/3 = P(B) P(A) 1/2 Example (10) _ _ Given P(A) = 0.60, P(B) = 0.40, P(A�B) = 0.24, find P(A/B), P(A�B), P(A/B), P(B/A), P(B). What is the relation between A and B? Solution Given P(A) = 0.60, P(B) = 0.40, and P(A�B) = 0.24. Then P(A�B) 0.24 P(A/B) = = = 0.60 P(B) 0.40 P(A�B) = P(A) + P(B) – P(A�B) = 0.60 + 0.40 – 2024 = 0.76 _ _ P(A�B) P(A) - P(A�B) 0.60 - 0.24 0.36 P(A/B) = = = = 0.60 P(B) 1 - P(B) 1 - 0.4 0.60 P(B�A) 0.24 = = 0.40 P(B/A) = 0.60 P(A) _ P(B) = 1 – P(B) = 1 – 0.40 = 0.60 The event A and B are independent as P(A/B) = 0.60 = P(A) P(B/A) = 0.40 = P(B) and P(A�B) = 0.24 = (0.60) (0.40) = P(A) P(B)

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Handout 04 Probability Theory

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Example (11) Let A and B be two events associated with an experiment. Suppose that P(A) = 0.5 and P(A�B) = 0.6. Find P(B) if (i) A and B mutually exclusive? (ii) A and B independent? (iv) P(A/B) = 0.4 Solution when A and B are mutually exclusive, then P(A�B) = P(A) + P(B) i.e. 0.60 = 0.50 + P(B) � P(B) = 0.10 When A and B are independent, we have P(A�B) = P(A) + P(B) – P(A�B) = P(A) + P(B) – P(A) P(B) = P(A) + P(B) [1 – P(A)] or 0.60 = 0.50 + P(B) [1 – 0.50] or 0.10 = 0.5 P(B) or P(B) = 0.20 P(A�B) P(A) + P(B) - P(A�B) (iii) P(A/B) = = P(B) P(B) 0.50 + P(B) - 0.60 or 0.40 = P(B) or 0.40P(B) = P(B) – 0.10 or 0.10 = 0.60 P(B) � P(B) = 1/6 = 0.17 Example (12) Indicate whether each of the following statements is true or false, indicate why. (i) If P(A/B) = 0, then A and B are mutually exclusive. (ii) If P(A/B) = 0, then A and B are independent. (iii) If P(A/B) = P(B/A), then P(A) = P(B). (iv) If A and B are independent, then P(A) = P(B). Solution (i) (ii) (iii) (iv)

the statement is true. The statement is false. If A and B are independent, then P(A/B) = P(A) the statement is true. The statement is false “Independent” does not mean that two events have equal probabilities.

Example (13) An urn contains 10 white and 3 black balls. Another contains 3 white and 5 black balls. Two balls are transferred from first urn and placed into second and then one ball is taken from the latter. What is the probability that it is a white ball? Solution Let A be the event that two balls are drawn from the first urn and transferred into second urn. Then A can occur in the following three mutually exclusive ways: A1 = 2 white balls A2 = 1 white ball and I blackball

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Handout 04 Probability Theory

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A3 = 2 black balls 10 9 15 Thus P(A1) = � = 13 12 26 10 3 5 P(A2) = � = 13 12 26 3 2 1 and P(A3) = � = 13 12 26 The second urn after having transferred 2 balls from the first urn, contains (i) 5 white and 5 black balls (2 white balls transferred) (ii) 4 white and 6 black balls (1 white and 1 black ball transferred) (iii) 3 white and 7 black balls (2 black balls transferred) Let W represent the event that a white ball is drawn from the second urn after having transferred 2 balls from the first urn. Then P(W) = P(W�A1) + P(W�A2) + P(W�A3) 5 15 15 Now P(W�A1) = � = 10 26 52 4 5 4 = P(W�A2) = � 10 26 52 3 1 3 P(W�A3) = � = 10 26 260 Hence the required probability is 15 4 3 98 P(W) = + + = = 0.3769 52 52 260 260 Example (14) A card is drawn from a deck of ordinary playing cards. What is the probability that it is a diamond, a face card or a king? Solution Let

A = the card drawn is diamond B = the card drawn is a face card and C = the card drawn is a king. Then we need P(A�B�C) = P(A) + P(B) + P(C) – P(A�B) – P(B�C) – P(A�C) + P(A�B�C) 13 12 4 Now P(A) = , P(B) = , P(C) = 52 52 52 13 3 3 P(A�B) = P(A) P(B/A) = � = 52 13 52 4 12 4 P(B�C) = P(B) P(C/B) = � = 52 12 52 1 13 1 � = P(A�C) = P(A) P(C/A) = 52 13 52 P(A�B�C) = P(A) P(B/A) P(C/A�B) 13 3 1 1 = � � = 52 13 3 52 Hence we get 13 12 4 3 4 1 1 22 P(A�B�C) = + + - - + = = 0.423 52 52 52 52 52 52 52 52

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Handout 04 Probability Theory

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Example (15) Three urns of the same appearance are given as follows: Urn A contains 5 red and 7 white balls Urn B contains 4 red and 3 white balls Urn C contains 3 red and 4 white balls An urn is selected at random and a ball is drawn from that urn. (i) What is the probability that the ball drawn is red? (ii) If the ball drawn is red, what is probability that it came from urn A? Solution Here we first select one of the three urns and then we draw a ball, which is either red (R) or white (W). In other words, we perform a sequence of two experiments. This process is described by the probability tree diagram (as shown on the right) , in which each branch of the tree give the respective probability. 1 5 5 Now the probability of selecting urn A, for instance, and then a red ball (R) is � = 3 2 36 because the probability that any particular path of the tree occurs is, by the multiplication law, the product of the probability of each branch of the path. Now the probability of drawing a rd ball is given by the relation P(R) = P(A) P(R/A) + P(B) P(R/B) + P(C) P(R/C) As there are three mutually exclusive paths leading to the drawing of a red ball. 1 5 1 4 1 3 Hence P(R) = � + � + � 3 12 3 7 3 7 119 = = 0.4722 12 Here we need the probability that the urn A is selected, given that ball drawn is red, that is, P(P/A). By definition, (A�R) P(A/R) = P(R) But P(A�R) = probability that earn A is selected and a red ball is drawn is 1 5 5 = � = 3 12 36 5/36 35 Hence P(A/R) = = 119/252 119 = 0.294 Example (16) In a study involving a manufacturing process, the probability was 0.10 that a part tested was defective and a probability that a part was produced on machine A was 0.30. Given that a part was produced on machine A, there is 0.15 probability that it is defective. (a) What is the probability that a part tested is both defective and produced by machine A? (b) If a part is found to be defective, what is the probability that it came from machine A? (c) Is finding a defective part independent of its being produced on machine A? Explain. (d) What is the probability of the part being either defective or produced by machine A? (e) Are the events “a defective part” and “produced by machine A” mutually exclusive events? Explain. Hints: Define the events; D: a part tested is defective, E: a part is produced by machine A Given information: P(D) = 0.10, P(E) = 0.30, P(D/E) = 0.15

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Handout 04 Probability Theory

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Find P(D�E), P(E/D). Check P(D/E) = P(D) and P(E/D) = P(E). Also check For mutually exclusive, check P(D�E) = 0

P(D�E)

Example (17) The probability that Mr. Khan will get an offer on the first job he applied for is 0.5 and the probability that he will get an offer on the second job he applied for is 0.6. He thinks that the probability that he will get an offer on both jobs is 0.15. (a) Define the events involved and use probability notation to show the probability information. (b) What is the probability that Mr Khan gets an offer on the second job given that he receives an offer for the first job. (c) What is the probability that Mr Khan gets an offer on at least one of the jobs he applied for. (d) What is the probability that Mr Khan gets an offer only of the job he applied for? (e) Are the jobs offers independent? Explain. Exercise (1)

(2) (3)

(4) (5) (6)

A ball is drawn at random from a box containing 6 red, 4 white and 5 blue balls. (i) Determine the probability that the ball drawn is (a) red, (b) white, (c) blue, (d) not red and (e) red or white. (ii) Find the probability that they are drawn in the order red, white and blue if each ball is (a) replaced (ii) not replaced. A fair die is rolled twice. Find the probability of getting a 4, 5 or 6 on the first rolling or and a 1, 2, 3 or 4 on the second rolling. (solve with both methods directly and using multiplication law) Consider the experiment of tossing a fair die and define the following events A = {observe an even number} B = {observe a number less than or equal to 4} Are events A and B are independent? A box contains 15 items, 4 of which are defective and 11 are good. Two items are selected. What is the probability that the first is good and second is defective? An urn contains 10 white and 3 black balls. Another contains 3 white and 5 black balls. Two balls are transferred from first urn and placed into second and then one ball is taken from the latter. What is the probability that it is a white ball? Examine whether each of the following statements is correct. Explain. a. P(A) = 0.40, P(A�B) = 0.25 b. P(A) = 0.40, P(A�B) = 0.60, so that A and B are independent. c. P(A) = 0.20, P(B) = 0.60, P(A�B) = 0.12, the events A and B are independent. d. If P(A) = 0.80, P(B) = 0.60, the events A and B cannot be mutually exclusive.

Assignment (1) In a study involving manufacturing process the probability is 0.10 that a part is tested turns out to be defective and the probability that a part was produced on machine A is 0.30. Given that a part was produced on machine A there is a 0.15 probability that it is defective. (f) What is the probability that a part tested is both defective and produced by machine A? (g) If a part is found defective, what is the probability that it came from machine A? (h) Is finding a defective part independent of its being produced on machine A? (i) What is the probability of the part being defective or produced by machine A? (j) Are the events ‘a defective part’ and ‘produced by machine A’ mutually exclusive events? Explain.

Muhammad Naeem, Assistant Professor Department of Mathematics

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