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# Probability +2

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### Probability +2

1. 1.  You've probably heard people say things like: 7/3/2013 2VSR Teen mother  The chance of rain tomorrow is 75%.  Teen mothers who live with their parents  He won the lottery! All of these statements are about probability. We see words like "chance", "less likely", "probably" since we don't know for sure something will happen, but we realise there is a very good chance that it will.
2. 2. 7/3/2013VSR 3  Probability is a concept which numerically measures the degree of uncertainty and therefore is certainty of the occurrence of events.  Probability had its origin in the 16th century when an Italian physician and mathematician Jerome Cardon (1501 -1576) wrote the first book on the subject. “Book of Games of Chance”. Subsequently, the theory of probability was developed by Bernoulli, De-Moivar, Fisher and others.
3. 3.  The mathematicians is basically concerned with drawing conclusions (or inference) from experiments involving uncertainties.  For these conclusions and inferences to be reasonably accurate, an understanding of probability theory is essential.  In this section, we shall develop the concept of probability with equally likely outcomes 7/3/2013VSR 4
4. 4. 7/3/2013VSR 5 Experiment This is any process of observation or procedure that  can be repeated (theoretically) an infinite number of times and  has a well-defined set of possible outcomes.
5. 5. 7/3/2013VSR 6 Sample space This is the set of all possible outcomes of an experiment. Event This is a subset of the sample space of an experiment Consider the following illustrations: The set of all event is the power set of S , denoted By 2s .
6. 6. 7/3/2013VSR 7 Experiment 1: Tossing a coin. Sample space: S = {Head or Tail} Experiment 2: Tossing a coin twice. S = {HH, TT, HT, TH} Experiment 3: Throwing a die. S = {1, 2, 3, 4, 5, 6}
7. 7. 7/3/2013VSR 8 Experiment 4: Two items are picked, one at a time, at random from a manufacturing process, and each item is inspected and classified as defective or non-defective. S = {NN, ND, DN, DD} where N = Non-defective D = Defective Some events: E1 = {only one item is defective} = {ND, DN} E2 = {Both are non-defective} = {NN}
8. 8. 7/3/2013VSR 9 Definition of a Probability Suppose an event A can happen in m ways out of a total of n possible equally likely ways. Then the probability of occurrence of the event (called its success) is denoted by n m Sn An AP )( )( )( →no. of favourable cases →no. of total outcomes of the experiment
9. 9. 7/3/2013VSR 10 The probability of non-occurrence of the event (called its failure) is denoted by n m n mn EP 1)(  Notice the bar above the E, indicating the event does not occur. Thus, = 1 In words, this means that the sum of the probabilities in any experiment is 1.
10. 10. 7/3/2013VSR 11 * Odds in favour of A )( )( AP AP * Odds in against A )( )( AP AP  Addition theorem for two or more events. i.e., P(A or B) = P(A) + P(B) – P(A B) . If A and B are mutually exclusive events then , P(A or B) = P(A) + P(B) .
11. 11. 7/3/2013VSR 12 * In this chapter, we shall discuses the concept of Conditional probability of event. Baye's theorem, Multiplication rule of probability and random variable and its probability, Binomial distribution etc. Conditional Probability If E and F are two events associated with the same sample space of a random experiment, E given that F has occurred, is given by 0)(, )( )( )( FP FP FEP EorFP
12. 12. 7/3/2013VSR 13 • Properties of conditional probability. i) P(S or F) = P(F/F) = 1. ii) If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0 then, P[(A B) or F] = P(A or F) + P(B or F) – P(A B) or F)
13. 13. 7/3/2013VSR 14 P(E F) = P(E).P(F/E)= P(F).P(E/F) . Provided P(E) and P(F) ≠ 0 # More than two events E, F and G, then by multiplication rule of probability P(E F G) = P(E).P(F/E).P(G/EF). Where EF = E F.
14. 14. 7/3/2013VSR 15 The given events are said to be Equally Likely, if none of them is expected to occur in preference to the other. Two events are said to be Independent, if the occurrence of one does not depend upon the other. Hence events E and F are independent event if P(EF) = P(E). P(F).
15. 15. 7/3/2013VSR 16 # For two independent events E and F, the addition theorem becomes, P(E or F) = P(E) + P(F) – P(E).P(F) Or P(E or F) ).().(1)(1 FPEPFEP
16. 16. 7/3/2013VSR 17 Let's consider "E1 and E2" as the event that "both E1 and E2 occur". If E1 and E2 are dependent events, then: P (E1 and E2) = P (E1) × P (E2 | E1) If E1 and E2 are independent events, then: P (E1 and E2) = P (E1) × P (E2)
17. 17. 7/3/2013VSR 18 For three dependent events E1, E2, E3, we have P(E1 and E2 and E3) = P(E1) × P(E2 | E1) × P(E3 | E1 and E2) For three independent events E1, E2, E3, we have P(E1 and E2 and E3) = P(E1) × P(E2) × P(E3)
18. 18. 7/3/2013VSR 19 Two or more events are said to be mutually exclusive if the occurrence of any one of them means the others will not occur (That is, we cannot have 2 events occurring at the same time). Thus if E1 and E2 are mutually exclusive events, then P(E1 and E2) = 0.
19. 19. 7/3/2013VSR 20 Suppose "E1 or E2" denotes the event that "either E1 or E2 both occur", then (a) If E1 and E2 are not mutually exclusive events: P(E1 or E2) = P(E1) + P(E2) − P(E1 and E2) We can also write: P(E1 ∪ E2) = P(E1) + P(E2) − P(E1 ∩ E2) A diagram for this situation is as follows. We see that there is some overlap between the events E1 and E2. The probability of that overlap portion is P(E1 ∩ E2).
20. 20. 7/3/2013VSR 21 (b) If E1 and E2 are mutually exclusive events: P(E1 or E2) = P(E1) + P(E2) Our diagram for mutually exclusive events shows that there is no overlap:
21. 21. 7/3/2013VSR 22 Exhaustive event. A set of events is said to be exhaustive if the performance of the experiment results in the occurrence of at least one of them. If a set of events E1, E2,……En , then for exhaustive events P(E1 E2 ….. En ) = 1. If E1, E2,……En are mutually exclusive and exhaustive events and any events E is said to be compound events, if 0)(,).( ()( 1 1 n i i i i n i i EifP E E PEP EEPEP
22. 22. 7/3/2013VSR 23 • The events E1, E2,……En represent a partition of the sample space S if they are pair wise disjoint, exhaustive and have non-zero probabilities. a) Ei Ej = , i ≠ j , i, j = 1, 2, 3, ……..n. b) Ei E2 …….. En = S c) P( Ei ) > 0 for all i = 1, 2, ………n.
23. 23. 7/3/2013VSR 24 P(A) = P(E1).P(A/E1) + P(E2).P(A/E2) + …….+ P(En).P(A/En)
24. 24. 7/3/2013VSR 25 Let E1, E2 ......En are n non empty events which constitute a partition of sample space S, then n j jj ii EAPEP EAPEP AEP 1 1 )/().(( )/().( )/( for any i = 1, 2, 3, ………n.
25. 25. 7/3/2013VSR 26 P(r) , the probability of r successes, is given by: r p rn qC rnrP )( Where p = the probability of successes. q = the probability of failure. p + q = 1 , i.e., q = 1 – p . Here n, p, q are called the parameters of Binomial Distribution.
26. 26. 7/3/2013VSR 27 Mean of binomial distribution Mean = np Variance V(X) = E(X2) – (E(X))2 . Mean of probability distribution = XP (X) Variance = X2P(X) – (mean)2 S.D. = X2P(X) – (mean)2
27. 27. 7/3/2013VSR 28 Variance = npq Standard Deviation ( ) npqDS ).(. Recurrence Formula ).(.. )1( )( )1( rP q p r rn rP
28. 28. 7/3/2013VSR 29 n i nni PxPxpxPixxE 1 2211 .......)( x1,x2 ……xn are possible values of the random variable x with probabilities P1 , P2 …..Pn respectively.
29. 29. 7/3/2013VSR 30
30. 30. 7/3/2013VSR 31 Q.1.A fair die is thrown, what is the probability that either an odd number or a number greater than 4 will up. (K.U) Let S be the sample space throwing of the die S = {1, 2, 3, 4, 5, 6}. i.e., n(S) = 6 Let A be the event of getting an odd number B be the number greater than 4 . A = {1, 3, 5} ; B = {5, 6} ; also A B = {5} n(A) = 3; n(B) = 2 ; n(A B) = 1 6 1 )( )( )( 3 1 6 2 )( )( )( 2 1 6 3 )( )( )( Sn BAn BAP Sn Bn BP Bn An AP )()()()( BAPBPAPBAP 3 2 6 123 6 1 3 1 2 1 The required probability = 2/3.
31. 31. 7/3/2013VSR 32 Q.2. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a 6. Find the probability that it is actually 6. (AICBSE) Let A be the man report that it is a 6. Let E1 be the event “6 has occurred” and E2 be that “6 has not occurred” P(E1 ) = 1/6; P(E1 ) = 1–P(E1 ) = 1–1/6 = 5/6 also P(A/ E1 ) = ¾ . [ i.e., speak 3 out of 4] P(A/ E2 ) = 1-P(A/ E1 ) = 1-3/4 = ¼
32. 32. 7/3/2013VSR 33 )/().()/().( )/().( )/( 2211 11 1 EAPEPEAPEP EAPEP AEP Formulae 8 3 8 24 8 1 24 8 8 1 25 5 8 1 8 1 4 1 . 6 5 4 3 . 6 1 4 3 . 6 1
33. 33. 7/3/2013VSR 34 Q3. An insurance company insured 2000 Scooter drivers, 3000 Car drivers and 4000 Truck drivers. The probabilities of their meeting with an accident respectively 0.04, 0.06 and 0.15. One of the insured persons meets with an accident, find the probability that he is a Car driver. (AICBSE) Let E1 , E2 , E3 , be the Scooter, Car and Truck drivers respectively. Let A be person that meets with an accident
34. 34. 7/3/2013VSR 35 9 4 400030002000 4000 )( 9 3 400030002000 3000 )( 3 2 400030002000 2000 )( 3 2 1 EP EP EP 100 15 15.0)/( ; 100 6 06.0)/(; 100 4 04.0)/( 3 21 EAP EAPEAP
35. 35. 7/3/2013VSR 36 Required probability ) 3 /(). 3 () 2 /(). 2 () 1 /(). 1 ( ) 2 /(). 2 ( )/ 2 ( EAPEPEAPEPEAPEP EAPEP AEP 43 9 60188 18 100 15 . 9 4 100 6 . 9 3 100 4 . 9 2 100 6 . 9 3
36. 36. 7/3/2013VSR 37 Q.4. Gowrave and Sowrave appear for an interview for two vacancies. The probability of Gowrave’s selection is 1/3 and that of Sowrave’s selection is 1/5. Find the probability that only one of them is selected. (CBSE) Let A be Gowrave’s selection B be the Sowrave’s selection. P(A) = 1/3 ; P(B) = 1/5. )( AP Probability that Gowrave not selected = 1 – P(A) = 1 – 1/3 = 2/3.
37. 37. 7/3/2013VSR 38 )(BP Probability that Sowrave not selected = 1 – P(B) = 1 – 1/5 = 4/5 Probability that only one of them is selected 5 2 15 2 15 4 5 1 . 3 2 5 4 . 3 1 )().()().( BPAPBPAP
38. 38. 7/3/2013VSR 39 Q.5. In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts of their output 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the total production and if found to be defective. Find the probability that it is manufactured by the machine B. (CBSE) Let E1 , E2 and E3 be the events of manufacturing the bolt by machine A, B and C respective.
39. 39. 7/3/2013VSR 40 Let A be the event that bolt drawn is defective 100 2 )/(; 100 4 )/(; 100 5 )/(.,. 321 EAPEAPEAPei Required probability 69 28 345 140 100 40 . 100 2 . 100 35 . 100 4 100 25 . 100 5 100 35 100 4 )()./()()./()()./( )()./( )/( 332211 22 2 EPEAPEPEAPEPEAP EPEAP AEP
40. 40. 7/3/2013VSR 41 Q.6. A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Find the probability that the number is divisible by 5 . (CBSE) Number of favourable cases m = four digit numbers, which are divisible by 5 = 6. The no. of exhaustive cases n = 4C4 = 4! =24. Required probability = m/n = 6/24 = ¼
41. 41. 7/3/2013VSR 42 Q.7. A lot contains 50 defective and 50 non- defective bulbs. Two bulbs are drawn at random, one by one, with replacement. The events A, B and C are defined as the first bulbs is defective, the second bulbs is non- defective, the two bulbs are both defective or non-defective, respectively. Determine whether A, B and C are pair wise independent. Sample space S = {DD, DN, ND, NN} . Where D = Defective bulb and N = Non-defective bulb
42. 42. 7/3/2013VSR 43 Thus A = {DD, DN} , B = {DN, NN} , C = {DD, NN}. P(A) = 2/4 = ½ , P(B) = 2/4 = ½ , P(C) = 2/4 = ½ . P(A B) = P(DN) = ¼ ; P(B C) = P(NN) = ¼ P(A C) = P(DD) = ¼ Now P(A). P(B) = ½ ½ = ¼ = P(A B) i.e., A and B are independent Similarly B and C and also A and C are independent Hence A, B, C are pair wise independent.
43. 43. 7/3/2013VSR 44 Q.8. An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white. Let W1 be : Ball drawn is white in the first draw. W2 be : Ball drawn is white in the second draw. and B1 be : Ball drawn is black in the first draw. Now, P(W2) = P(W1) . P(W2 / W1) + P(B1) P(W2 / B1)
44. 44. 7/3/2013VSR 45 nm m knmnm knmm knmnm mnmkm knm km nm m knm km nm m ))(( )( ))(( .. 2
45. 45. 7/3/2013VSR 46 Q.9. A card is drawn out from a well shuffled pack of 52 cards. If E is the event “the card drawn out is a king or queen” and F is the event “the card drawn out is a queen or an ace”, find the probability P(E/F). (CBSE) P(F) = P( card is queen or ace) = 8/52 = 2/13. P(E F) = P(card is queen) = 4/52 = 1/13
46. 46. 7/3/2013VSR 47 Q.10. If P(A) = 3/5 and P(B) = 1/5 find P(A B) if A and B are independent events. Given that A and B are independent events. P(A B) = P(A).P(B) = (3/5) (1/5) = 3/25
47. 47. 7/3/2013VSR 48 Q.11. If P(A) = 1/2 and P(B) = p ,P(A B) = 3/5 if A and B are mutually exclusive events. Given that A and B are mutually exclusive events. P(A B) = P(A) + P(B) i.e., 3/5= ½ + p p = 3/5 – ½ = 1/10.
48. 48. 7/3/2013VSR 49 Q.12. A husband and his wife appear for an interview for two posts. The probability of husband’s selection is 1/7 and that of wife’s selection is 1/5 . What is the probability that only one of them is selected.? Let A and B be the events of husband and wife respectively. Here P(A) = 1/7 and P(B) = 1/5 5 4 5 1 1)( 7 6 7 1 1)( BP andA
49. 49. 7/3/2013VSR 50 Here A and B are independent events
50. 50. 7/3/2013VSR 51 Q.13. A problem in Mathematics is given to three students whose chances of solving it are ½, 1/3, ¼. What is the probability in the following cases ? 1) That the problem is solved. (Kerala) 2) only one of them solved correctly (CBSE) 3) at teast one of them may solve it. Let A, B, C be the three event when the problem in maths is solved by the three students.
51. 51. 7/3/2013VSR 52 4 3 4 1 1)(; 3 2 3 1 1)(; 2 1 2 1 1)( . 4 1 )(; 3 1 )(; 2 1 )( CPBPAP CPBPAP 1) The probability that the problem is solved = Probability that the problem is solved by at least one student. 4 3 4 1 1 4 3 3 2 2 1 1 )().().(1 CPBPAP
52. 52. 7/3/2013VSR 53 2) The probability that only one solves it correctly )().().()().().()().().( CPBPAPCPBPAPCPBPAP 24 11 12 1 8 1 4 1 4 1 3 2 2 1 4 3 3 1 2 1 4 3 3 2 2 1
53. 53. 7/3/2013VSR 54 3) The probability that atleast one of them may solve the problem 4 3 4 1 1 4 3 3 2 2 1 1 )().().(1 CPBPAP
54. 54. 7/3/2013VSR 55 Q.14. In a bolt factory, machines A, B and C manufacture respectively 25%, 35%, 40% of the total. Of their output 5, 4 and 2% are defective. A bolt is drawn at random from the product. 1). What is the probability that the bolt drawn is defective? 2) If the bolt drawn is found to be defective, find the probability that it is a product of machine B?.
55. 55. 7/3/2013VSR 56 100 2 )/(; 100 4 )/(; 100 5 )/( 100 40 )(; 100 35 )(; 100 25 )( CDPBDPADandP CPBPAHereP Where D denotes defective bolt. 0345.0 100 2 . 100 40 100 4 . 100 35 . 100 5 . 100 25 )/()()/()()/()()().1 CDPCPBDPBPADPAPDP
56. 56. 7/3/2013VSR 57 2) By Baye’s theorem 69 28 345 140 100 2 . 100 40 100 4 . 100 35 100 5 . 100 25 100 4 . 100 35 )/().()/().()/().( )/().( )/( CDPCPBDPBPADPAP BDPBP DBP
57. 57. 7/3/2013VSR 58 Q.15. If the probability that person A will be alive in 20 years is 0.7 and the probability that person B will be alive in 20 years is 0.5, what is the probability that they will both be alive in 20 years? These are independent events, so P(E1 and E2) = P(E1) P(E2) = 0.7 0.5 = 0.35
58. 58. 7/3/2013VSR 59 Q.16. A fair die is tossed twice. Find the probability of getting a 4 or 5 on the first toss and a 1, 2, or 3 in the second toss. P(E1) = P(4 or 5) = 2/6 = 1/3 P(E2) = P(1, 2 or 3) = 3/6 = 1/2 They are independent events, so P(E1 and E2) = P(E1) P(E2) = 1/3 1/2 = 1/6
59. 59. 7/3/2013VSR 60 Q.17. It is known that the probability of obtaining zero defectives in a sample of 40 items is 0.34 whilst the probability of obtaining 1 defective item in the sample is 0.46. What is the probability of (a) obtaining not more than 1 defective item in a sample? (b) obtaining more than 1 defective items in a sample? (a) Mutually exclusive, so P(E1 or E2) = P(E1) + P(E2) = 0.34 + 0.46 = 0.8 (b) P(more than 1) = 1 − 0.8 = 0.2
60. 60. 7/3/2013VSR 61 Q.18. Find the mean and variance of the random variable X, where probability distribution is given by the following table: X -2 -1 0 1 2 3 P(X) 0.10 0.20 0.30 0.20 0.15 0.05 Mean = XP(X) - 2 0.10 + -1 0.20 + 0 0.30 + 1 0.20 + 2 0.15 + 3 0.05. = 0.25. Variance = X2P(X) – (mean)2 = - 22 0.10 + -12 0.20 + 02 0.30 + 12 0.20 + 22 0.15 + 32 0.05 – (0.25)2 = 1.85 – 0.0625 = 1.7875
61. 61. 7/3/2013VSR 62 Q.19. The SD of a binomial distribution (q + p)16 is 2 , its mean is ? SD = npq = 2 ., squaring npq = 2 . But n = 16 ; we know q = 1 – p 16p(1 – p) = 4 ; 4p(1 – p) = 2 4p2 – 4p + 1 = 0 (2p – 1)2 = 0 ; p = ½ We know mean = np = 16 ½ = 8.
62. 62. 7/3/2013VSR 63 Q.20. In a binomial distribution mean = 5 and variance = 4 , then the number of trials is ? Mean = np = 5. ; Variance = npq = 4 . npq/np = 4/5 q = 4/5. p = 1 – q = 1 – 4/5 = 1/5 Now, n p = 5 i.e., n 1/5 = 5 n = 25
63. 63. 7/3/2013VSR 64 Q.21.How many words can be formed from the letter of the word “COMMITTEE”. We have the formulae !!! ! 321 nnn n Total no. of letters = 9 ; M = 2, T = 2, E = 2 3 321 )!2( !9 !2!.2!.2 !9 !!! ! nnn n
64. 64. 7/3/2013VSR 65 Q.22. In a leap year , the probability of having 53 Friday or Saturday is ? In a non-leap year there are 365 days, 52 complete weeks and 1 day. That can be Monday, Tuesday, Wednesday, Thursday, Friday and Saturday. P(Friday) = 1/7 ; P( Saturday) = 1/7 The required probability = 1/7 + 1/7 = 2/7 .
65. 65. 7/3/2013VSR 66 Q.23. A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a Jack and an Eight. P(Jack) = 4/52 ; P(8) = 4/52 Multiplication rule P(Jack and Eight) = 4/52 4/52 = 16/2704 = 1/169.
66. 66. 7/3/2013VSR 67 Q.23. Obtain binomial distribution, if n = 6, p = 1/5 We have p = 1/5 and n = 6. We know q = 1 – p = 1 – 1/5 = 4/5 . Binomial distribution = (q + p)n =
67. 67. 7/3/2013VSR 68 Q.24. The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. What is the probability that out of 5 such bulbs (i) None ; (ii) not more than one ; (iii) at least one will fuse after 150 days of use. Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials. It is given that, p = 0.05 ; q = 1 – p = 1 – 0.05 = 0.95. X has a binomial distribution with n = 5 and p = 0.05
68. 68. 7/3/2013VSR 69 (i) P (none) = P(X = 0) Using this formulae
69. 69. 7/3/2013VSR 70 (ii) P (not more than one) = P(X ≤ 1)
70. 70. 7/3/2013VSR 71 (iii) P (at least one) = P(X ≥ 1)
71. 71.  The End of Probability  The End of Probability  The End of Probability  The End of Probability  The End of Probability  The End of Probability  The End of Probability  The End of Probability  The End of Probability 7/3/2013VSR 72