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  1. 1. ProbabilityAn experiment is a process - natural or set updeliberately - that has an observable outcome. Inthe deliberate setting, the word experiment andtrial are synonymous. An experiment has a randomoutcome if the result of the experiment cant bepredicted with absolute certainty. An event is acollection of possible outcomes of an experiment. Anevent is said to occur as a result of an experiment ifit contains the actual outcome of that experiment.Individual outcomes comprising an event are said tobe favorable to that event. Events are assigned ameasure of certainty which is called probability (ofan event.)Quite often the word experiment describes anexperimental setup, while the word trial applies to
  2. 2. actually executing the experiment and obtaining an outcome. P(E) = A formal theory of probability has been developed in the 1930s by the Russian mathematician A. N. Kolmogorov. The starting point is the sample (or probability) space - a set of all possible outcomes. Lets call it Ω. For the set Ω, a probability is a real-valued function P defined on the subsets of Ω:(*) P: 2Ω → [0, 1] Thus we require that the function be non-negative and that its values never exceed 1. The subsets of Ω for which P is defined are called events. 1-element events are said to be elementary. The function is
  3. 3. required to be defined on the empty subset Φ andthe whole set Ω:(1) P(Φ) = 0, P(Ω) = 1.This says in particular that both Φ and Ω are events.The event Φ that never happens is impossible andhas probability 0. The event Ω has probability 1 andis certain or necessary.If Ω is a finite set then usually the notions of animpossible event and an event with probability 0coincide, although it may not be so. If Ω is infinitethen the two notions practically never coincide. Asimilar dichotomy exists for the notions of a certainevent and that with probability 1. Examples will begiven shortly.The probability function (or, as it most commonlycalled, measure) is required to satisfy additional
  4. 4. conditions. It must be additive: for two disjointevents A and B, i.e. whenever A ∩ B = Φ,(2) P(A∪B) = P(A) + P(B),which is a consequence of a seemingly more generalrule: for any two events A and B, their union A∪Band intersection A∩B are events and(2) P(A∪B) = P(A) + P(B) - P(A ∩ B).Note, however, that (2) can be derived from (2).Indeed, assuming that all the sets involved areevents, events A - B and A ∩ B are disjoint as are B -A and A ∩ B. In fact, all three events A - B, B - A,and A ∩ B are disjoint and the union of the three isexactly A∪B. We have, P(A)+P(B)= (P(A - B) + P(A ∩ B)) + (P(B - A) +
  5. 5. P(A ∩ B)) = P(A∪B) + P(A ∩ B)Some properties of P follow immediately from thedefinition. For example, denote the complement of anevent A, A = Ω - A. It is also called a complementaryevent. Then, naturally, A and A’ are disjoint, A ∩ A’ = Φ,and 1 = P(Ω) = P(A) + P(A’). In other words, (4) P(A) = 1 - P(A’). if B = A∪C for disjoint A and C, then P(B) = P(A∪C) = P(A) + P(C) ≥ P(A). In other words, if A is a subset of B, A B, then
  6. 6. (5) P(B) ≥ P(A).Probability is a monotone function - the fact thatjibes with our intuition that a larger event, i.e. anevent with a greater number of favorable outcomes,is more likely to occur than a smaller event.Disjoint events do not share favorable outcomes and,for this reason, are often called incompatible or,even more assertively, mutually exclusive.Finite Sample SpacesTossing a coin. The experiment is tossing a coin (orany other object with two distinct sides.) The coinmay land and stay on the edge, but this event is soenormously unlikely as to be considered impossibleand be disregarded. So the coin lands on either oneor the other of its two sides. One is usually calledhead, the other tail. These are two possible
  7. 7. outcomes of a toss of a coin. In the case of a singletoss, the sample space has two elements thatinterchangeably, may be denoted as, say, {Head, Tail}, or {H, T}, or {0, 1}, ...Rolling a die. The experiment is rolling a die. Acommon die is a small cube whose faces showsnumbers 1, 2, 3, 4, 5, 6 one way or another. Thesemay be the real digits or arrangements of anappropriate number of dots, e.g. like theseThere are six possible outcomes and the samplespace consists of six elements: {1, 2, 3, 4, 5, 6}.
  8. 8. Many random variables may be associated with thisexperiment: the square of the outcome f(x) = x2,with values from {1, 4, 9, 16, 25, 36},Drawing a card. The experiment is drawing a cardfrom a standard deck of 52 cards. The cards are oftwo colors - black (spades and clubs) and red(diamonds and hearts), four suits (spades, clubs,diamonds, hearts), 13 values (2, 3, 4, 5, 6, 7, 8, 9,10, Jack, Queen, King, Ace). (Some decks use 4colors, others use different names. For example, aJack may be called a Knave. We shall abbreviate thenamed designations as J, Q, K, A.) There are 52possible outcomes with the sample space {2♠, 2♣, 2♦, 2♥, 3♠, 3♣, 3♦, 3♥, ..., A♠, A♣,
  9. 9. A♦, A♥}.Of course, if we are only interested in the color of adrawn card, or its suite, or perhaps the value, thenit would be as natural to consider other samplespaces: {b, r}, {♠, ♣, ♦, ♥} or {2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A}.Choosing a birthday. The experiment is to select asingle date during a given year. This can be done, forexample, by picking a random person and inquiringfor his or her birthday. Disregarding leap years forthe sake of simplicity, there are 365 possiblebirthdays, which may be enumerated
  10. 10. {1, 2, 3, 4, ..., 365}.Tossing two coins. The experiment is tossing twocoins. One may toss two coins simultaneously, or oneafter the other. The difference is in that in thesecond case we can easily differentiate between thecoins: one is the first, the other second. If the twoindistinguishable coins are tossed simultaneously,there are just three possible outcomes, {H, H}, {H,T}, and {T, T}. If the coins are different, or if theyare thrown one after the other, there are fourdistinct outcomes: (H, H), (H, T), (T, H), (T, T),which are often presented in a more concise form:HH, HT, TH, TT. Thus, depending on the nature ofthe experiment, there are 3 or 4 outcomes, with thesample spaces
  11. 11. Indistinguishable coins {{H, H}, {H, T}, {T, T}}. Distinct coins {HH, HT, TH, TT}Rolling two dice. The experiment is rolling two dice.If the dice are distinct or if they are rolledsuccessively, there are 36 possible outcomes: 11, 12,..., 16, 21, 22, ..., 66. If they are indistinguishable,then some outcomes, like 12 and 21, fold into one.There are 6×5/2 = 15 such pairs giving the totalnumber of possible outcomes as 36 - 15 = 21. In thefirst case, the sample space is {11, 12, ..., 16, 21, 22, ..., 66}.
  12. 12. When we throw two dice we are often interested notin individual numbers that show up, but in their sum.The sum of the two top numbers is an example of arandom variable, say Y(ab) = a + b (where a, b rangefrom 1 through 6), that takes values from the set {2,3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. It is also possible tothink of this set of a sample space of a randomexperiment. However, there is a point in workingwith random variables. It is often a convenience tobe able to consider several random variables relatedto the same experiment, i.e., to the same samplespace. For example, besides Y, we may beinterested in the product (or some other function) ofthe two numbers.Infinite Discrete Sample SpacesFirst tail. The experiment is to repeatedly toss a coin untilfirst tail shows up. Possible outcomes are sequences of H
  13. 13. that, if finite, end with a single T, and an infinite sequence of H: {T, HT, HHT, HHHT, ..., {HHH...}}. As we shall see elsewhere, this is a remarkable space that contains a not impossible event whose probability is 0. One random variable is defined most naturally as the length of an outcome. It draws values from the set of whole numbers augmented by the symbol of infinity: {1, 2, 3, 4, ..., ∞}. Mutually exclusive events: In the experiment of rolling a die,a sample space is S={1,2,3,4,5,6}. Consider A `an odd numberappears’ and B `an even number appears’ as A={1,3,5} , B={2,4,6} A∩B=∅ Exhaustive events: Consider the experiment of rolling a die, asample space is S={1,2,3,4,5,6}.Let’s define the following Events A={1,2,3}, B={3,4} C={5,6} , then AUBUC=S OR Ei ∩ Ej = ∅for i≠jFOR CLASS ---XI
  14. 14. Question:21(ncert) In a class of 60 students, 30 opted for NCC,32 opted for NSS and 24 opted for both NCC and NSS. If One of these students is selected at random, find theprobability that(i) The student opted for NCC or NSS.(ii) The student opted neither NCC nor NSS. (iii) The student opted for NSS but not NCC. Solution: P(A) =30/60, P(B) = 32/60, P(A∩B) = 24/60 , wherethe event that student choose NCC be taken by A andThe student choose NSS be taken as B. (i) P(AUB) = 19/30 (ii) P(A’∩B’) = P(AUB)’ = 1 – P(P(AUB)=11/30 (De Morgan’slaw) (iii) P(A’∩B) = P(B) – P(A∩B) ( Only B) = 8/15 – 6/15 = 2/15. If A, B, C ARE THREE EVENTS ASSOCIATED WITH A RANDOMEXPERIMENT( Addition theorem fo three events) P(A B C) = P(A) + P(B) + P(C) – P(B∩C) – P(A∩B) – P(A∩C) +P(A∩B∩C) Question5 ( misc.): Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that (a) You both enter the same section. (b) You both enter the different section.
  15. 15. Answer: (a) required probability = = 17/33 [ C(n,r) = , n! = n(n-1)(n-2)……3.2.1 ] (b) required probability = = 16/33 or 1 – 17/33 = 16/33. Question 6 (misc.): Three letters are indicated to three persons and an envelope is addressed to each of them. The letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the prob. That at least one letter is in it’s proper envelope. Answer: Let the envelope be denoted by E1, E2, E3 are the corresponding letters by L1, L2, L3 . Total outcomes = 3! = 6 There are 4 favourable cases when at least one letter is in proper envelope = 4/6 = 2/3. E1 E2 E3 L1 L3 L2 L3 L2 L1 L2 L1 L3 L1 L2 L3 L2 L3 L1 L3 L1 L2 2 cases when no letter goes in correct envelope 1 case when 2 letters & hence all are in correct 3 cases when 1 letter is in correct & 2 letters in wrongenvelope.
  16. 16. Question 9(misc.): If 4- digit nos. greater than or equal to 5000are randomly formed from the digits 0,1,3,5 and 7, what is the prob. Of forming a number divisible by5 when (i) the digits may be repeated, (ii) the repetition of digits isnot allowed.Answer: 5 or 7 in the extreme left position i.e., in thousand place (i) when digits may be repeated , let 4 places in a 4 digit number be marked as below: I II III IV 1 5 5 5 ways If place I is filled by 5, then each of the places II,III,IV can be filled in 5 ways and hence 5x5x5=125 ways Similarly if place number I is filled by 7, then total no. of ways = 125 , therefore n(S) =125+125=250 Now to find favourable cases , when the number is divisible by 5 I II III IV (2ways 5,7) 5 5 (0 or 5)2 ways favourable outcomes = 100 , then req. prob. = 100/ 250=2/5 (ii) No repetition 5 is fixed at I place and II,III,IV can be filled in (4)(3)(2) = 24 ways Similarly when 7 is fixed in place at I , then total ways = 24+24=48ways favourable outcomes = 18 ways as I II III IV 5 - - 0 7 - - 0 7 - - 5 3 2
  17. 17. Total no. of ways = 6+6+6 = 18, so req. pob. = 18/48=3/8.Question (misc.): A combination lock on a suitcase has 3wheels each labeled with nine digitsfrom 1 to 9 . If an opening combination is a particularsequence of three digits with no repeats, what is the prob. Ofa person guessing the right combination?Answer: 1/p(9,3) , p(n,r) =SOME USEFUL RESULTS: 1. P(A∩B’) = P(A-B) = P(A) – P(A∩ B) = P( Only A) 2. P(A’∩ B’) = P(AUB)’ = 1 – P(AUB) and P(A’UB’) = 1 –P(A∩B) 3. P(at least one) = 1 – P(None) = 1 – P(0)4. For any two events A and B, P(A∩B) ≤P(A)≤P(AUB)≤P(A)+P(B)Question: 1 A box contains 100 bolts and 50 nuts, It is giventhat 50% bolt and 50% nuts are rusted. Two Objects are selected from the box at random. Find theprobability that both are bolts or both are rusted. Answer: A= event of getting a bolt , B= event of getting arusted item and A ∩B= event of getting a rusted bolt. Required probability = P(A)+P(B)-P(A ∩B) = + - = 260/447.Question:2 A pair of dice is rolled. Find the probability ofgetting a doublet or sum of number to be at least 10. Solution: Total no. = 36 , A = event of getting a doublet i.e.,(1,1), (2,2), (3,3), (4,4), (5,5), (6,6) P(A) = 1/6 , P(B) = 6/36 = 1/6 {(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)+, P(A∩B) = 1/18
  18. 18. P(AUB) = 5/18.**Question:3 A pair of dice is rolled. Find the probability thatthe sum of the number on the two faces is (i) neither divisible by 3 nor by 4 ? (ii) is either a multiple of3 or 2?Answer: A = event that sum of numbers on the two faces isdivisible by 3 = *(1,2),(2,1),(1,5),(5,1)…..(5,4),(6,6)+ = 12 , B= sum divisible by 4 = 9 *(1,3), (2,2), (2,6)……(4,4),(6,6)+ P(A) = 1/3, P(B) = ¼, P(A∩B) = 1/36 , P(AUB) = 5/9 , then(i) P(A’∩B’) = 1 – P(AUB) = 4/9. (ii) A = sum of numbers on two faces multiple of 2 , B =multiple of 3 P(A) = 18/36 = ½ , P(B) = 12/36 = 1/3 P(AUB) = 2/3.**Question:4 Of the students attending a lecture, 50% couldnot see what was written on the board and 40% could not hearwhat the lecture was saying. Most unfortunate 30% fell intoboth of these categories. WhatIs the probability that a student picked at random was able tosee and hear satisfactorily?Answer: Let A = event that a student can see satisfactorily , B= student can hear satisfactorily P(A’) =50% = ½ , P(B’) = 40% = 2/5 , P(A) =1- p(A’) = ½ ,P(B) 1 – P(B’) = 3/5 P(A’∩B’) = 30% = 3/10 , P(AUB)’ = 1 – P(AUB) = 3/10 ⇨P(AUB)=7/10. P(A∩B) = 2/5.Question:5 Two dice are tossed once. Find the prob. Of getting“an even number on the first die or a total of 2”.[Hint A = even number on first die , B =total of 2 Find P(AUB) = P(A) + P(B) = 19/36.]Question:6 Three squares of chess board are selected atrandom. Find the porb. Of getting 2 squares of one Color and other of a different color .
  19. 19. Answer Total 64 squares in chess board , 32 are of white and32 of black colour ,selecting as 1w & 2b or 1b &2w Req. prob. = = 16/21.Question:7 Three of the six vertices of a regular hexagon arechosen at random. What is the prob.That the triangle with these vertices is equilateral?Answer: Total no. of triangles c(6,3) = 20 ( since no threepoints are collinear). Of these only ∆ ACE; ∆BDF are equilateral ∆’s. so req. prob. = 2/20=1/10. B C A D F EQuestion:8 If A, B, C are three mutually exclusive andexhaustiveEvents of an experiment such that 3P(A) = 2P(B) = P(C).Then find P(A).Answer: use P(A)+P(B)+P(C)=1 as mutually exclusive andexhaustive events, p(A)=2/11Question: 9 Two unbiased dice are thrown. Find the prob. Thatneither a doublet nor a total of 10 will appear.Answer: (5,5) is only common event , P(neither a doubletnor a total of 10) = 1-P(doublet or total 10) = 1 – 2/9 = 7/9.Question:10 In a class of 100 students, 60 drink tea, 50 drinkcoffee and 30 drink both. A student from this class is selectedat random. Find the prob. That the student takes (i) at leastone of 2 drinks (ii) only one of two drinks.Answer: You can use venn diagram also , (i) P(TUC)=4/5 ,
  20. 20. (ii) P(event of taking only tea or only coffee)= P(EUF)=P(E)+P(F)=1/2 = exactly one drink. FOR CLASS --XII conditional ProbabilityIf E and F are two events associated with the samesample space of a random experiment, the conditionalProbability of the event E given that F has occurred,i.e., P(E/F) is given by P(E/F) = providedP(F)≠0.PROPERTIES:(I) Let E & F be events of a sample space S of anexperiment, then we have P(S/F) = P(F/F) = 1(II) If A & B are any two events of a sample space Sand F is an event of S such that P(F)≠0, then P((AUB)/F) = P(A/F) + P(B/F) – P((A∩B)/F) ,for disjoint events ⇨ P((A∩B)/F) =0.(IIII) P(E’/F) = 1 – P(E/F).EXAMPLE(ncert): Consider the experiment of tossing a coin. Ifthe coin shows head, toss it again but if it shows tail, then throwa die. Find the conditional prob. Of the event that `the die showsa number greater than 4’ given that `there is at least one tail’.
  21. 21. SOLUTION: S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5),(T,6)} E ={(T,5), (T,6)} , F ={(H,T), (T,1), (T,2), (T,3), (T,4), (T,5),(T,6)+ & E∩F =*(T,5), (T,6)+ P(F) = ¼+1/12+1/12+1/12+1/12+1/12+1/12 = ¾ [ P(H,T)= ½.1/2 =1/4 , P(T,1) = ½ .1/6 =1/12]P(E∩F) = 1/12+1/12= 1/6 , then P(E/F) = =(1/6)/(3/4) = 2/9. Question13(ncert): An instructor has a question bankconsisting of 300 easy true/false questions, 200 difficulttrue/false Questions , 500 easy multiple choice questions and 400 difficultmultiple choice questions. If a question is selected at randomfrom the question bank, what is the prob. That it will be an easyquestion give it is a multiple choice question?Answer: E denotes easy questions, D = difficult , T = true/falsequestions and M =multipleTotal no. of questions = 1400, no. of easy multiple choicequestions = 500P(E∩M) = 500/1400, no. of multiple choice questions500+400=900P(M) = 900/1400 , P(E/M) = P(E∩M)/P(M) = 5/9.Question: A can solve 90% of the problems given in a book andB can solve 70%. What is the prob. That at least one of them willsolve the problem, selected at random from the book?
  22. 22. Answer: Use P(AUB)= 1 – P(A’)P(B’), then answer is .97, eventsare indep.]Question: A can hit a target 4 times in 5 shots, B 3 times in 4shots, and C 2 times in 3 shots. Calculate the prob. That (i) A, B, C all may hit (ii) B, C may hit and A may lose (iii)any two of A, B and C will hit the target (iv) non of them will hit the target.Answer: Hint: A, B, C are indep. (i) P(A∩B∩∩C) =P(A).P(B).P(C) =4/5.3/4.2/3, (ii) find P(A’∩B∩C)= 1/10 (iii) P(A∩B∩C’)U P(A’∩B∩C)U P(A∩B’∩C)= 13/30 (IV)P(A’∩B’∩C’)= 1/60, same as part (i)-Question: Three persons A, B, C throw a die in succession till onegets a `six’ and wins the game. Find their respective prob. Of winning , if A begins.Answer: A wins if gets `six’ in Ist or 4th or 7th …. Throw ofunbiased die , then P(A)=1/6 and P(A’)=5/6 A will get 4th throw if he fails in 1st , B fails in in second and Cfails in 3rd throw. Prob. Of winning of A in 4th throw= P(A’∩A’∩A’∩A) =(5/6)3 .1/6 Similarly, prob. Of winning of A in 7th throw = (5/6)6 . 1/6and so on… Prob. Of winning of A 1/6+(5/6)3 .1/6 + (5/6)6. 1/6 +…….=
  23. 23. = 36/91B wins if gets a six in 2nd or 5th or 8th or……..throw. Prob. Of winning of B = (5/6)1/6+(5/6)4.1/6+(5/6)7.1/6+…… 30/91, for prob. Of C = 1 –[P(A)+P(B)]= 25/91.The probability of 7 when rolling two die is 1/6 (= 6/36)because the sample space consists of 36 equiprobableelementary outcomes of which 6 are favorable to the eventof getting 7 as the sum of two die. Denote this event A: P(A) = 1/6.Consider another event Bwhich is having at least one 2. There are still 36 elementaryoutcomes of which 11 are favorable to B; therefore, P(B) =11/36. We do not know whether B happens or not, but this isa legitimate question to inquire as to what happens if itdoes. More specifically, what happens to the probability of Aunder the assumption that B took place?The assumption that B took place reduces the set of possibleoutcomes to 11. Of these, only two - 25 and 52 - arefavorable to A. Since this is reasonable to assume that the 11elementary outcomes are equiprobable, the probability of Aunder the assumption that B took place equals 2/11. Thisprobability is denoted P(A|B) - the probability of A assumingB: P(A|B) = 2/11. More explicitly P(A|B) is called theconditional probability of A assuming B. Of course, for anyevent A, P(A) = P(A|Ω), where, by convention, Ω is theuniversal event - the whole of the sample space - for which
  24. 24. all available elementary outcomes are favorable.We see that, in our example, P(A|B) ≠ P(A). In general, thismay or may not be so.Retracing the steps in the example, P(A|B) = P(A∩B) / P(B),and this is the common definition of the conditionalprobability. The formula confirms to the earlier definitions: P(A|Ω) = P(A∩Ω) / P(Ω) = P(A)since P(Ω) = 1 and A∩Ω = A (because A⊂Ω.)For another example, lets look at the events associated withrolling a dice in the form of octahedron, a shape with eightfaces. The sample space naturally consists of 8 equiprobableoutcomes: Ω = {1, 2, 3, 4, 5, 6, 7, 8}.Let A be the event of getting an odd number, B is the eventgetting at least 7. Then P(A) = 4/8 = 1/2, P(B) = 2/8 = 1/4.A∩B = {7}, so that P(A|B) = P(A∩B)/P(B) = 1/2 = P(A).Observe also that P(B|A) = P(A∩B)/P(A) = 1/4 = P(B).On the other hand, define A+ = {1, 2, 3, 5, 7} and
  25. 25. A- = {3, 5, 7}.Then P(B|A+) = P(A+∩B)/P(A+) = 1/5 < 1/4 = P(B),whilst P(B|A-) = P(A-∩B)/P(A-) = 1/3 > 1/4 = P(B).So we see that, in general, there is no definite relationshipbetween the probability P(B) and the conditional probabilityP(B|A). They may be equal, or one of them may be greaterthan the other. In the former case the events are said to beindependent.Bayes TheoremBayes theorem (or Bayes Law and sometimes Bayes Rule) isa direct application of conditional probabilities. Theprobability P(A|B) of "A assuming B" is given by the formula P(A|B) = P(A∩B) / P(B) [it is known as multiplication rule of prob.] which for our
  26. 26. purpose is better written as P(A∩B) = P(A|B)·P(B). The left hand side P(A∩B) depends on A and B in asymmetric manner and would be the same if we started withP(B|A) instead: P(B|A)·P(A) = P(A∩B) = P(A|B)·P(B).This is actually what Bayes theorem is about:(1) P(B|A) = P(A|B)·P(B) / P(A).Most often, however, the theorem appears in a somewhatdifferent form(1) P(B|A) = P(A|B)·P(B) / (P(A|B)P(B) + P(A|B)P(B)),where B is an event complementary to B: B∪B = Ω, theuniversal event. (Of course also B∩B = Φ, an empty event.)This is because
  27. 27. A = A ∩ (B ∪ B) = (A∩B) ∪( A∩B)and, since A∩B and A∩B are mutually exclusive, P(A) = P(A∩B ∪ A∩B) = P(A∩B) + P(A∩B) = P(A|B)P(B) + P(A|B)P(B).More generally, for a finite number of mutually exclusive andexhaustive events Hi (i = 1, ..., n), i.e. events that satisfy Hk ∩ Hm = Φ, for k ≠ m and H1 ∪ H2 ∪ ... ∪ Hn = Ω,Bayes theorem states that P(Hk|A) = P(A|Hk) P(Hk) / ∑i P(A|Hi) P(Hi),where the sum is taken over all i = 1, ..., n.Question: (ncert) A labrotary blood test is 99% effective in
  28. 28. detecting a certain disease when it is in fact , present. However , the test also yields a false positive result for0.5% of the healthy person tested(i.e., if a healthy person istested , then, with prob. .005, the test will imply he has thedisease). If 0.1% of the population actually has the disease,what is the prob. That a person has the disease given that histest result is positive?Answer: E , E’ are events that a person has disease and doesnot have, A = blood test is positive P(E) = 0.1% , P(E’) = .999, P(A/E) = .99 and P(A/E’)=.005 P(E/A) = 22/133.Question(NCERT) A card from a pack of 52 cards is lost.From the remaining cards of the pack, two cards are drawnand found to be both dimond. F ind the prob. Of the lostcard being a diamond.Answer: E1 , E2 are events that lost card is diamond and
  29. 29. not, A denotes a diamond (lost card) P(A/E1)=C(12,2)/C(51,2) = (12 /(51 ) , P(A/E2) = C(13,2)/C(51,2)=13 /51 , P(E1/A) = 11/50Conditional Probability and Independent Events P(A|B) = P(A∩B) / P(B), P(B|A) = P(A∩B) / P(A). The events A and B are said to be independent provided P(A|B) = P(A), or, which is the same P(B|A) = P(B). Neither the probability of A or B is affected by the occurrence (or a occurrence) of the other event. A symmetric way of expressing the same fact is this P(A∩B) = P(A) P(B).
  30. 30. Multiplication rule of prob. For more than two events:P(E∩F∩G) = P(E)P(F/E)P(G/(E∩F)= P(E)P(F/E)P(G/EF) Independent Events and Independent Experiments1. Independent experiments: When two events E & F are such that the prob. Of occurrence of one of them is not affected by occurrence of the other. The trials - the experiments - may or may not affect the outcomes of later trials. If they do, the experiments are called dependent; otherwise, they are independent.2. Events may or may not be independent; two events, A and B, are independent iff P(A∩B) = P(A) P(B).3. Some times there a confusion b/w indep. Events & mutually exclusive events. Indep. Is related to prob. Of events Whereas mutually exclusive is defined in terms of events (subset of sample space).Indep. events, may have common
  31. 31. outcome but mutually never have outcome common, so they are different in meaning Nonzero prob. In two indep. Events cannot be nonzero prob. In mutually exclusive events , converse is also not true. If A & B are two indep. Events, then the prob. Of occurrenceof at least one of A & B [P(AUB)] is given by 1 – p(A’)P(B’)Consider tossing a coin three times in a row. Since each of thethrows is independent of the other two, we consider all 8 (= 23)possible outcomes as equiprobable and assign each theprobability of 1/8. Here is the sample space of a sequence ofthree tosses: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.There are 28 possible events, but we are presently interested in,say, two: A = {HHH, HTH, THH, TTH} and B = {HHH, HHT, THH, THT}.
  32. 32. A is the sequence of tosses in which the third one came upheads. B is the event in which heads came up on the secondtoss. Since each contains 4 outcomes out of the equiprobable 8, P(A) = P(B) = 4/8 = 1/2.The result might have been expected: 1/2 is the probability ofthe heads on a single toss. Are events A and B independentaccording to the definition? Indeed they are. To see that,observe that A ∩ B = {HHH, THH},the event of having heads on the second and third tosses. P(A ∩B) = 2/8 = 1/4. Further, lets find the conditional probabilityP(A|B): P(A|B) = P(A ∩ B) / P(B) = 1/4 / 1/2 = 1/2 = P(A).
  33. 33. So that P(A|B) = P(A) and, according to the definition, events Aand B are independent, as expected.Bernoulli trial is an experiment with only two possible outcomesthat have positive probabilities p and q such that p + q = 1. Theoutcomes are said to be "success" and "failure", and arecommonly denoted as "S" and "F" or, say, 1 and 0.For example, when rolling a die, we may be only interestedwhether 1 shows up, in which case, naturally, P(S) = 1/6 andP(F) = 5/6. If, when rolling two dice, we are only interestedwhether the sum on two dice is 11, P(S) = 1/18, P(F) = 17/18.The Bernoulli process is a succession of independent Bernoullitrials with the same probability of success. One importantquestion about a succession of n Bernoulli trials is theprobability of k success.Since the individual trials are independent, we are talking of theproduct of probabilities of successes and failures. Such a
  34. 34. product is independent of the order in individual successes andfailures come about. For example, P(SSFSF) = P(SFFSS) = P(FFSSS) = p3q2.In general, the probability of k successes in n trials is denotedb(k; n, p) and is equal to b(k; n, p) = C(n, k)pkqn - k,where C(n, k) is the binomial coefficient n choose k. Observethat, by the binomial formula, ∑b(k; n, p) over k from 0 to n isexactly 1: ∑b(k; n, p) = ∑C(n, k)pkqn - k = (p + q)n = 1.As a function of k, b(k; n, p) is known as the binomialdistribution and plays an important role the theory ofprobabilities.Binomial distribution is the distribution of a total number ofsuccesses in a given number of Bernoulli trials. The commonnotation is b(k; n, p), where k is the number of successes, n is
  35. 35. the number of trials, p is the probability of success. We knowthat b(k; n, p) = C(n, k) pk(1 - p)n - k.The result for the expected value np = ∑ k b(k; n, p) might havebeen anticipated given the interpretation of the probability as arelative frequency.Note that the expected value np is always different from themost likely value (n + 1) p, provided of course p ≠ 0. ASSIGNMENT CLASS XI PROBABILITY 1. A pair of dice is rolled. Consider the following events: A : the sum is greater than 8, B : 2 occurs on either die , C : the sum is atleast 7 and a multiple of 3. Which pair of events are mutually exclusive? 2. A card is drawn at random from a deck of 52 playing cards. Find the probability that it is: (i) an ace (ii) a jack of hearts (iii) a three of clubs or a six of diamonds (iv) a heart
  36. 36. (v) any suit except heart (vi) a ten or a spade (vii) neither a fournor a club(viii) an honours card (ix) a face card (x) a spade or a face card3. Four cards are drawn at random from a pack of 52 cards. Findthe probability of getting:(i) all the four cards of the same suit (ii) all the four cards of thesame number(iii) one card from each suit (iv) all four face cards(v) two red cards and two black cards (vi) all cards of the samecolor (vii) getting four aces4. An urn contains 9 red, 7 white and 4 black balls. If two balls aredrawn at random, find probability that:(i) both the balls are red (ii) one ball is white(iii) the balls are of the same color (iv) one is white and other isred5. A five digit number is formed by the digits 1, 2, 3, 4, 5 withoutrepetition. Find the probability that thenumber is divisible by 4.6. One number is chosen from numbers 1 to 200. Find theprobability that it is divisible by 4 or 6.7. The letters of word “SOCIETY” are placed at random in a row.What is the probability that three vowelscome together?8. Fine the probability that in a random arrangement of the lettersof the word “UNIVERSITY” the two I’scome together.9. The letters of the word “MUMMY” are placed at random in arow. What is the chance that letters at theextreme are both M ?** 10. A bag contains 50 tickets numbered 1, 2, 3… ,50 of whichfive are drawn at random and arranged inascending order of magnitude x1 < x2 < x3< x4 <x5 . Find theprobability that x3= 30 .
  37. 37. ANSWERS: 1. A, B; B, C are mutually exclusive; but A, C are not.2.(i ) 1/13 (ii ) 1/52 (iii ) 1/26 (iv ) 1/4 (v ) 3/4 (vi ) 4/13 (vii ) 9/13(viii ) 4/13 (ix ) 3/13 (x ) 11/263. (i ) 44/4165 (ii ) 1/20825 (iii ) 2197/20825 (iv ) 99/54145 (v )325/833 (vi ) 92/833 (vii ) 1/2707254.(i ) 18/95 ( ii) 91/190 (iii ) 63/190 (iv ) 63/190 5. 1/5 , 6. P(A)=50/200, P(B)= 33/200 & P(A∩B)=16/200,P(AUB)= 67/200 , 7. (5!.3!)/7! = 1/7 , 8. 1/5 , 9. 3/10 , 10.X1,x2 < 30 , they come from tickets 1 to 29 & x4,x5>30 comefrom 20 (31to 50) REQ. PROB. =( 29C2.20C2)/50C5 = 551/15134.