Class 9

Class 9 February 19, 2016 Todays topic is Probability. In this topic we will be working on the same track as we. so we will first talk about theory and then do 25 - 26 questions to get a clear idea. again i would suggest you to stick with one thing that F avourableOptions T otalN umberOf Options . No matter how kind of question is asked stick with this definition. there are three types of questions in your paper. Depending on • Cards • dice • Head and tails Lets start with cards Now everybody knows there are 13 Spades, 13 Cubs, 13 Hearts and 13 diamonds in a pack of cards. Now if i say there is a pack of 52 cards, you pick one card. What is the probability that it comes out to be spade? 13 52 . We will stick with basic definition. We have a pack of 52 cards. We have to select a card. In how many ways can we s elect card out of 52 ? 52 C1 . this signifies the total number of ways in which you can select 1 card out of 52 cards. which is total number of ways. Now favourable ways are you want to select 1 card outof 13 card 13 C1 so 13 1 probability is 52 C C1 Now we talked about and means multiplication and OR means addition. These words will also be used in probability What if i say there is a pack of 52 cards, you select one card what is the probaility that it is either spade OR diamond. 13 13 C1 C1 52 C + 52 C 1 1 Now lets see the inportamce of AND If i say there is a pack of 52 cards i select 2 of them what is the probability 1 of them is spade and other is diamond ? just think about the question not the answer Has the order being given to you ?NO What i mean by order is, 1st can be spade and next diamond or 1st diamond next spade. What is the probability 1

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C1 52 C 1

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C1 C1 1 × 51 C C1 + 52 C1 × 51 C1 now either this can happen or that can happen. What does OR signifies ? ideally you should do both the things seprately and add them but it got quite cumbersome regards to calculations so rather than this i would suggest you to do it in a short way if i have to select 2 cards out of 52 cards, in how many ways can we do that ? 52 C2 now favourable what you want? 1 spade 13 C1 AND 1 diamond 13 C1 . you can do in either way but doing this way is simpler because in case of three or more arguments it would become cumbersome. so if the order is not stated then do it collectively but of the order is given like first is spade and the second is diamond then you would do it separately then come the question of dice, if i say i have one dice in one hand and one dice in another hand if i look at left hand what will be in the top ? Anything from 1-6 . if i look at the left hand anything can be on it from 1-6. if there is 1 on the left side how many are possible for the left hand ? any of the 6 when there is 2 in the left hand how many are possible for the left? any of the 6 if 3 on the left how many are possible for the left any of the six so how many possible outcomes ? 36 if i say my first digit expresses the outcome of first digit and my second hand expresses the outcome of second hand then these are the options possible 11 12 13 14 15 16 similarly if 2 is on the first then 21 22 23 24 25 26 and so on. so how many possible outcome are possible? 36 as far as dice is concerned don’t waste time on the denominator it would always be 36 now if say i through two dice what is the probaility that the sum of the digits is exactly 5? possible outcome are 14 41, 23 32. typically in dice questions 14 is different from 41. so favourable cases are 4 total are 36 answer is 4/36 similarly if i would have asked that the sum is 3 then the possible outcomes are 12 and 21. and the answer would be 2/36 what if i sat sum is less then or equal to 5? if 1 on the first dice how many outcome are possible on the second dice? 1,2,3,4 if 2 is on the first dice then how many outcome are possible in the second dice? 123 similarly with 3 we have 12 and with 3 we have 4 so total number of ways are 10 if i role 2 dice what is the probaility that sum is less then or equal to 6 there is one critical keypoint in the dice question just remember. 12 is different from 21 next is the topic on head and tails the most important topic in your paper. if i say toss a coin 6 times what is the probability of getting exactly 2 heads ? 1 26 if i toss a coin six times how many outcomes are possible, HT HT HT HT HT HT=26 26 signifies total number of possible ways. What are the favourable ways. now we talked about the word BEGIN. wHEN WE SAID HOW MANY 5 ALPHABET WORDS CAN BE formed using these alphabets 5! then we said the word digit. it was 5! 2! if you think on the same track when you said exactly 2 heads remaining has to

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be tails. HHTTTT if you analyse thy can rearrange among themselves we will 6! so our final answer will be FT get favourable outcome 2!×4! i would like you to think in the context of alphabetic problems. similarly if i ask what is the probability of exactly 3 heads when i toss 6 coins ? f rac6!3!×3! 26 QUESTIONS: 1 2 7 8 9 10 11 12 13 –> RED=3 BLUE=4 GREEN=5 In probability although you have to calculate favourable as well as total in most of the questions weather you calculate favourable or total number of options first it does not make much difference but in unconventional questions if you calculate total first it would become easier to calculate favourable after. here you have to select 3 red out of total 12 so 3 3 probability is 3 CC12 similarly the b part, ball cant be blue. it can all be red [4] it can all be green [5] so total favourable = 8 C3 . We will not use the approach of 1- something because in this case you have to consider all 3 cases, when 1 ball is blue, 2 balls are blue and 3 balls are blue. 2)total ways 5 2C3 favourable are 4 C3 7)1 2 3 4 5 6 7 8 9 10, 10 people are standing in a straight line so that 3 particular people don’t come together. DO it by (total - favourable)/Total 8) 9)1 2 4 8 16 32 Total 6 C2 10)remember these 3 values they are fairly common 25 = 32 ; 29 = 512 ; 21 0 = 1024 if we toss a coin 5 times the possible outcomes are 25 if we toss a coin 10 times we get the possible number of outcomes to be 21 0. If there are 5 true false questions how many possible outcomes 25 . if there are n trues false questions how many possible outcomes? 2n in how many ways can you mark a question true? 1 all have to be TTTT.... 1 2n f rac11000 QUESTIONS: 17 18 22 23 24 25 21 19 ****************************************** 17) B=25 G=17 y=16 what is the probability that first one is blue 25 C1 /58 C1 × 16 C1 /57C1 just remeber 1 has been picked you have to pick other from the remaining 57. 18) Probability to rain in a day = .8.. so probability of not raining in a particular day = .2. The probability of event happen is .2 then the probability of event not happening is 1-.2=.8 now if a Person goes on Saturday, what is the probability that he comes on Monday? probability he comes on Monday means he stays there on saturday and sunday and he comes back on Monday. Which is equal to saying it does not rain on Saturday it does not rain on Sunday and it rains on Monday. so what is the probability it does not rain on saturday AND does not rain on sunday AND it rains on Monday .8 × .8 × .2 something like that is asked in 22 also

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22) he says probability of a particular song to come on a particular channel is .3. so probability of not coming that particular song is .7. now he says a person flicks the channel in the order A B C. Whatwill be the probability he reach channel C? probability to reach channel C means he does not hit on channel A and he does not hit on channel B so he reach channel C. so answer is .7 * .7 please notice the difference he does not hit channel C. He reached channel C now weather he hits channel C or not thats a different matter. A part i to only reach channel C wearas B part is to hit channel C. now what is a fair coin ? a fair coin is when there is equal probability of H and T which is 0.5. Now if the probability is not 0.5 then that is what you called a biased coin. if its a biased coin then you have to learn a specific formula ReqP rob = P robOf HeadN umberOf Heads ×P roOf T ailsN umberOf T ails ×T T CN umberOf Heads (1) You can use either number of heads or tails but not both this formula is universal formula which is valid for fair as well as biased coins. so lets check it what is the probability of exactly 3 heads when i toss a coin 6 times ? 1 1 (2) ( )3 × ( )3 ×6 C3 2 2 in your paper the word biased is not used. normaly the question starts with the probability of coin head is 0.5 to guide you that it is a fair coin. if it is different then 0.5 the its a biased coin. 19) it says probability of a head is 0.6 that means it is a biased head. what is the probability of hitting at least 5 heads. At least means there can be 5(1T) heads or 6(0T) heads (.6)5 × (.4) ×6 C5 similarly for the next one (.6)6 × (.4)0 ×6 C6 23)You select a point inside the larger square. if you analyse you can select any point inside the larger square. so what are the total points? total points are all the points inside the larger square. favourable are the points which lie inside the shaded region. if you add up all the points inside the square it will form area of the square because are is accumulation of all the points. favourable will be all the points inside the shaded region. so if we add all the favourable points they will cme out to be shaded region so though he is asking probability it turns out to be simple geometric problem that is shaded area divided by total area. thats the basic idea apart from it the figures will go on changing the regions will go on changing. shaded is the favourable part total is the complete part. Now remember we talked when ever the geometric figure are given try to break them in symmetric parts for convenient. now if you join the opoosite vertices you will realise the figure consists of 16 triangles. so total = 16 shaded =4 probability is 4/16. 24) 25)remember x2 + y 2 = r2 represents the equation of a circle with centre at 4

(0,0) and radius r. so equation x2 + y 2 = 12 tells us what ? circle with radius 1. here x2 + y 2 ≤ 1 represents all the points inside circle whic is equalt o the π π area of circle. side 2 = 4 QUESTIONS: 6 26 10 11 12 13 14 5 4 he says n varies from 1 to 99 what is the probability that N(N+1) is divisible by 3 . now probability is what favourable over total i said first calclate the total bunber of ways then calculating the favourable becomes easy that time i was specially refering to these kind of unconventional type of questions. now when n is 1 we get the first term 1*2 if you put n=2 you get the next term 2*3 and so on if you analyse when you put n=1 you get first term if you put n=2 you get second term if youput n=3 you get third term similarly if you put n=99 you will get 99th term so total terms will be 99. so total number of terms will be = 99. now he is talking about multiple of 3. when you are talking about multiple of 3 i would suggest you to make a group of 3 terms. had he asked you a multiple of 4 you would make a group of 4 trems. in b part he is talking about multiple of 6 terms so you would make a group of 6 terms but why you would make a group of 3 term what is the purpose of making a group of 3 terms. purpose is it will be a cyclic property you dont have check each and every group. If you analyse in first group how many terms are multiple of 3 ? 2 terms in every group there will be two terms. that two will be the typical last two terms. the properties followed in first group wil be followd in all the groups. Total number of groups will be 99/3=33. Each group contains two multiples of 3 so total fvourable options are 66 and our probability is 66/99 now again when n varies from 1 -99 what is the probability that n(n+1)(n+2) is a multiple of 6. again wgen n is 1 we have first term when n is 2 we have second term and so on. when n is 99 we have the 99th term. so total terms are 99. again you will make group of 6. you wil realise that in each group all the terms are favourable. so probability is 99/99. 26) two cubes numbered 1 -6. you realise if he is saying cube numbered 1 6 he is talking about dice. What is the probability when rolling two dice the difference is greater then 2. If 1 comes on the first dice which possibilities are allowed on 2nd 4,5,6... for 2 on first 5,6 for 3 on first 6. 4 on first 1 , 5 on first 1,2 6 on first 1,2,3 y C1 2 11) Total options = y+30 C1 ; Favourable options = y C1 Given y+30 C1 < 5 C3 ×3 C1 12 C 4 7 C1 ×63 C1 13) 70 C2

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