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10/12/2015 Bankers Adda: Banking Pathway 2015: Quant Study Notes & Quiz (Permutation & Combination and Probability)
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WEDNESDAY, 8 JULY 2015
Banking Pathway 2015: Quant Study Notes &Quiz (Permutation & Combination andProbability)Dear Reader , In continuance to provide study notes(Short Tricks ) today we are ProvidingNotes on Permutation& combination and Probability. Which is very crucialtopic for upcoming competitive examinations.Here we also Providing someconcept clearing quiz for analysing yourself. Time management also verycrucial part for gain good marks in exam. so use wisely your time. These Study notes are provided by one of our ardent BA reader Insomniac.We wish you good luck for future.
Permutation& Combination and Probability:Permutation and Combination are not that important for the purpose of examBecause Question are rarely asked from This Topic but We have to learnthem anyway because Question of probability can't be solved withoutlearning permutation and combination. So will give you all a little hint aboutwhat is permutation and what is combination and then we will move on toProbability.
But Before That Just Look at A very Important Concept Without Which Youcan't solve a single Question of permutation/combination or probability.And that Factorial Notation.It's represented by (!) and it is read as Factorial.So if i write 5! it will be read as Five Factorial.And what it means ? It means to simply multiply all the numbers indecreasing order till 1.Like if i write 6! it means 6*5*4*3*2*1 = 720Or 7! = 7*6*5*4*3*2*1 = 5040For Fast Calculation You all must learn the value of factorial till 10.Just Learn these values1! = 12! = 23!= 64! = 245! = 1206! = 7207! = 50408! = 403209! = 36288010! = 3628800
Well Before I Start Explaining Permutation and combination one thing i wantto tell and that is It's the easiest topic that you will find in maths. Mostpeople are unable to understand it and that's why people think it's complexand all type of misconceptions but trust me it's the easiest topic in the wholemathematics and It's not actually even maths, It's less about Calculation andmore about Logical Thinking. Well We all can't Calculate Fast but we all canthink fast.
So what is permutation? In Simple words it's arrangement or No. of ways things can be arranged.Suppose there are 3 words ABC and if it's asked How many ways thesethree can be arranged then all yu or What are the no. of permutationsPossible. Then all you have to do is Arrange this things in as many ways it'sPossible.Let's try to arrange them now. SO There is ABC, ACB, BAC, BCA, CAB,CBA Are there any more ways these can be arranged ? try it ? No These arethe all possible arrangements. So The answer to the above Question will be6. That is ABC can be arranges in different ways.
Now there were only 3 alphabets What if there were more like You have toArrange ABCDEFGHI. Now for 3 alphabets it was easy you easily arrangedthem But Arranging these 9 letters will take you days and even then you willnot be able to get a certain answer.
So what we should do here. No need to worry our mathematicians were
genius they created a very simple formula for that.And Formula is like this. N Different things can be arranged in N! ways.So in above Question there were 9 alphabets so the no. of possibelearrangements will be 9! = 362880.
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arrangements will be 9! = 362880.So that was out basic concept Now let's move on to another basic concept.
So in the above questions It was Asked in how many ways ABCDEFGHICan be arranged. In this question they were asking the possiblearrangements of all the 9 Alphabets, They can also Ask In how many ways 4alphabets from above 9 alphabets can be arranged.
In such type of Questions there is another formula Which is very very veryimportant because it will be used in almost every question.So the formula is Out of n things r things can be arranged in nPr ways. and
nPr = n!/(nr)! So in the above Question it is asked that in how many ways 4 alphabetsfrom the total 9 alphabets can be arranged.So apply the formula nPr = 9P4 = 9!/(94)! = 9*8*7*6*5*4*3*2*1/5*4*3*2*1=9*8*7*6 = 3024.
Now there is a trick to easily calculate nPr by which you won't have to doany division work.Like if it say 9P3 then you just have to multiply Starting from 9 in decreasingorder till the next 2 digit i.e 9P3 = 9*8*7. Why we multiply till 7 only ? that isbecause the value of r is 3 and total multiplication should contain the value ofr.Another example if it 7P2 then you will just do 7*6[ 2 number because r = 2ok]if it's 7P4 then the answer will be 7*6*5*3[ 4 no. because value of r=4]So If it's 10P5 then the value will be 10*9*8*7*6 [ 5 digit because value of r =5]
I think you understand my point now. Now move on to the cases.Actually there are infinite cases in Permutation and Combination 100's ofdifferent type of question can be formed So i will only discuss the cases thatare important for the exam, And if you have any problem in any other casethen you can ask me personally.
Case 1 Simple Arrangement Case well all words are unique.
By UNIQUE i mean all alphabets are differentIn how many ways the letters of the word ROCKET can be arrnged.very Simple just count the no. of words in ROCKET that will be 6So number of arrangements will be n! that will be 6!
CASE 2 Arrangement When All the words are not UNIQUE.That means some words are repeated.Like No. of possible arrangements of word TITANICNow In this case you Just have to find the total possible ways first withouteven thinking about Repeated words and then after that You will divide thatwith the numbers of times a Word is repeated.So in the above Question Total alphabets = T = 2, I = 2. A= 1 C =1 N =1Total 7 So Permutations will be 7! and Now you will divide It by No. oftimes A word is repeated SO T is repeated 2 times and I is repeated twotimes So divde 7! by these 2. So final Answer will be 7!/(2!*2!)
Let's See another Example. In how many ways the letters of the wordRUNNING Can be arranged.So total no. of alphabets in the above Words = 7No. of words that are repeated = N = 3 times repeated.So the solution will be Total permutation divided by no. of times a word isrepeated and that will be 7!/3! that will be your answer.
Case 3 Arrangement Some Words are always together and SomeWords and Never together.No of possible arrangements of the words LAYERING When Vowels arealways together.In this case what we do Is we consider the no. of Vowels as 1 singlealphabet That [AEI] is a one single alphabet In that way they will always betogether and the rest words are LYRNG.So the total no. of alphabets will be 6 ? Why 5 Alphabets are LYRNG and[AEI] is One alhpabet remember so The total alphabet will be 6And no. of possible arrangements will be 6!But but the question is not complete yet [AEI] Though considered as 1alphabet but stil the words AEI can change places within itself Like AEI italso can be AIE or EIA. So there are 3 words so no. of total arrangementsthat they can do within itself will be 3!So our final answer will be 6!*3![ that is because 6! is the no. of possibleways when AEI are together and And multiplied by 3! because AEI canchange places within themselves in 3! possible Ways]
If it was asked that VOWELS in LAYERING are never together that what wewill do ?
This Question can't be solved directly.In order to solve this We will have to FIND the total no. of arrangements ofthe word LAYERING and then Subtract the no. of arrangemnts in which AEIare Always together.So no. of possible arrangements of LAYERING will be 8!
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So no. of possible arrangements of LAYERING will be 8!And We already Solved that when AEI are always together the no. ofpossible ways are 6!*3! So no. of possible ways when AEI are never together will be 8! 6!*3!
Now i told you that there are many more cases but that are really notimportant I am explaining these cases because they are important and helpypu while solving Probability.
Now We should move on to the next Topic That Is Combination. Now youknow that Permutation means Arrangement or no. of possible ways A thingcan be arranged.What is the meaning of Combination.Combination is a simple act of Choosing or Selection.Like When it is asked What are no. of possible ways Word TITAN can bearrange You have to find The Permutation.But if it is asked what are no. of possible ways You can Select 2 alphabetfrom the word TITAN, It means you have to find Combination.The act of selection or Choosing is called COMBINATION.Now you all must know what is nPr so it's time to move towards nCrLike nPr = n!/(nr)!
nCr is somewhat simillar but that is just an extra r! in the denominator So nCr = n!/[(nr)!*r!]
nCr means r things has to be selected out of n things.Like in the above Question No. of possible ways 2 alphabets can be selectedfrom the word TITANSo total no. of alphabets n = 5no. of alphabets which we have to select r = 2So the answer will be 5C2 = 5!/(52)!*2! = 5!/3!*2! = 5*4/2*1 = 10
Now i told you have to calculating nPr in a simple way Just like that we canalso calculate nCr in a simple way All you have to do is Follow the method ofnPr and In division you have to also multiply in increasing order from 1Like 6C3 = 6*5*4/1*2*3And 9C2 = 9*8/1*2and 10C4 = 10*9*8*7/1*2*3*47C5 = 7*6*5*4*3/1*2*3*4*5
This much knowledge of combination is enough for solving the Questions ofProbability.So without wasting Time just move on to our main Topic ie Probability.
Probability
So what is Probability ?
Probability is Just the chances have happening of an event. Like what arethe chances that You will Become a PO or An Income Tax Inspector or aClerk. What are the chances that you will find the love of you life (Thatchance of that is very rare)These all chances are just the game of Probability. Our Life is Also The sumof all these chances, the chances we take Like What are the chances thatyou will study after 12 instead of gossiping on whatsapp.So how de we find the probability of happening of an event. In mathematicalterms probability = Number of favourable Outcomes/ Total outcomesNo. of favourable outcomes means the outcomes which we want.Total outcomes Means the total possible outcomes (That's the reason westudied Permutation and Combination so that we can find total outcomes]Let me give you a very realistic example. What is the probability that You willBecome a PO in SBI ?So We have to find the favourable outcomes here That will be the No. ofPosts in SBI[ because if you get any of the post in the total post you will bea PO]So total no. of Posts In SBI this time is 2000And what are the total outcomes or What are the total no. of Applicants =20,00,000 So what is the probability that You will be 1 of them Simple Probability ofYou getting selected = favourable Outcomes/ total outcomes =2,000/20,00,000 = 1/1000 That is Your Chances. Or in other words 1 in a thousand Aspirant canbecome a PO in SBI.So i think Now you have the basic Idea what is PROBABILITY.
So now Lets Move On to Questions.But before that.VERY VERY VERY VERY VERY IMPORTANT AND = Multiplication(*)OR = Addition (+)If anywhere and I mean Anywhere you see a question which say what is theprobabilty of getting X or Y, It simply means that you have to find probabilty
of X and Probability of Y and ADD them, The word OR means AdditionAlways Keep in Mind that.And if It is asked what is the prbability of getting X and Y, It simply meansthat you have to find the probability of X and Y and Multiply them, The wordAND means Multiplication Always remember that.
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AND means Multiplication Always remember that.At least = Minimum We require [ Or kam se kam Kitna hone chahiye Usasejyada bhi ho sakta hai but usase kam nahi hona chahiye]Example If we want at least 2that measn Minimum we need 2 We can have3 or 4 or 5 It doesn't matter but Should not be less than 2.At Most = Maximum We Require[ Jyada se Jyada Kitna ho sakta hai, UsaseKam ho sakta hai farak nahi padta but usase jyada ahi hona chahiye]Example if we want AT MOST 2 That means we can have 2 we can have 1and we can have 0 also any less value it doesn't But we can can't haveanything greater than 2.These cases will be more clear to you when we will solve some Questions.
Questions related to Balls. Case 1: Normal CaseThere are Total 5Red, 3Blue and 2 Green bals In a Bag, Two balls aretaken out at random What is the probability that i)2 Balls will be Green.ii 2balls will be REDiii) 2 balls will be BLUE.i) What is the probability that 2 balls are taken out at random from a bag andboth balls are Green.So Calculate First the favourable Outcomes. That is how many ways 2 ballscan be taken out from a bag which have 2Green balls = 2C2 = 1Now calculate Total Outcomes. That is how many ways 2 balls can be takenout from the bag containing total 10 balls [ 5red + 3Blue + 2 Green = Total10] = 10C2 = 10*9/1*2 = 45So probability = favourable outcomes/total outcomes = 1/45
ii) What is the probability that 2 balls are taken out from bag and both areRED So Calculate First the favourable Outcomes. That is how many ways 2 ballscan be taken out from a bag which have 5Red balls = 5C2 = 5*4/1*2 = 10Now calculate Total Outcomes. That is how many ways 2 balls can be takenout from the bag containing total 10 balls [ 5red + 3Blue + 2 Green = Total10] = 10C2 = 10*9/1*2 = 45So probability = favourable outcomes/total outcomes = 10/45 = 2/9
iii) What is the probability that 2 balls are taken out from bag and both areBlue.So Calculate First the favourable Outcomes. That is how many ways 2 ballscan be taken out from a bag which have 3BLUE balls = 3C2 = 3*2/1*2 = 3Now calculate Total Outcomes. That is how many ways 2 balls can be takenout from the bag containing total 10 balls [ 5red + 3Blue + 2 Green = Total10] = 10C2 = 10*9/1*2 = 45 So probability = favourable outcomes/total outcomes = 3/45 = 1/15.
CASE 2 AND Case.There are Total 5Red, 3Blue and 2 Green bals In a Bag, Three balls aretaken out at random What is the probability that i) 2 balls are Red and 1 ball is Greenii) 2balls are Blue and 1 ball is Green
i) What is the probabilty that 3 balls are taken out and out of those 3balls 2 balls are red and 1 is green.In this questions there are 2 events i.e getting 2 red ball and getting 1 greenballSo first we have to calulate the seperate probabilities first.So no. of ways 2 Red balls can be selected out of total 5 balls = 5C2 =5*4/1*2 = 10So no. of ways 1 Green ball can be selected out of total 2 balls = 2C1 = 2So Favourable Outcomes i.e No. of ways 2 Red balls AND 1 Green Ball canBe Selected = 10*2 = 20[ Whenver you see and Just Multiply it ]And Total No. of Outcomes i.e Selecting 3 balls out of total 10 balls = 10C3= 10*9*8/1*2*3 = 120So probability of getting 2 Red and 1 green Balls = favourable outcomes/totaloutcomes = 20/120 = 1/6
ii) What is the probabilty that 3 balls are taken out and out of those 3balls 2 balls are BLUE and 1 is GREENSo no. of ways 2 BLUE balls can be selected out of total 3 balls = 3C2 =3*2/1*2 = 3So no. of ways 1 Green ball can be selected out of total 2 balls = 2C1 = 2So Favourable Outcomes i.e No. of ways 2 BLUE balls AND 1 Green Ballcan Be Selected = 3*2 = 6[ Whenver you see and Just Multiply it ]And Total No. of Outcomes i.e Selecting 3 balls out of total 10 balls = 10C3= 10*9*8/1*2*3 = 120So probability of getting 2 BLUE and 1 green Balls =favourableoutcomes/total outcomes = 6/120 = 1/20
Case 3: OR CASE
There are Total 5Red, 3Blue and 2 Green bals In a Bag, 2 balls are taken outat random What is the probability that 2 balls are Red or 2 balls are BlueIn this questions there are 2 events i.e getting 2 red ball or getting 2 BlueballsSo first we have to calulate the seperate probabilities first.
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10/12/2015 Bankers Adda: Banking Pathway 2015: Quant Study Notes & Quiz (Permutation & Combination and Probability)
http://www.bankersadda.com/2015/07/bankingpathway2015quantstudynotes.html 5/11
So first we have to calulate the seperate probabilities first.So no. of ways 2 Red balls can be selected out of total 5 balls = 5C2 =5*4/1*2 = 10So no. of ways 2 BLUE balls can be selected out of total 3 balls = 3C2 =3*2/1*2 = 3So Favourable Outcomes i.e No. of ways 2 RED balls OR 2 BLUE Balls canBe Selected = 10 + 3 = 13[ Whenver you see OR Just ADD it ]Now calculate Total Outcomes. That is how many ways 2 balls can be takenout from the bag containing total 10 balls [ 5red + 3Blue + 2 Green = Total10] = 10C2 = 10*9/1*2 = 45So the probability of getting 2 Red ball or 2 Blue balls = favourableoutcomes/total outcomes = 13/45
CASE 4 AT LEAST CASEThere are Total 5Red, 3Blue and 2 Green bals In a Bag, Three balls are takenout at random What is the probability that At least 2 Balls are RED.Now What i Told you In At Least Case You have to select at least 2 meansYou can have all 3 balls red But at least 2 balls should be RED means wewill have to find the probability of getting 2 red balls OR 3 red balls.So there are 2 cases here 1st case is when we get 2 red balls and 1 ball canbe of any other colourand 2nd case is when we get all 3 balls as red.1st caseSo no. of ways 2 Red balls can be selected out of total 5 balls = 5C2 =5*4/1*2 = 10And No. ways 1 ball can be selected out of rest 5 balls = 5C1 = 5 Our Favourable outcomes I.e getting 2 red balls and 1 ball of any colour =5*10 = 50 [ And case so multiply ]And Total No. of Outcomes i.e Selecting 3 balls out of total 10 balls = 10C3= 10*9*8/1*2*3 = 120So Probability of getting 2 red balls and 1 ball of any other colour = favourable outcomes/total outcomes = 50/120 = 5/122nd case When we get all 3 balls as red.So no of ways 3 red balls can be selected out of total 5 red balls i.e also ourfavourable outcome = 5C3 = 5*4*3/1*2*3 = 10And Total No. of Outcomes i.e Selecting 3 balls out of total 10 balls = 10C3= 10*9*8/1*2*3 = 120So probability of getting 3 red balls =favourable outcomes/total outcomes =10/120 = 1/12
Now Either Case One will happen OR Case 2 will happen. that means eitherwe will get 2red balls and 1 other ball or we will get all 3 red balls So As ialready explained that In OR case Probabilities gets added so we will justadd the probability To get the final probability.So when 3 balls are taken out at random the probability that at least 2 ballsare green = 1/12 + 5/12 = 6/12 = 1/2
Case 5 At MOST CASEThere are Total 5Red, 3Blue and 2 Green bals In a Bag, Three balls are takenout at random What is the probability that At Most 2 Ball is RED.So as i told you all in case of AT most We can have any number less thanBut not greater than That means We can Have 2 Red balls out of 3 balls andWe also can 1 red ball out of 3 balls and we can also have 0 red balls but wecan't have More than 2 Red ball. That means all 3 balls can't be RED.So we will solve same like the last case.No. Of ways 2 red balls and 1 others balls can be selected = 5C2*5C1 =10*5 = 50No. Of ways 1 red balls and 2 others balls can be selected = 5C1*5C2 =5*10 = 50Now Of ways Balls are selected that there are NO red balls That means Allthree balls are of Other Colours = 5C3 = 10Total No. Of Outcomes = 10C3 = 10*9*8/1*2*3 = 120So Probability will be (50+10 +50)/120 = 110/120 = 11/12
Quiz :
Time: (45 minutes)
1. A bag has six red marbles and six blue marbles. If two marbles aredrawn randomly from the bag, what is the probability that they will bothbe red?A) 1/2 B) 11/12 C) 5/12 D) 5/22 E)1/3
2. There are five students in a study group: two finance majors and threeaccounting majors. If two students are chosen at random, what is theprobability that they are both accounting students?
A) 3/10 B) 2/5 C) 1/5 D) 3/5 E) 4/5
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E) 4/5
3. At a certain business school, 400 students are members of the sailingclub, the wine club, or both. If 200 students are members of the wineclub and 50 students are members of both clubs, what is the probabilitythat a student chosen at random is a member of the sailing club?A) 1/2 B) 5/8 C) 1/4 D) 3/8 E) 3/5
4. A bag contains 3 red marbles, 3 blue marbles, and 3 green marbles. Ifa marble is randomly drawn from the bag and a fair, sixsided dice istossed, what is the probability of obtaining a red marble and getting 6from dice?A. 1/15 B. 1/6 C. 1/3 D. 1/4 E. 1/18
5. A letter is randomly select from the word "STUDIOUS". What is theprobability that the letter be a U?A. 1/8 B. 1/4 C. 1/3 D. 1/2 E. 3/8
6. In how many different ways can the letters of the word'MATHEMATICS' be arranged so that the vowels always come together?A.124045B.20890C. 133156D. 120960E. None of these
7. How many 4letter words with can be formed out of the letters of theword, 'LOGARITHMS', if repetition of letters is not allowed?A. 400B. 4050C. 5040D. 5773E. None of these
8. In a group of 6 boys and 4 girls, four children are to be selected. Inhow many different ways can they be selected such that at least one boyshould be there?A. 156B. 209C. 193D. 245E. None of these
9. In a bag, there are 8 red, 7 blue and 6 green balls. One ball is pickedup randomly. What is the probability that it is neither red nor green?A. 3/91B. 1/3C. 3/7D. 7/15E. None of these
10. One card is drawn at random from a pack of 52 cards. What is theprobability that the card drawn is a face card (Jack, Queen and Kingonly)?A. 3/13B. 1/13C. 7/52D. 9/13E. None of these
Thank you Insomniac
Answers:
1.DProbability that both are red marbles = 6/12 x 5/11 = 5/22
2. AProbability of first student to be accounting student =3/5Probability of second student to be accounting student =2/4 = 1/2Probability that both students to be accounting students =3/5 x 1/2 =3/10