Through The Eyes Of An Aspirant - Probability Theory Decoded - Ankan Sengupta, FMS Delhi


  • FMS Delhi (2015 - 2017), CAT 2014 - 99.79 percentile


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    Probability itself doesn’t pose much of a threat to aspirants. But a probability question, clubbed with Number theory or permutation-combination, when is dished out to aspirants, sometimes it can really break a havoc. Geometrical probability is another portion to cause discomfort too.  So I’d be segregating this chapter in mainly three parts - General probability theory (including conditional probability), Binomial probability and Geometrical probability.

    Probability in general is defined as the extent to which an event is likely to occur, measured by the ratio of the favourable cases to the whole number of cases possible. So mathematically we can write it as P(A) = s(A)/n, where P(A)=probability of occurrence of the event A,s(A)=No. of events in favour of A,n=total no. of events=sample space.This is called the Classical Definition of Probability.

    But if trials are to be repeated a great number of times under essentially the same condition then the limit of the ratio of the number of times that an event happens to the total number of trials, as the number of trials increases indefinitely is called the probability of the happening of the event. It is assumed that the limit exists and finite uniquely. Symbolically p (A) = p = lim (n -> infinity) m/n,  provided it is finite and unique.This definition is more general in nature and is called Statistical Defnition of Probability.

    The two definitions are apparently different but both of them can be reconciled the same sense.

    1) General Probability- General probability theory includes questions of probability that are clubbed with other topics from Number theory, Algebra or PnC and questions on conditional probability. So I’d be discussing a few questions regarding these topics in this space. The first question is a simplistic probability question which only incorporates general probability concept. I’ve solved it in an unconventional way in method 1 and with a self-developed shortcut in method 2.

    Find the probability that 3 random points selected from circumference of circle lie on the same semi-circle

    Approach 1:

    Since we're here concerned with circumference,so just consider a clock.suppose there are marks at 3,6,9 and 12 positions..now for choice of first 2 points there is no restriction.So choose 2 points at 12 and 3 positions..Now there are 4 gaps in between those marked positions.Check only between 6 and 9 there cannot be the 3rd point to satisfy our criteriaSsince I considered those 2 points randomly,so we can randomise it throughout the domain.Answer is 1-1/4=3/4 only.

    Approach 2:

    n/2^(n-1)..put n=3..it is the general formula for this question,n=no. of points under consideration.This is an empirical formula without any proof.

    Now a problem which deals with number theory.

    Let N be a set of all 4 digit perfect squares. Find the probability if a number whose sum of tens digit and units digit is equal to 7 is selected from the set

    a) 12/68

    b) 18/68

    c) 13/68

    d) None of the above

    4 digit square roots will range from 32 to 99, so total 68 numbers
    now, check for those number from 1 to 30, which give last two digits sum as 7, so found it as 4, 5,15,19,25, (one should know that any 5 square would end up in 25.
    so favourable numbers are 35,45,.....,95,(7 such numbers)
    (50-4)^2,(50+4)^2,(100-4)^2(3 such numbers)
    (50+19)^2,(100-19)^2( 2 such numbers)
    total,7 + 3 + 2 = 12.
    P=12/68.

    A question now which clubs probability with algebra.

    Each coefficient of ax^2+bx+c=0 is determined by throwing a dice. Find probability of its roots being real.

    For real roots,b²-4ac>=0,where 1< =a,b,c< =6.
    this cannot happen for b=1
    b=2 gives a=c=1--- 1 soln.
    b=3 gives a=c=1, a=2,c=1, a=1,c=2---3 soln
    b=4 gives 8 soln
    b=5 gives 14 soln.
    b=6 gives 17 cases
    Total=43 cases
    total cases =no. of ways of a*no. of ways of b*no. of ways of c=6*6*6=216
    Ans=43/216

    Now I’d discuss one of my most favourite problems here. This is mainly a question of PnC.I’m modifying it in terms of probability.

    A dice can be coloured with at most 2 different colours. Find the probability that it is coloured with exactly 2 colours.

    This problem can be solved by making cases.In that you’ve to consider symmetry of the figure which incorporates a complex calculation.

    So I’d be approaching it differently with a technique derived from graph theory mainly,Polya’s enumeration theorem.

    How many ways are there to color the sides of a 3-dimensional cube with upto t colors, up to rotation of the cube?

    Answer is,

    So,In our problem denominator has t=2 and numerator is t=1 subtracted from t=2.

    When t=2,total cases=10 and when t=1 total cases=1.

    So answer is (10-1)/10=9/10.

    A problem on conditional probability will conclude the theory of general probability aptly.

    Your neighbor has 2 children. He picks one of them at random and comes by your house; he brings a boy named Joe (his son). What is the probability that Joe’s sibling is a brother ?

    Consider the experiment of selecting a random family having two children and recording whether they are boys or girls. Then, the sample space is S = {BB,BG,GB,GG}, where, e.g., outcome “BG” means that the first-born child is a boy and the second-born is a girl. Assuming boys and girls are equally likely to be born, the 4 elements of S are equally likelythat the The event, F, that the neighbor has two boys (i.e., Joe has a brother) is the set F = {BB}.Now we are given the event E′ that “your neighbor randomly chose one of his 2 children, and that chosen child is a son”

    We want to compute P(F|E ′ ) = P(F ∩ E′ ) P(E′) = P({BB}) P(E′ |{BB})P({BB}) + P(E′ |{BG})P({BG}) + P(E′ |{GB})P({GB}) + P(E′ |{GG})P({GG})

    = 1/4 /[1(1/4) + (1/2)(1/4) + (1/2)(1/4) + 0(1/4)] = 1/4 /1/2 = 1/ 2 .

    2) Binomial Proability:

    We consider binomial probability when we’re dealing with finite number of caes and only 2 outcomes are possible.Say,X=the variable,total no. of trials=n.p=probability of having a favourable case,q=1-p=probability of having an unfavourable event. Then,if we’re to calculate probability that r events happened ,then P ( X = r ) = nCr * pr * q (n-r)

    Now we’d consider an example.

    out of 5 students like Maths.Say total 5^9 professors took data on 10 students to check those students like Maths or not,find how many of them would report only 2 like Maths.

    A.So,here r=2.n=10.p=2/5,q=1-2/5=3/5.

    P(X=r=2)=10C2*(2/5)^2*(3/5)^8=45*2^2*3^8/5^10.

    Therefore out of 5^9 professors only P*5^9=2^2*3^10 would report the given.

    3) Geometrical Probability:

    This is the third variant of probability. I’ll be providing a few examples regarding how to deal with problems of this kind.

    Ankan and Rajanya agreed to meet at Shiva temple between 10.00 to 11.00.Each one of them arrives there and stays for exactly 6 minutes.What’s the probability that they’ll meet?

    A. The probability that they do not meet is represented by the total of the areas of the 2 outer triangles in the figure given below,which is 0.81.So probability that they’ll meet=1-0.81=0.19.

    A driver goes out of the city to a place which is 100 Km away from the city and comes back in the same day. With a full tank he can drive upto 250Km.Today he forgot to fuel up his car.What is the probability that he’ll run out of fuel?

     I’d be concluding my article now.

    Keep practising and happy learning.


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