DSE 2012- Question 42-44, 48 and 55-56.

[Darth Vader]

Ok, so a lot of confusion regarding how to go about all of these questions. Question 55-56 would require some context so I'll paste the link as well as the questions.

QUESTION 42. A fair coin is tossed until a head comes up for the 1st
time. The probability of this happening on an odd-numbered toss is
(a) 1/2
(b) 1/3
(c) 2/3
(d) 3/4
Answer: (c)

QUESTION 43. An experiment has 10 equally likely outcomes. Let A and
B be two non-empty events of the experiment. If A consists of 4 outcomes,
then the number of outcomes B must have so that A and B are independent,
is
(a) 4
(b) 3 or 9
(c) 6
(d) 5 or 10
Answer: (d)
QUESTION 44. Consider the system of equations
x + y = 0
x + y = 0
; ;  and  are i.i.d random variable. Each of them takes value 1 and 0
with equal probability.

Statement A: The probability that the system of equations has a unique
solution is 3/8.

Statement B: The probability that the system of equations has at least one
solution is 1.
(a) Both the statements are correct
(b) Both the statements are false
(c) Statement A is correct but B is false
(d) Statement B is correct but A is false


Q 48. A rectangle has its lower left hand corner at the origin and
its upper right hand corner on the graph of f(x) = x^2 + (1/x^2). For which x
is the area of the rectangle minimized?
(a) x= 0
(b) x = Infinity
(c) x = (1/3)^1/4
(d) x = 2^1/3

For 55 and 56:
http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/2012-option-a.pdf

[maahi]

48) we have to minimize d area of d rectangle area= xy =x(x^2 +1/x^2) . now minimise it .

[wolverine]

q42)  P(H)=P(T)=1/2

so prob of getting head in odd numbered toss wud be

p(H)+p(TTH)+p(TTTTH)+p(TTTTTTH)...so on

= 1/2 +1/2*1/2*1/2 + 1/2*1/2*1/2*1/2*1/2..so on...

 its a infinite gp with a=1/2 and r=(1/2)^2

sum of infinte  gp= a/(1-r)= (1/2)/(1-(1/2)^2)= 2/3

[wolverine]

.44)    ax+by=0
        Cx+dy =0
       
        P(0)=P(1)=1/2


      we can check the consistency of the above eqn using AX=0
      |A|=ad-bc eqn (1)

       
       a,b,c,d are iid rv
       we will create the sample space and find the joint prob

       a    d      P(A).P(d)
       0    1        1/4
       1    0        1/4
       0    1        1/4
       1    1        1/4
       

       b    c      P(b).P(c)
       0    1        1/4
       1    0        1/4
       0    1        1/4
       1    1        1/4

      from eqn 1 if system of eqn has unique soln ad-bc is nt equal to zero

     case 1 a=0,d=1 gives ad=0...b=1=c gives bc=1  ad-bc nt equal to zero
            so P=1/4.1/4=1/16

     case 2 and 3, when a=1,0 and d= 0,1  resp..ad=0 and b=c=1 gives bc=1...
             p=1/4.1/4+1/4.1/4=2/16


     case 4,5,6    when a=1,d=1..ad=1   b=0,1,0 and c=1,0,1 resp giving bc=0
                   p=1/4.1/4 + 1/4.1/4 +1/4.1/4= 3/16


             adding all the six cases which gives the probability of a unique soln =6/16 =3/8  

[Ayushya Kaul]

Thanks!

Any idea about q 55-56?

[Ayushya Kaul]

Also how do we show that there's certainty of a solution?

[Ayushya Kaul]

Thanks! Any idea about q55-56?

[Ankit Agrawal]

Q 55)

Let x represent mark up

F(u,A)= (A/u) - 1

F(u,A) = 1/(1+x)

A/u -1 = 1/(1+x)

Re arranging the above equation,

u = A(1+x)/(2+x)-----(1)

Hence, this is the answer to q 55.

For Q 56) Use equation 1, and Simply substitute the variables with any number, you will get the answer.

I put:
x=1
A=90
Y(gamma)= 10%

So if you put these numbers in eq 1, you get:

2/3(90)=60

Now if you increase A by Y, i.e., 90 by 10%, the corresponding increase in u, i.e. 60, is also 10%, i.e. Y.

So if the workers productivity increases at a constant rate Y, then the rate of unemployment also increases by Y.

Hence answer is (a)

[Ayushya Kaul]

Hi Ankita,
In 56, how can you say "u" also increases by "gamma"? Shouldn't it be a fraction of "gamma"?

[vandita24x7]

QUESTION 43. An experiment has 10 equally likely outcomes. Let A and
B be two non-empty events of the experiment. If A consists of 4 outcomes,
then the number of outcomes B must have so that A and B are independent,
is
(a) 4
(b) 3 or 9
(c) 6
(d) 5 or 10

pls Help

[Sumit]

[vandita24x7]

y shud x be <=4 ?

[Sumit]

bcoz intersection btw two events means probability of happening of both the events simultaneously...so x should be <= 4 bcoz p(A)=4...( In other words, maximum elements that  could be common btw A & B will be less than or equal to 4..)

[maahi]

can u xplain d part where you chose x=2 nd 4 n rejected  1 ? i m nt able to read d picture clearly

[Ankit Agrawal]

No it does not increase by a 'fraction' of gamma. Try using numbers in place of variables. It makes your work a whole lot easier.

P.S. My name is not Ankita. It's Ankit:-P

[Sumit]

@Maahi: we reject 1 n 3 bcoz outcomes are always in the integer form. suppose if we take x=1 the P(B)=2.5/10 which can't be possible bcoz outcome can take only integer values from 1 to 10 only...not 2.5 which means 2 n a half outcome.

[maahi]

ok thanks a lot :-)

[Ayushya Kaul]

Ok, great.
Ha, apologies for the mistake.
Cheers!

[Erika]

could you plz tell me abt Q 40. My logic is as follows but doesnt quite satisfy.

P(Green)=P(Red)= 0.5  as the balls are placed back each time

is it that we would expect to lose as much money by drawing a red ball, as much we earn by picking a green ball; as the prob of doing the two are the same??

and rs 10-10=0??

[Sinistral]

maahi wrote

48) we have to minimize d area of d rectangle area= xy =x(x^2 +1/x^2) . now minimise it .

the 2 ends of the diagonal of the rectangle are given (one at the origin and other on the curve). My doubt is: with a given diagonal we can make the area of the rectangle as small as possible, just by decreasing the angle between the diagonal and one of the sides of the rectangle. so it shouldn't matter at which point on the curve the other end of the diagonal lies. the minimum area can always be made zero.
what am I missing?