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+x^(-2).For which x is the area of the rectangle minimised?
a. x=0
b. x=infinity
c. x=(1/3)^(1/4)
d. x=2^(1/3)

Q.1 Two women and four men are to be seated randomly around a circular table.find the probability that the women are not seated next to each other.
a. 1/2
b. 1/3
c. 2/5
d. 3/5

@Noel
Total no of ways =5!
Lets make the men seat first.4 men can be seated in 4! ways.Now remaining 2 women can be seated in 5P2 ways.
therfore prob is (4!*5P2)/5!=2/5

@Noel i made a silly error...here it goes
Total=5!ways
4 men can be seated in 3! ways.now there are 4 gaps and these 4 gaps can be filled by women in 4P2 ways.
Therefore prob is (3!*4P2)/5! =3/5

Thanks ben10
@ron thanks again..i'm a bit weak at probability..can you please explain how the men can be seated in 3! ways and then how can there be 4 places for the women after you've seated all the men?.thank you

hi Noel, since its circular arrangement if u assume first the 4 men are seated, they can be seated in 3! ways. The two women cannot sit together, they have to be seated between men. There are 4 possible places(Imagine a chair between two men). The two women can sit in any of this 4 places in 4C2*2! ways. You get total possible ways as 3!*4P2

On Fri, Apr 11, 2014 at 12:20 AM, Noel [via Discussion forum] <[hidden email]> wrote:

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+x^(-2).For which x is the area of the rectangle minimised?
a. x=0
b. x=infinity
c. x=(1/3)^(1/4)
d. x=2^(1/3)

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Q.2 Suppose X1,X2,...,Xn are observed completion times of an experiment with values in [0,1].Each of these random variables is uniformly distributed on [0,1].If Y is the maximum observed completion time,then the mean of Y is
a. [n/(n+1)]^2
b. n/[2*(n+1)]
c. n/(n+1)
d. 2n/(n+1)

Y = max{X(1), X(2), ... , X(n)} where X(i)s are i.i.d U[0, 1].
To find E(Y). Let us first find the CDF of Y.
for 0 < y < 1,
F(y)
= Pr(Y <= y)
= Pr(max{X(1), X(2), ... , X(n)} <= y)
= Pr(X(1) <= y, X(2) <= y, ... , X(n)} <= y)
= Pr(X(1) <= y)Pr(X(2) <= y) ... Pr(X(n)} <= y) [Because X(i)s are independent]
= y^n [Because X(i)s are U[0, 1]]