Stats doubt

[Shefali]

Suppose that X and Y have a continuous joint distribution
for which the joint p.d.f. is defined as follows:
f (x, y) =cy^2 for 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1,
                       0 otherwise.
Determine (a) the value of the constant c;
 (b) Pr(X +Y >2);
(c) Pr(Y < 1/2);

how to do part b and c...hw do we set limits for such quest?pls help..

[Granpa Simpson]

Got c as (3/2)..??

[Shefali]

yeah 3/2..

[Granpa Simpson]

Now put the limits of dx integral as [0,1] and dy integral as [0,2-x] and solve for the joint pdf...!!!!

[Shefali]

how did u find the lts??pls help me..i dont knw hw to approach it

[Granpa Simpson]

let me check once..i think i am doing mistake somewhere..!!!!

[Granpa Simpson]

yes the limits for dy will be [2-x,1]....look at it in this manner first define x in the interval [0,1], now X+Y>2, so Y>2-X so the minimum value that Y can take is 2-X and the maximum value that it can take is 1, thus the limits for dy will be [2-X,1]...!!!!!

[mrittik]

Not like that Subhayu....actually our mind itself playing a part of Monopoly and we are all in cosmological allocation of knowledge by Pareto Improvement created by Amit Sir....what do u think? but how then monopoly in place of competitiveness? and moreover we are all focused on 'On Myself, And Other which is purely mimetism by Rene Girard

[Shefali]

oky i got this part but ans is 3/8,its not cuming!

[Akshay Jain]

[Granpa Simpson]

Sorry typing error set limits as [0,2] and [2-x,1]....and Akshay's method is much easier...

[Shefali]

thnxxx akshay i got it...
bt ans is 3/8

[Akshay Jain]

@shefali...let me chek...there must be some calculation error....

[Akshay Jain]

I got the mistake shefali.....in the 4th line of solution in 2nd pic it shud be (2-x)^3 instead of (2-x)^2.

[aakash]

@subhayu
x limit should be 1 to 2 ?

[Granpa Simpson]

limit of x shud be [0,2]

[ashutosh]

@akshay can you please see if this method is wrong ?

[Tanushree]

Can some one help me with detailed answers for

Q1. 10 outcomes of a random variable were recorded. The sample mean is 0 and sample variance is 4. It is discovered that two outcomes were recorded incorrectly: one outcome was -5 instead of -6 and another was 11 instead of 12. What is correct variance?

a) 4            b) 3.6              c) 7.4                 d) 5.2

Q2. Amit has a box with 6 red and 3 green balls. Amita has a box with 4 red and 5 green balls. Amit randomly draw a ball and put it into Amita's box and Amita too randomly pics up a ball from her box. What 's the probability that the ball drawn by the two were of diff colors.

a) 1/3         b) 2/15       c) 4/15          d) 7/15

Q3. A traffic light on the way to the University is red 40% of the time. What's the probability of getting red  light (i) two days in a row, (ii) two out if three days.
 
a) 0.16 & 0.20     b) 0.16 & 0.29     c) 0.24 & 0.29      d) none of the above

Which book should I go for cost functions, production function and market sums?

[Akshay Jain]

Yes ashutosh....this method is also correct.....if u hav put the limits in a accurate manner dn it doesnt matter whether u integrate 1st w.r.t. X or Y....
double integration is just the game of getting the limits right.....
use diagrams for domains wherever possible.....it makes it easier

I think I made part b solution quite complicated.....

[Homer Simpson]

Hey Tanushree, I got these answers -

2) (d)

For different colors, Amrita has to pick red when green is added and pick green when red is added.

First Amit has to pick a ball - it can either come out red or green. Respective probabilities of that are 6/9 and 3/9 accordingly. Now Amita has to pick a ball but we know that a ball has been added, right? So her sample space is now 10. Thus P(G) = 5/10 and P(R) = 4/10

Thus, P(both pick different colors) = (6/9)(5/10) + (3/9)(4/10) = 7/15

3) (b)

These are independent events. So two days a row would be 0.40x0.40 = 0.16
and in 3 days, if it happens two days a row then = 3x(0.60)(0.16) = GRR, RGR, RRG = 0.288 = approx 0.29

Please correct if i am wrong. thank you!