Discussion Problem_(28)

[duck]

Q1) The odds against X solving a problem is 8:6 and odd in favour of Y solving a problem is 14:16.
What is the probability that the problem will be solved if they both try.

Q2) Suppose, X has the pdf:
f(x) = 3x^2 , 0<x<1.
      = 0 otherwise.

Consider a random rectangle whose sides are "x" and "1-x". Determine the expected value of the area of rectangle.

[Mauli]

q1)
since the pr(x) winning and  pr(y) winning are independent  of each other.
we have pr(x) = 6/14 pr(x') = 8/14 where..x=winning
                                                         x'= not winning
similarly for y =
pr(y)=14/30 and pr(y')= 16/30
required pr = pr(x)*pr(y') +pr(x')*pr(y)+pr(x)*pr(y)=73/105

q2)
integral over 0 to 1
3x^2 *x*x*(1-x) dx = 1/10
please cnfrm wthr my methods and answers are correct.

[rohan]

i) 73/105
ii) 3/20

[laracroft]

1) 73/105

[laracroft]

2) 3/20

[Sumit]

Q1. 73/105.
Q2. 3/20.

[Ayushya Kaul]

Could you please shed light on how to solve the second question?

[Sumit]

Expected Value Of random variable X=Integral(limit -infinity to +infinity)xP(x) dx..
where Pdf Given P(x)=3x^2 for 0<x<1..

Now Area of rectangle =x(1-x)
As we can see area of the Rectangle is the function of random variable x.

The E[x(1-x)]=Integral(limit zero to one)x(1-x)*3x^2 dx=3/20.

[duck]

Great.. :)
Correct Answers:
1) 73/105.
2) 3/20

[Ayushya Kaul]

Thanks Sumit :)