let us assume that y = 3x – 5 and y = 2x – 4 are lines of regression of x on y and y on x respectively,therefore
but sign of is same as and
therefore
The required condition for concurrency is a3(b1c2 - b2c1) + b3(c1a2 - c2a1) + c3(a1b2 - a2b1) = 0
Here a1 = l, a2 =m, a3 = n and b1 = m, b2 = n, b3 = l and c1 = n, c2 = l and c3 = m
Substituting the values we get
n(ml - n2) + l(nm - l2) +m(ln - m2) = 0
This implies l3 + m3 + n3 - 3lmn = 0
That is (l + m+ n)(l2 + m2 + n2 -lm -mn - nl) = 0
This implies l + m + n = 0
p:
q:I get first class
the contrapositive of . hence the answer is If I do not get a first class , then
After completing the square,we get
so center is (0,0) and radius is 5 units.
As the normal should pass through the centre of the circle, center should satisfy the equation of normal
so 3x – 4y = 0 is the equation of normal as (0,0) satisfy this equation.

Total number of ways of getting even number is 3
Out of these 3 even number we have to get either 2 or 4 which can be done in 2 ways.
So required probability is 2/3
