Integers x and y are chosen at random so that -5

Question

Integers x and y are chosen at random so that  -5\leq{x}\leq{5}  and  -5\leq{y}\leq{5} . What is the probability that  (x+y)^{2}  >  x^{2}+y^{2} ?

(A)  \frac{73}{124}

(B)  \frac{43}{59}

(C)  \frac{50}{121}

(D)  \frac{34}{73}

(E)  \frac{19}{35}

sujoykrdatta · Answer

(x+y)² > x² + y² 
=> x² + y² + 2xy > x² + y²
=> 2xy > 0
=> xy > 0

So both x and y are positive or both are negative

Possible cases: 
A) x = -5,-4,-3,-2,-1; combines with y = -5,-4,-3,-2,-1 => total 5*5 = 25 cases

B) x = 5,4,3,2,1; combines with y = 5,4,3,2,1 => total 5*5 = 25 cases

Thus, 25+25 = 50 cases (Favourable)

Total cases: There are 11 values of x (-5,-4,-3,+2,+1,0,1,2,3,4,5) that combine with 11 values of y; giving 11*11 = 121 cases

Probability = 50/121

Posted from my mobile device

Integers x and y are chosen at random so that -5<=x<=5 and