X , Y ∼ U [ 0 , 1 ], what is the probability that t 2 + X t + Y = 0 has a real root.Let the continuous random variables X,Y be independent of each other and uniformly distributed on [0,1] i.e. X , Y ∼ U [ 0 , 1 ], what is the probability that t 2 + X t + Y = 0 has a real root?I am trying to solve it with Geometric probability models, knowing X 2 − 4 Y ≥ 0. But I think the real answer should be related with r.v. and the properties of uniform distribution. What is the right way to solve the problem?

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X  ,  Y  ∼  U  [  0  ,  1  ], what is the probability that       t    2    +  X  t  +  Y  =  0 has a real root.Let the continuous random variables X,Y be independent of each other and uniformly distributed on [0,1] i.e.   X  ,  Y  ∼  U  [  0  ,  1  ], what is the probability that       t    2    +  X  t  +  Y  =  0 has a real root?I am trying to solve it with Geometric probability models, knowing       X    2    −  4  Y  ≥  0. But I think the real answer should be related with r.v. and the properties of uniform distribution. What is the right way to solve the problem?

Hamza Conrad · Accepted Answer

Step 1As you said, the probability to have a real root is   P  (      X    2    −  4  Y  ≥  0  )Step 2you can calculate it integrating   ∫      ∫                  x        2            ≥      4      y        f  (  x  ,  y  )  d  x  d  y being   f  (  x  ,  y  )  =  1 the double integral is equivalent to the integration area   P  (      X    2    ≥  4  Y  )  =      ∫    0    1              x      2        4    d  x  =      1    12

Let the continuous random variables X,Y be independent of each other and uniformly distributed on [0,1] i.e. X,Y∼U[0,1], what is the probability that t^2+Xt+Y=0 has a real root?

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