This example is from the book first course in probability Example 4a.Let X denote a random variable that takes on any of the values −1, 0, and 1 with respective probabilities P { X = − 1 } = .2 P { X = 0 } = .5 &gt; P { X = 1 } = .3.Solution.Let Y = X 2 . Then the probability mass function of Y is given by P { Y = 1 } = P { X = − 1 } + P { X = 1 } = .5 P { Y = 0 } = P { X = 0 } = .5What rule is used to compute P { Y = 1 }? When Y = X 2 Why addition is used ?

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This example is from the book first course in probability Example 4a.Let   X denote a random variable that takes on any of the values −1, 0, and 1 with respective probabilities   P  {  X  =  −  1  }  =  .2    P  {  X  =  0  }  =  .5  &amp;gt;    P  {  X  =  1  }  =  .3.Solution.Let   Y  =      X    2  . Then the probability mass function of   Y is given by  P  {  Y  =  1  }  =  P  {  X  =  −  1  }  +  P  {  X  =  1  }  =  .5  P  {  Y  =  0  }  =  P  {  X  =  0  }  =  .5What rule is used to compute   P  {  Y  =  1  }? When   Y  =      X    2   Why addition is used ?

Noemi Flores · Accepted Answer

P  (  Y  =  1  )  =  P  (      X    2    =  1  )  =  P  (  X  =  ±  1  )  =  P  (  X  =  1  )  +  P  (  X  =  −  1  )  .Addition can be used since events   {  X  =  1  } and   {  X  =  −  1  } are naturally disjoint.

Seamus Moran · Accepted Answer

This is by taking root over 1   P  (      X    2    =  1  )  =  P  (  X  =  ±  1  ). But how did you derive this part   P  (      X    2    =  1  )  =  P  (  X  =  1  )  +  P  (  X  =  2  ).

This example is from the book first course in probability Example 4a. Let X denote a random

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