Question

Suppose that the genders of the three children of a certain family are soon to be revealed. Outcomes are thus triples of "girls" (g) and "boys" (b), which we write gbg, bbb, etc. For each outcome, let R be the random variable counting the number of girls in each outcome. For example, if the outcome is gbb, then R9gbb)=1. Suppose that the random variable X is defined in terms of R as follows: X=2R^2-4R-2. The values of X are thus:

Answer 1

Complete Question

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Answer:

  Value  x of X     -3    -7     -15

          $P_X (x)$          $\frac{1}{2}$      $\frac{3}{8}$       $\frac{1}{8}$

   

Step-by-step explanation:

From the question we are told that  

     The values of X are  $X = -3 , -7 , -15$

     The total number of outcomes is  n =  8

The probability distribution function of X is evaluated as follow

   $p(X = -3 ) = \frac{N_{-3}}{n}$

Where  $N{-3}$ is the number of time  X = -3  occurred and from the table the value is  $N _{-3} = 4$

     Therefore

       $p(X = -3 ) = \frac{4}{8}$

        $p(X = -3 ) = \frac{1}{2}$

Now  

    $p(X = -7 ) = \frac{N_{-7}}{n}$

Where $N_{-7} = 3$ from table  

So  

     $p(X = -7 ) = \frac{3}{8}$

Also  

    $p(X = -15 ) = \frac{N_{-15}}{n}$

    $p(X = -15 ) = \frac{1}{8}$