The discrete random variable X has the probability distribution function shown below. Find $E(X).$ [closed]












1












$begingroup$


when $x = 0$, $ Pr[X = x] = 2/9,$



when $x = 1$, $Pr[X = x] =4/9,$



when $x = 2$ , $Pr[X = x]= 1/3$






probability







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$endgroup$



closed as off-topic by copper.hat, Adrian Keister, JimmyK4542, Shailesh, KReiser Dec 4 '18 at 1:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, JimmyK4542, Shailesh, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    Why not try computing using the formula $sum_k x_k p_k$?
    $endgroup$
    – copper.hat
    Dec 3 '18 at 21:09
















1












$begingroup$


when $x = 0$, $ Pr[X = x] = 2/9,$



when $x = 1$, $Pr[X = x] =4/9,$



when $x = 2$ , $Pr[X = x]= 1/3$






probability







share|cite|improve this question











$endgroup$



closed as off-topic by copper.hat, Adrian Keister, JimmyK4542, Shailesh, KReiser Dec 4 '18 at 1:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, JimmyK4542, Shailesh, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    Why not try computing using the formula $sum_k x_k p_k$?
    $endgroup$
    – copper.hat
    Dec 3 '18 at 21:09














1












1








1





$begingroup$


when $x = 0$, $ Pr[X = x] = 2/9,$



when $x = 1$, $Pr[X = x] =4/9,$



when $x = 2$ , $Pr[X = x]= 1/3$






probability







share|cite|improve this question











$endgroup$




when $x = 0$, $ Pr[X = x] = 2/9,$



when $x = 1$, $Pr[X = x] =4/9,$



when $x = 2$ , $Pr[X = x]= 1/3$






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 21:10









greedoid

44k1155109




44k1155109










asked Dec 3 '18 at 21:06









C. KamathC. Kamath

61




61




closed as off-topic by copper.hat, Adrian Keister, JimmyK4542, Shailesh, KReiser Dec 4 '18 at 1:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, JimmyK4542, Shailesh, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by copper.hat, Adrian Keister, JimmyK4542, Shailesh, KReiser Dec 4 '18 at 1:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, JimmyK4542, Shailesh, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    Why not try computing using the formula $sum_k x_k p_k$?
    $endgroup$
    – copper.hat
    Dec 3 '18 at 21:09














  • 1




    $begingroup$
    Why not try computing using the formula $sum_k x_k p_k$?
    $endgroup$
    – copper.hat
    Dec 3 '18 at 21:09








1




1




$begingroup$
Why not try computing using the formula $sum_k x_k p_k$?
$endgroup$
– copper.hat
Dec 3 '18 at 21:09




$begingroup$
Why not try computing using the formula $sum_k x_k p_k$?
$endgroup$
– copper.hat
Dec 3 '18 at 21:09










1 Answer
1






active

oldest

votes


















1












$begingroup$

$$ E(X) = 0cdot P(X=0)+1cdot P(X=1)+2cdot P(X=2)$$ $$= 0+1cdot {4over 9}+ 2cdot {1over 3} = {10over 9}$$






share|cite|improve this answer











$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $$ E(X) = 0cdot P(X=0)+1cdot P(X=1)+2cdot P(X=2)$$ $$= 0+1cdot {4over 9}+ 2cdot {1over 3} = {10over 9}$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      $$ E(X) = 0cdot P(X=0)+1cdot P(X=1)+2cdot P(X=2)$$ $$= 0+1cdot {4over 9}+ 2cdot {1over 3} = {10over 9}$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $$ E(X) = 0cdot P(X=0)+1cdot P(X=1)+2cdot P(X=2)$$ $$= 0+1cdot {4over 9}+ 2cdot {1over 3} = {10over 9}$$






        share|cite|improve this answer











        $endgroup$



        $$ E(X) = 0cdot P(X=0)+1cdot P(X=1)+2cdot P(X=2)$$ $$= 0+1cdot {4over 9}+ 2cdot {1over 3} = {10over 9}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 21:13

























        answered Dec 3 '18 at 21:09









        greedoidgreedoid

        44k1155109




        44k1155109















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