Independent event probability












0












$begingroup$


Quoted from a popular book,




For instance, if we flip a fair coin twice, knowing whether the first
flip is Heads gives us no information about whether the second flip is
Heads. These events are independent.



On the other hand, knowing whether the first flip is Heads certainly
gives us information about whether both flips are Tails. (If the first
flip is Heads, then definitely it’s not the case that both flips are
Tails.) These two events are dependent.



Mathematically, we say that two events E and F are independent if the
probability that they both happen is the product of the probabilities
that each one happens:



P(E,F) = P(E)P(F)


In the example above, the probability of “first flip Heads” is 1/2,
and the probability of “both flips Tails” is 1/4, but the probability
of “first flip Heads and both flips Tails” is 0.




QUESTION: In the example we say 



P(first flip Heads) = 1/2

P(both flips Tails) = 1/4

P(first flip Heads and both flips Tails) = 1/2 * 1/4 (WHY NOT THIS?)

P(first flip Heads and both flips Tails) = 0 (CORRECT ANSWER)


Although it makes logical sense to put zero, but doesn't explain why doesn't it fit the formula






probability







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The formula doesn't apply because of course these events aren't independent.
    $endgroup$
    – Mason
    Dec 12 '18 at 4:37


















0












$begingroup$


Quoted from a popular book,




For instance, if we flip a fair coin twice, knowing whether the first
flip is Heads gives us no information about whether the second flip is
Heads. These events are independent.



On the other hand, knowing whether the first flip is Heads certainly
gives us information about whether both flips are Tails. (If the first
flip is Heads, then definitely it’s not the case that both flips are
Tails.) These two events are dependent.



Mathematically, we say that two events E and F are independent if the
probability that they both happen is the product of the probabilities
that each one happens:



P(E,F) = P(E)P(F)


In the example above, the probability of “first flip Heads” is 1/2,
and the probability of “both flips Tails” is 1/4, but the probability
of “first flip Heads and both flips Tails” is 0.




QUESTION: In the example we say 



P(first flip Heads) = 1/2

P(both flips Tails) = 1/4

P(first flip Heads and both flips Tails) = 1/2 * 1/4 (WHY NOT THIS?)

P(first flip Heads and both flips Tails) = 0 (CORRECT ANSWER)


Although it makes logical sense to put zero, but doesn't explain why doesn't it fit the formula






probability







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The formula doesn't apply because of course these events aren't independent.
    $endgroup$
    – Mason
    Dec 12 '18 at 4:37
















0












0








0


1



$begingroup$


Quoted from a popular book,




For instance, if we flip a fair coin twice, knowing whether the first
flip is Heads gives us no information about whether the second flip is
Heads. These events are independent.



On the other hand, knowing whether the first flip is Heads certainly
gives us information about whether both flips are Tails. (If the first
flip is Heads, then definitely it’s not the case that both flips are
Tails.) These two events are dependent.



Mathematically, we say that two events E and F are independent if the
probability that they both happen is the product of the probabilities
that each one happens:



P(E,F) = P(E)P(F)


In the example above, the probability of “first flip Heads” is 1/2,
and the probability of “both flips Tails” is 1/4, but the probability
of “first flip Heads and both flips Tails” is 0.




QUESTION: In the example we say 



P(first flip Heads) = 1/2

P(both flips Tails) = 1/4

P(first flip Heads and both flips Tails) = 1/2 * 1/4 (WHY NOT THIS?)

P(first flip Heads and both flips Tails) = 0 (CORRECT ANSWER)


Although it makes logical sense to put zero, but doesn't explain why doesn't it fit the formula






probability







share|cite|improve this question









$endgroup$




Quoted from a popular book,




For instance, if we flip a fair coin twice, knowing whether the first
flip is Heads gives us no information about whether the second flip is
Heads. These events are independent.



On the other hand, knowing whether the first flip is Heads certainly
gives us information about whether both flips are Tails. (If the first
flip is Heads, then definitely it’s not the case that both flips are
Tails.) These two events are dependent.



Mathematically, we say that two events E and F are independent if the
probability that they both happen is the product of the probabilities
that each one happens:



P(E,F) = P(E)P(F)


In the example above, the probability of “first flip Heads” is 1/2,
and the probability of “both flips Tails” is 1/4, but the probability
of “first flip Heads and both flips Tails” is 0.




QUESTION: In the example we say 



P(first flip Heads) = 1/2

P(both flips Tails) = 1/4

P(first flip Heads and both flips Tails) = 1/2 * 1/4 (WHY NOT THIS?)

P(first flip Heads and both flips Tails) = 0 (CORRECT ANSWER)


Although it makes logical sense to put zero, but doesn't explain why doesn't it fit the formula






probability




probability






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 4:24









VishnudevVishnudev

1033




1033








  • 2




    $begingroup$
    The formula doesn't apply because of course these events aren't independent.
    $endgroup$
    – Mason
    Dec 12 '18 at 4:37
















  • 2




    $begingroup$
    The formula doesn't apply because of course these events aren't independent.
    $endgroup$
    – Mason
    Dec 12 '18 at 4:37










2




2




$begingroup$
The formula doesn't apply because of course these events aren't independent.
$endgroup$
– Mason
Dec 12 '18 at 4:37






$begingroup$
The formula doesn't apply because of course these events aren't independent.
$endgroup$
– Mason
Dec 12 '18 at 4:37












3 Answers
3






active

oldest

votes


















1












$begingroup$

You are misinterpreting the expression a little bit.



When we say both flip heads, we are referring to
$$P(H_1,H_2)=P(H_1)P(H_2)$$
which we can directly apply the formula.



But when we take it one step further with "first flip head and both flip tails" we are finding



$$P(H_1wedge(T_1wedge T_2))$$



for now the 'terms' $H_1$ and $(T_1wedge T_2)$ are not independent since we are talking about the same first coin which can only show one side. But we can rewrite the expression to
$$P((H_1wedge T_1)wedge T_2))$$
which makes the same combinatorial sense, but the two 'terms' are now independent so we can evaluate
$$P((H_1wedge T_1)wedge T_2))=P(H_1wedge T_1)P( T_2)$$
and clearly
$$P(H_1wedge T_1)=0$$
I hope this explains :)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the answer and it explained the situation flawlessly. I won't ever forget this.
    $endgroup$
    – Vishnudev
    Dec 12 '18 at 8:02



















2












$begingroup$

You have already given yourself the answer.



The events $TT$ and $Hcdot$ are dependent. Independence of two events $E,F$ would mean that





  • $P(E|F)= frac{P(E cap F)}{P(F)}= P(E)$.


But in your case you have
$$P(TT | Hcdot) = 0 neq P(TT)$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Let's say event $E$ is first flip Heads and event $F$ is both flips Tails.



    begin{align*}
    P(E) &= 1/2 \
    P(F) &= 1/4 \
    P(E,F) &= 0 \
    P(E,F) &neq P(E) cdot P(F) \
    end{align*}



    It doesn't fit the formula for independence because the two events are not independent.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      You are misinterpreting the expression a little bit.



      When we say both flip heads, we are referring to
      $$P(H_1,H_2)=P(H_1)P(H_2)$$
      which we can directly apply the formula.



      But when we take it one step further with "first flip head and both flip tails" we are finding



      $$P(H_1wedge(T_1wedge T_2))$$



      for now the 'terms' $H_1$ and $(T_1wedge T_2)$ are not independent since we are talking about the same first coin which can only show one side. But we can rewrite the expression to
      $$P((H_1wedge T_1)wedge T_2))$$
      which makes the same combinatorial sense, but the two 'terms' are now independent so we can evaluate
      $$P((H_1wedge T_1)wedge T_2))=P(H_1wedge T_1)P( T_2)$$
      and clearly
      $$P(H_1wedge T_1)=0$$
      I hope this explains :)






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks for the answer and it explained the situation flawlessly. I won't ever forget this.
        $endgroup$
        – Vishnudev
        Dec 12 '18 at 8:02
















      1












      $begingroup$

      You are misinterpreting the expression a little bit.



      When we say both flip heads, we are referring to
      $$P(H_1,H_2)=P(H_1)P(H_2)$$
      which we can directly apply the formula.



      But when we take it one step further with "first flip head and both flip tails" we are finding



      $$P(H_1wedge(T_1wedge T_2))$$



      for now the 'terms' $H_1$ and $(T_1wedge T_2)$ are not independent since we are talking about the same first coin which can only show one side. But we can rewrite the expression to
      $$P((H_1wedge T_1)wedge T_2))$$
      which makes the same combinatorial sense, but the two 'terms' are now independent so we can evaluate
      $$P((H_1wedge T_1)wedge T_2))=P(H_1wedge T_1)P( T_2)$$
      and clearly
      $$P(H_1wedge T_1)=0$$
      I hope this explains :)






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks for the answer and it explained the situation flawlessly. I won't ever forget this.
        $endgroup$
        – Vishnudev
        Dec 12 '18 at 8:02














      1












      1








      1





      $begingroup$

      You are misinterpreting the expression a little bit.



      When we say both flip heads, we are referring to
      $$P(H_1,H_2)=P(H_1)P(H_2)$$
      which we can directly apply the formula.



      But when we take it one step further with "first flip head and both flip tails" we are finding



      $$P(H_1wedge(T_1wedge T_2))$$



      for now the 'terms' $H_1$ and $(T_1wedge T_2)$ are not independent since we are talking about the same first coin which can only show one side. But we can rewrite the expression to
      $$P((H_1wedge T_1)wedge T_2))$$
      which makes the same combinatorial sense, but the two 'terms' are now independent so we can evaluate
      $$P((H_1wedge T_1)wedge T_2))=P(H_1wedge T_1)P( T_2)$$
      and clearly
      $$P(H_1wedge T_1)=0$$
      I hope this explains :)






      share|cite|improve this answer









      $endgroup$



      You are misinterpreting the expression a little bit.



      When we say both flip heads, we are referring to
      $$P(H_1,H_2)=P(H_1)P(H_2)$$
      which we can directly apply the formula.



      But when we take it one step further with "first flip head and both flip tails" we are finding



      $$P(H_1wedge(T_1wedge T_2))$$



      for now the 'terms' $H_1$ and $(T_1wedge T_2)$ are not independent since we are talking about the same first coin which can only show one side. But we can rewrite the expression to
      $$P((H_1wedge T_1)wedge T_2))$$
      which makes the same combinatorial sense, but the two 'terms' are now independent so we can evaluate
      $$P((H_1wedge T_1)wedge T_2))=P(H_1wedge T_1)P( T_2)$$
      and clearly
      $$P(H_1wedge T_1)=0$$
      I hope this explains :)







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 12 '18 at 4:37









      Karn WatcharasupatKarn Watcharasupat

      3,9742526




      3,9742526












      • $begingroup$
        Thanks for the answer and it explained the situation flawlessly. I won't ever forget this.
        $endgroup$
        – Vishnudev
        Dec 12 '18 at 8:02


















      • $begingroup$
        Thanks for the answer and it explained the situation flawlessly. I won't ever forget this.
        $endgroup$
        – Vishnudev
        Dec 12 '18 at 8:02
















      $begingroup$
      Thanks for the answer and it explained the situation flawlessly. I won't ever forget this.
      $endgroup$
      – Vishnudev
      Dec 12 '18 at 8:02




      $begingroup$
      Thanks for the answer and it explained the situation flawlessly. I won't ever forget this.
      $endgroup$
      – Vishnudev
      Dec 12 '18 at 8:02











      2












      $begingroup$

      You have already given yourself the answer.



      The events $TT$ and $Hcdot$ are dependent. Independence of two events $E,F$ would mean that





      • $P(E|F)= frac{P(E cap F)}{P(F)}= P(E)$.


      But in your case you have
      $$P(TT | Hcdot) = 0 neq P(TT)$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        You have already given yourself the answer.



        The events $TT$ and $Hcdot$ are dependent. Independence of two events $E,F$ would mean that





        • $P(E|F)= frac{P(E cap F)}{P(F)}= P(E)$.


        But in your case you have
        $$P(TT | Hcdot) = 0 neq P(TT)$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          You have already given yourself the answer.



          The events $TT$ and $Hcdot$ are dependent. Independence of two events $E,F$ would mean that





          • $P(E|F)= frac{P(E cap F)}{P(F)}= P(E)$.


          But in your case you have
          $$P(TT | Hcdot) = 0 neq P(TT)$$






          share|cite|improve this answer









          $endgroup$



          You have already given yourself the answer.



          The events $TT$ and $Hcdot$ are dependent. Independence of two events $E,F$ would mean that





          • $P(E|F)= frac{P(E cap F)}{P(F)}= P(E)$.


          But in your case you have
          $$P(TT | Hcdot) = 0 neq P(TT)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 4:43









          trancelocationtrancelocation

          12.3k1826




          12.3k1826























              1












              $begingroup$

              Let's say event $E$ is first flip Heads and event $F$ is both flips Tails.



              begin{align*}
              P(E) &= 1/2 \
              P(F) &= 1/4 \
              P(E,F) &= 0 \
              P(E,F) &neq P(E) cdot P(F) \
              end{align*}



              It doesn't fit the formula for independence because the two events are not independent.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let's say event $E$ is first flip Heads and event $F$ is both flips Tails.



                begin{align*}
                P(E) &= 1/2 \
                P(F) &= 1/4 \
                P(E,F) &= 0 \
                P(E,F) &neq P(E) cdot P(F) \
                end{align*}



                It doesn't fit the formula for independence because the two events are not independent.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let's say event $E$ is first flip Heads and event $F$ is both flips Tails.



                  begin{align*}
                  P(E) &= 1/2 \
                  P(F) &= 1/4 \
                  P(E,F) &= 0 \
                  P(E,F) &neq P(E) cdot P(F) \
                  end{align*}



                  It doesn't fit the formula for independence because the two events are not independent.






                  share|cite|improve this answer









                  $endgroup$



                  Let's say event $E$ is first flip Heads and event $F$ is both flips Tails.



                  begin{align*}
                  P(E) &= 1/2 \
                  P(F) &= 1/4 \
                  P(E,F) &= 0 \
                  P(E,F) &neq P(E) cdot P(F) \
                  end{align*}



                  It doesn't fit the formula for independence because the two events are not independent.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 4:40









                  clayclay

                  767415




                  767415






























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