Independent Event Complements












1












$begingroup$


I have the following homework assignment that I've already finished, but am confused on whether I've gotten right/wrong, and was hoping someone could help explain so I understand the problem better.




An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian projet is successful and B be the event that the European project is successful. Suppose that A and B are independent events with P(A)=0.4 and P(B)=0.7



a.) If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning.



b.) What is the probability that at least one of the two projects will be successful?



c.) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?"




Here is what I've gotten for each part:



$$P(A cap B) = P(A)P(B) = (.4)(.7) = .28 $$



a.) $P(B^c) = 1 - 0.7 = 0.3 $



b.) $P(A cup B) = P(A) + P(B) - P(A cap B) = .4 + .7 - .28 = .82$



c.) $P(A) - P(A cap B) = .4 - .28 = .12 $
$.12/.82 = .146$



I am confused in that the two events are independent of each other and the book states that for part a the answer should be .126 instead of what I got. Am I doing these problems correctly or am I committing some error?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Part a) is correctly done.
    $endgroup$
    – André Nicolas
    Feb 21 '14 at 17:32










  • $begingroup$
    So do I need to get $P(B^c | A^c)$ or not for part a? I originally thought I was correct and there may be a chance the book is wrong, but I don't see how to get 0.126 as the answer for part a
    $endgroup$
    – Valrok
    Feb 21 '14 at 17:35












  • $begingroup$
    You are right, $B^c$ and $A^c$ are independent. But if you really really feel like it, you can calculate. We have $Pr(B^c|A^c)=frac{Pr(B^ccap A^c)}{Pr(A^c)}$. After some grinding you will get $0.3$.
    $endgroup$
    – André Nicolas
    Feb 21 '14 at 17:41










  • $begingroup$
    Ah, I guess I may have stumbled into one of those few times that the book is incorrect with part a. For the other parts my answers were the same as the book's but I wasn't sure if there may have been other mistakes that I had not noticed.
    $endgroup$
    – Valrok
    Feb 21 '14 at 17:53
















1












$begingroup$


I have the following homework assignment that I've already finished, but am confused on whether I've gotten right/wrong, and was hoping someone could help explain so I understand the problem better.




An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian projet is successful and B be the event that the European project is successful. Suppose that A and B are independent events with P(A)=0.4 and P(B)=0.7



a.) If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning.



b.) What is the probability that at least one of the two projects will be successful?



c.) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?"




Here is what I've gotten for each part:



$$P(A cap B) = P(A)P(B) = (.4)(.7) = .28 $$



a.) $P(B^c) = 1 - 0.7 = 0.3 $



b.) $P(A cup B) = P(A) + P(B) - P(A cap B) = .4 + .7 - .28 = .82$



c.) $P(A) - P(A cap B) = .4 - .28 = .12 $
$.12/.82 = .146$



I am confused in that the two events are independent of each other and the book states that for part a the answer should be .126 instead of what I got. Am I doing these problems correctly or am I committing some error?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Part a) is correctly done.
    $endgroup$
    – André Nicolas
    Feb 21 '14 at 17:32










  • $begingroup$
    So do I need to get $P(B^c | A^c)$ or not for part a? I originally thought I was correct and there may be a chance the book is wrong, but I don't see how to get 0.126 as the answer for part a
    $endgroup$
    – Valrok
    Feb 21 '14 at 17:35












  • $begingroup$
    You are right, $B^c$ and $A^c$ are independent. But if you really really feel like it, you can calculate. We have $Pr(B^c|A^c)=frac{Pr(B^ccap A^c)}{Pr(A^c)}$. After some grinding you will get $0.3$.
    $endgroup$
    – André Nicolas
    Feb 21 '14 at 17:41










  • $begingroup$
    Ah, I guess I may have stumbled into one of those few times that the book is incorrect with part a. For the other parts my answers were the same as the book's but I wasn't sure if there may have been other mistakes that I had not noticed.
    $endgroup$
    – Valrok
    Feb 21 '14 at 17:53














1












1








1





$begingroup$


I have the following homework assignment that I've already finished, but am confused on whether I've gotten right/wrong, and was hoping someone could help explain so I understand the problem better.




An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian projet is successful and B be the event that the European project is successful. Suppose that A and B are independent events with P(A)=0.4 and P(B)=0.7



a.) If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning.



b.) What is the probability that at least one of the two projects will be successful?



c.) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?"




Here is what I've gotten for each part:



$$P(A cap B) = P(A)P(B) = (.4)(.7) = .28 $$



a.) $P(B^c) = 1 - 0.7 = 0.3 $



b.) $P(A cup B) = P(A) + P(B) - P(A cap B) = .4 + .7 - .28 = .82$



c.) $P(A) - P(A cap B) = .4 - .28 = .12 $
$.12/.82 = .146$



I am confused in that the two events are independent of each other and the book states that for part a the answer should be .126 instead of what I got. Am I doing these problems correctly or am I committing some error?






probability







share|cite|improve this question











$endgroup$




I have the following homework assignment that I've already finished, but am confused on whether I've gotten right/wrong, and was hoping someone could help explain so I understand the problem better.




An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian projet is successful and B be the event that the European project is successful. Suppose that A and B are independent events with P(A)=0.4 and P(B)=0.7



a.) If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning.



b.) What is the probability that at least one of the two projects will be successful?



c.) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?"




Here is what I've gotten for each part:



$$P(A cap B) = P(A)P(B) = (.4)(.7) = .28 $$



a.) $P(B^c) = 1 - 0.7 = 0.3 $



b.) $P(A cup B) = P(A) + P(B) - P(A cap B) = .4 + .7 - .28 = .82$



c.) $P(A) - P(A cap B) = .4 - .28 = .12 $
$.12/.82 = .146$



I am confused in that the two events are independent of each other and the book states that for part a the answer should be .126 instead of what I got. Am I doing these problems correctly or am I committing some error?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 '16 at 22:41









Did

247k23223460




247k23223460










asked Feb 21 '14 at 17:22









ValrokValrok

2933926




2933926












  • $begingroup$
    Part a) is correctly done.
    $endgroup$
    – André Nicolas
    Feb 21 '14 at 17:32










  • $begingroup$
    So do I need to get $P(B^c | A^c)$ or not for part a? I originally thought I was correct and there may be a chance the book is wrong, but I don't see how to get 0.126 as the answer for part a
    $endgroup$
    – Valrok
    Feb 21 '14 at 17:35












  • $begingroup$
    You are right, $B^c$ and $A^c$ are independent. But if you really really feel like it, you can calculate. We have $Pr(B^c|A^c)=frac{Pr(B^ccap A^c)}{Pr(A^c)}$. After some grinding you will get $0.3$.
    $endgroup$
    – André Nicolas
    Feb 21 '14 at 17:41










  • $begingroup$
    Ah, I guess I may have stumbled into one of those few times that the book is incorrect with part a. For the other parts my answers were the same as the book's but I wasn't sure if there may have been other mistakes that I had not noticed.
    $endgroup$
    – Valrok
    Feb 21 '14 at 17:53


















  • $begingroup$
    Part a) is correctly done.
    $endgroup$
    – André Nicolas
    Feb 21 '14 at 17:32










  • $begingroup$
    So do I need to get $P(B^c | A^c)$ or not for part a? I originally thought I was correct and there may be a chance the book is wrong, but I don't see how to get 0.126 as the answer for part a
    $endgroup$
    – Valrok
    Feb 21 '14 at 17:35












  • $begingroup$
    You are right, $B^c$ and $A^c$ are independent. But if you really really feel like it, you can calculate. We have $Pr(B^c|A^c)=frac{Pr(B^ccap A^c)}{Pr(A^c)}$. After some grinding you will get $0.3$.
    $endgroup$
    – André Nicolas
    Feb 21 '14 at 17:41










  • $begingroup$
    Ah, I guess I may have stumbled into one of those few times that the book is incorrect with part a. For the other parts my answers were the same as the book's but I wasn't sure if there may have been other mistakes that I had not noticed.
    $endgroup$
    – Valrok
    Feb 21 '14 at 17:53
















$begingroup$
Part a) is correctly done.
$endgroup$
– André Nicolas
Feb 21 '14 at 17:32




$begingroup$
Part a) is correctly done.
$endgroup$
– André Nicolas
Feb 21 '14 at 17:32












$begingroup$
So do I need to get $P(B^c | A^c)$ or not for part a? I originally thought I was correct and there may be a chance the book is wrong, but I don't see how to get 0.126 as the answer for part a
$endgroup$
– Valrok
Feb 21 '14 at 17:35






$begingroup$
So do I need to get $P(B^c | A^c)$ or not for part a? I originally thought I was correct and there may be a chance the book is wrong, but I don't see how to get 0.126 as the answer for part a
$endgroup$
– Valrok
Feb 21 '14 at 17:35














$begingroup$
You are right, $B^c$ and $A^c$ are independent. But if you really really feel like it, you can calculate. We have $Pr(B^c|A^c)=frac{Pr(B^ccap A^c)}{Pr(A^c)}$. After some grinding you will get $0.3$.
$endgroup$
– André Nicolas
Feb 21 '14 at 17:41




$begingroup$
You are right, $B^c$ and $A^c$ are independent. But if you really really feel like it, you can calculate. We have $Pr(B^c|A^c)=frac{Pr(B^ccap A^c)}{Pr(A^c)}$. After some grinding you will get $0.3$.
$endgroup$
– André Nicolas
Feb 21 '14 at 17:41












$begingroup$
Ah, I guess I may have stumbled into one of those few times that the book is incorrect with part a. For the other parts my answers were the same as the book's but I wasn't sure if there may have been other mistakes that I had not noticed.
$endgroup$
– Valrok
Feb 21 '14 at 17:53




$begingroup$
Ah, I guess I may have stumbled into one of those few times that the book is incorrect with part a. For the other parts my answers were the same as the book's but I wasn't sure if there may have been other mistakes that I had not noticed.
$endgroup$
– Valrok
Feb 21 '14 at 17:53










3 Answers
3






active

oldest

votes


















0












$begingroup$

Indeed you got part (a) right, while the explanation to your solution would be that $A^C$ and $B^C$ are independent.



Note that even though it is quite intuitive that "$A$ and $B$ are independent $rightarrow$ $A^C$ and $B^C$ are independent", I think it isn't completely trivial.

(You can find some proofs here, though I bet you would succeed in proving it by yourself.)






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    c) (p(A) - p(A n B))/(p(A) + p(B) - p(A n B))



    With your numbers



    .4 - (.4 * .7) / (.4 + .7 - (.4 * .7))






    share|cite|improve this answer









    $endgroup$





















      -3












      $begingroup$

      (a) This question belongs to "conditional model"
      But since A and B are independent
      You may take directly p(B not)=0.3



      Reason: independent means



      P(A .B)=P(A).P(B)






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        For the next problem (b)at least one of the two projects will be success. = 1- P(no project will success). = 1- P(Anot).P(Bnot)=1 -(0.6)(0.3)=0.82 answer
        $endgroup$
        – siva naga kumar
        Jun 3 '16 at 4:57













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Indeed you got part (a) right, while the explanation to your solution would be that $A^C$ and $B^C$ are independent.



      Note that even though it is quite intuitive that "$A$ and $B$ are independent $rightarrow$ $A^C$ and $B^C$ are independent", I think it isn't completely trivial.

      (You can find some proofs here, though I bet you would succeed in proving it by yourself.)






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Indeed you got part (a) right, while the explanation to your solution would be that $A^C$ and $B^C$ are independent.



        Note that even though it is quite intuitive that "$A$ and $B$ are independent $rightarrow$ $A^C$ and $B^C$ are independent", I think it isn't completely trivial.

        (You can find some proofs here, though I bet you would succeed in proving it by yourself.)






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Indeed you got part (a) right, while the explanation to your solution would be that $A^C$ and $B^C$ are independent.



          Note that even though it is quite intuitive that "$A$ and $B$ are independent $rightarrow$ $A^C$ and $B^C$ are independent", I think it isn't completely trivial.

          (You can find some proofs here, though I bet you would succeed in proving it by yourself.)






          share|cite|improve this answer









          $endgroup$



          Indeed you got part (a) right, while the explanation to your solution would be that $A^C$ and $B^C$ are independent.



          Note that even though it is quite intuitive that "$A$ and $B$ are independent $rightarrow$ $A^C$ and $B^C$ are independent", I think it isn't completely trivial.

          (You can find some proofs here, though I bet you would succeed in proving it by yourself.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 19 '18 at 8:35









          Oren MilmanOren Milman

          1747




          1747























              -1












              $begingroup$

              c) (p(A) - p(A n B))/(p(A) + p(B) - p(A n B))



              With your numbers



              .4 - (.4 * .7) / (.4 + .7 - (.4 * .7))






              share|cite|improve this answer









              $endgroup$


















                -1












                $begingroup$

                c) (p(A) - p(A n B))/(p(A) + p(B) - p(A n B))



                With your numbers



                .4 - (.4 * .7) / (.4 + .7 - (.4 * .7))






                share|cite|improve this answer









                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  c) (p(A) - p(A n B))/(p(A) + p(B) - p(A n B))



                  With your numbers



                  .4 - (.4 * .7) / (.4 + .7 - (.4 * .7))






                  share|cite|improve this answer









                  $endgroup$



                  c) (p(A) - p(A n B))/(p(A) + p(B) - p(A n B))



                  With your numbers



                  .4 - (.4 * .7) / (.4 + .7 - (.4 * .7))







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 8 '16 at 19:55









                  LikeLike

                  1




                  1























                      -3












                      $begingroup$

                      (a) This question belongs to "conditional model"
                      But since A and B are independent
                      You may take directly p(B not)=0.3



                      Reason: independent means



                      P(A .B)=P(A).P(B)






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        For the next problem (b)at least one of the two projects will be success. = 1- P(no project will success). = 1- P(Anot).P(Bnot)=1 -(0.6)(0.3)=0.82 answer
                        $endgroup$
                        – siva naga kumar
                        Jun 3 '16 at 4:57


















                      -3












                      $begingroup$

                      (a) This question belongs to "conditional model"
                      But since A and B are independent
                      You may take directly p(B not)=0.3



                      Reason: independent means



                      P(A .B)=P(A).P(B)






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        For the next problem (b)at least one of the two projects will be success. = 1- P(no project will success). = 1- P(Anot).P(Bnot)=1 -(0.6)(0.3)=0.82 answer
                        $endgroup$
                        – siva naga kumar
                        Jun 3 '16 at 4:57
















                      -3












                      -3








                      -3





                      $begingroup$

                      (a) This question belongs to "conditional model"
                      But since A and B are independent
                      You may take directly p(B not)=0.3



                      Reason: independent means



                      P(A .B)=P(A).P(B)






                      share|cite|improve this answer











                      $endgroup$



                      (a) This question belongs to "conditional model"
                      But since A and B are independent
                      You may take directly p(B not)=0.3



                      Reason: independent means



                      P(A .B)=P(A).P(B)







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jun 3 '16 at 4:56

























                      answered Jul 7 '15 at 4:58









                      siva naga kumarsiva naga kumar

                      938




                      938












                      • $begingroup$
                        For the next problem (b)at least one of the two projects will be success. = 1- P(no project will success). = 1- P(Anot).P(Bnot)=1 -(0.6)(0.3)=0.82 answer
                        $endgroup$
                        – siva naga kumar
                        Jun 3 '16 at 4:57




















                      • $begingroup$
                        For the next problem (b)at least one of the two projects will be success. = 1- P(no project will success). = 1- P(Anot).P(Bnot)=1 -(0.6)(0.3)=0.82 answer
                        $endgroup$
                        – siva naga kumar
                        Jun 3 '16 at 4:57


















                      $begingroup$
                      For the next problem (b)at least one of the two projects will be success. = 1- P(no project will success). = 1- P(Anot).P(Bnot)=1 -(0.6)(0.3)=0.82 answer
                      $endgroup$
                      – siva naga kumar
                      Jun 3 '16 at 4:57






                      $begingroup$
                      For the next problem (b)at least one of the two projects will be success. = 1- P(no project will success). = 1- P(Anot).P(Bnot)=1 -(0.6)(0.3)=0.82 answer
                      $endgroup$
                      – siva naga kumar
                      Jun 3 '16 at 4:57




















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                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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