Probability of three sequential events












8












$begingroup$



A smuggler wants to transfer his smuggled goods from city A to city B. There are three police check posts between these two cities. Assume that there is no communication among the check posts. The probabilities of him being caught at these three stops are $0.7, 0.5$ and $0.3$ respectively. What is the probability that he successfully transfers his goods?




  • A] $0.105$

  • B] $0.5$

  • C] $0.245$

  • D] $0.045$




Is it as simple as calculating success in all three scenarios:
$0.3 * 0.5 * 0.7 = 0.105$. Is A correct answer ?






probability proof-verification







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Definitely looks correct to me
    $endgroup$
    – gt6989b
    Nov 25 '18 at 16:41






  • 1




    $begingroup$
    Yes, the probability of not being stopped by the three posts equals the probability of not being stopped by any post which is $(1-0.3)(1-0.5)(1-0.7)=0.105$ because they are independent.
    $endgroup$
    – BlackMath
    Nov 25 '18 at 18:17








  • 1




    $begingroup$
    It doesn't help this problem that the probability of success at the first post is the probability of failure at the third post, and vice versa. I would argue that makes this a poor problem. But no one here can do anything about that, OP included.
    $endgroup$
    – Teepeemm
    Nov 25 '18 at 19:37
















8












$begingroup$



A smuggler wants to transfer his smuggled goods from city A to city B. There are three police check posts between these two cities. Assume that there is no communication among the check posts. The probabilities of him being caught at these three stops are $0.7, 0.5$ and $0.3$ respectively. What is the probability that he successfully transfers his goods?




  • A] $0.105$

  • B] $0.5$

  • C] $0.245$

  • D] $0.045$




Is it as simple as calculating success in all three scenarios:
$0.3 * 0.5 * 0.7 = 0.105$. Is A correct answer ?






probability proof-verification







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Definitely looks correct to me
    $endgroup$
    – gt6989b
    Nov 25 '18 at 16:41






  • 1




    $begingroup$
    Yes, the probability of not being stopped by the three posts equals the probability of not being stopped by any post which is $(1-0.3)(1-0.5)(1-0.7)=0.105$ because they are independent.
    $endgroup$
    – BlackMath
    Nov 25 '18 at 18:17








  • 1




    $begingroup$
    It doesn't help this problem that the probability of success at the first post is the probability of failure at the third post, and vice versa. I would argue that makes this a poor problem. But no one here can do anything about that, OP included.
    $endgroup$
    – Teepeemm
    Nov 25 '18 at 19:37














8












8








8





$begingroup$



A smuggler wants to transfer his smuggled goods from city A to city B. There are three police check posts between these two cities. Assume that there is no communication among the check posts. The probabilities of him being caught at these three stops are $0.7, 0.5$ and $0.3$ respectively. What is the probability that he successfully transfers his goods?




  • A] $0.105$

  • B] $0.5$

  • C] $0.245$

  • D] $0.045$




Is it as simple as calculating success in all three scenarios:
$0.3 * 0.5 * 0.7 = 0.105$. Is A correct answer ?






probability proof-verification







share|cite|improve this question











$endgroup$





A smuggler wants to transfer his smuggled goods from city A to city B. There are three police check posts between these two cities. Assume that there is no communication among the check posts. The probabilities of him being caught at these three stops are $0.7, 0.5$ and $0.3$ respectively. What is the probability that he successfully transfers his goods?




  • A] $0.105$

  • B] $0.5$

  • C] $0.245$

  • D] $0.045$




Is it as simple as calculating success in all three scenarios:
$0.3 * 0.5 * 0.7 = 0.105$. Is A correct answer ?






probability proof-verification




probability proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 16:51









Maria Mazur

47.9k1260120




47.9k1260120










asked Nov 25 '18 at 16:38









anmolmoreanmolmore

416




416








  • 3




    $begingroup$
    Definitely looks correct to me
    $endgroup$
    – gt6989b
    Nov 25 '18 at 16:41






  • 1




    $begingroup$
    Yes, the probability of not being stopped by the three posts equals the probability of not being stopped by any post which is $(1-0.3)(1-0.5)(1-0.7)=0.105$ because they are independent.
    $endgroup$
    – BlackMath
    Nov 25 '18 at 18:17








  • 1




    $begingroup$
    It doesn't help this problem that the probability of success at the first post is the probability of failure at the third post, and vice versa. I would argue that makes this a poor problem. But no one here can do anything about that, OP included.
    $endgroup$
    – Teepeemm
    Nov 25 '18 at 19:37














  • 3




    $begingroup$
    Definitely looks correct to me
    $endgroup$
    – gt6989b
    Nov 25 '18 at 16:41






  • 1




    $begingroup$
    Yes, the probability of not being stopped by the three posts equals the probability of not being stopped by any post which is $(1-0.3)(1-0.5)(1-0.7)=0.105$ because they are independent.
    $endgroup$
    – BlackMath
    Nov 25 '18 at 18:17








  • 1




    $begingroup$
    It doesn't help this problem that the probability of success at the first post is the probability of failure at the third post, and vice versa. I would argue that makes this a poor problem. But no one here can do anything about that, OP included.
    $endgroup$
    – Teepeemm
    Nov 25 '18 at 19:37








3




3




$begingroup$
Definitely looks correct to me
$endgroup$
– gt6989b
Nov 25 '18 at 16:41




$begingroup$
Definitely looks correct to me
$endgroup$
– gt6989b
Nov 25 '18 at 16:41




1




1




$begingroup$
Yes, the probability of not being stopped by the three posts equals the probability of not being stopped by any post which is $(1-0.3)(1-0.5)(1-0.7)=0.105$ because they are independent.
$endgroup$
– BlackMath
Nov 25 '18 at 18:17






$begingroup$
Yes, the probability of not being stopped by the three posts equals the probability of not being stopped by any post which is $(1-0.3)(1-0.5)(1-0.7)=0.105$ because they are independent.
$endgroup$
– BlackMath
Nov 25 '18 at 18:17






1




1




$begingroup$
It doesn't help this problem that the probability of success at the first post is the probability of failure at the third post, and vice versa. I would argue that makes this a poor problem. But no one here can do anything about that, OP included.
$endgroup$
– Teepeemm
Nov 25 '18 at 19:37




$begingroup$
It doesn't help this problem that the probability of success at the first post is the probability of failure at the third post, and vice versa. I would argue that makes this a poor problem. But no one here can do anything about that, OP included.
$endgroup$
– Teepeemm
Nov 25 '18 at 19:37










2 Answers
2






active

oldest

votes


















4












$begingroup$

Let $A_i$ be an event $i$-th police stop him. We are interested in $$P(A_1'cap A_2'cap A_3')= P(A_1')P (A_2')P(A_3')= 0.3 cdot 0.5 cdot 0.7 = 0.105$$



(since $A_1, A_2, A_3$ are independant so are $A_1', A_2', A_3'$ ) so your answer is correct.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    You can think this in two ways.
    first way
    Not being caught on first check post and not being caught on second check post and not being caught on third post.
    i.e.
    $(1-P(A1))×(1-P(A2))×(1-P(A3) =
    0.3×0.5×0.7 = 0.105$



    second way
    Find P(being caught)



    Caught on first check post
    Or
    Not being caught on first post and caught on second post
    Or
    Not being caught on first post and Not being caught on second post and caught on third post.



    $P(A1) + (1-P(A1))(P(A2)) + (1- P(A1))(1- P(A2))(P(A3)
    =0.7 + 0.3×0.5 + 0.3×0.5×0.3
    =0.895$



    Now,
    P(not being caught) = 1- P(being caught)
    = 1- 0.895
    =0.105






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Let $A_i$ be an event $i$-th police stop him. We are interested in $$P(A_1'cap A_2'cap A_3')= P(A_1')P (A_2')P(A_3')= 0.3 cdot 0.5 cdot 0.7 = 0.105$$



      (since $A_1, A_2, A_3$ are independant so are $A_1', A_2', A_3'$ ) so your answer is correct.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Let $A_i$ be an event $i$-th police stop him. We are interested in $$P(A_1'cap A_2'cap A_3')= P(A_1')P (A_2')P(A_3')= 0.3 cdot 0.5 cdot 0.7 = 0.105$$



        (since $A_1, A_2, A_3$ are independant so are $A_1', A_2', A_3'$ ) so your answer is correct.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Let $A_i$ be an event $i$-th police stop him. We are interested in $$P(A_1'cap A_2'cap A_3')= P(A_1')P (A_2')P(A_3')= 0.3 cdot 0.5 cdot 0.7 = 0.105$$



          (since $A_1, A_2, A_3$ are independant so are $A_1', A_2', A_3'$ ) so your answer is correct.






          share|cite|improve this answer









          $endgroup$



          Let $A_i$ be an event $i$-th police stop him. We are interested in $$P(A_1'cap A_2'cap A_3')= P(A_1')P (A_2')P(A_3')= 0.3 cdot 0.5 cdot 0.7 = 0.105$$



          (since $A_1, A_2, A_3$ are independant so are $A_1', A_2', A_3'$ ) so your answer is correct.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 '18 at 16:44









          Maria MazurMaria Mazur

          47.9k1260120




          47.9k1260120























              4












              $begingroup$

              You can think this in two ways.
              first way
              Not being caught on first check post and not being caught on second check post and not being caught on third post.
              i.e.
              $(1-P(A1))×(1-P(A2))×(1-P(A3) =
              0.3×0.5×0.7 = 0.105$



              second way
              Find P(being caught)



              Caught on first check post
              Or
              Not being caught on first post and caught on second post
              Or
              Not being caught on first post and Not being caught on second post and caught on third post.



              $P(A1) + (1-P(A1))(P(A2)) + (1- P(A1))(1- P(A2))(P(A3)
              =0.7 + 0.3×0.5 + 0.3×0.5×0.3
              =0.895$



              Now,
              P(not being caught) = 1- P(being caught)
              = 1- 0.895
              =0.105






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                You can think this in two ways.
                first way
                Not being caught on first check post and not being caught on second check post and not being caught on third post.
                i.e.
                $(1-P(A1))×(1-P(A2))×(1-P(A3) =
                0.3×0.5×0.7 = 0.105$



                second way
                Find P(being caught)



                Caught on first check post
                Or
                Not being caught on first post and caught on second post
                Or
                Not being caught on first post and Not being caught on second post and caught on third post.



                $P(A1) + (1-P(A1))(P(A2)) + (1- P(A1))(1- P(A2))(P(A3)
                =0.7 + 0.3×0.5 + 0.3×0.5×0.3
                =0.895$



                Now,
                P(not being caught) = 1- P(being caught)
                = 1- 0.895
                =0.105






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  You can think this in two ways.
                  first way
                  Not being caught on first check post and not being caught on second check post and not being caught on third post.
                  i.e.
                  $(1-P(A1))×(1-P(A2))×(1-P(A3) =
                  0.3×0.5×0.7 = 0.105$



                  second way
                  Find P(being caught)



                  Caught on first check post
                  Or
                  Not being caught on first post and caught on second post
                  Or
                  Not being caught on first post and Not being caught on second post and caught on third post.



                  $P(A1) + (1-P(A1))(P(A2)) + (1- P(A1))(1- P(A2))(P(A3)
                  =0.7 + 0.3×0.5 + 0.3×0.5×0.3
                  =0.895$



                  Now,
                  P(not being caught) = 1- P(being caught)
                  = 1- 0.895
                  =0.105






                  share|cite|improve this answer











                  $endgroup$



                  You can think this in two ways.
                  first way
                  Not being caught on first check post and not being caught on second check post and not being caught on third post.
                  i.e.
                  $(1-P(A1))×(1-P(A2))×(1-P(A3) =
                  0.3×0.5×0.7 = 0.105$



                  second way
                  Find P(being caught)



                  Caught on first check post
                  Or
                  Not being caught on first post and caught on second post
                  Or
                  Not being caught on first post and Not being caught on second post and caught on third post.



                  $P(A1) + (1-P(A1))(P(A2)) + (1- P(A1))(1- P(A2))(P(A3)
                  =0.7 + 0.3×0.5 + 0.3×0.5×0.3
                  =0.895$



                  Now,
                  P(not being caught) = 1- P(being caught)
                  = 1- 0.895
                  =0.105







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 21 '18 at 14:53

























                  answered Nov 25 '18 at 17:28









                  kapil pundirkapil pundir

                  574




                  574






























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