Probability that a 5 occurs first

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Suppose we roll pair of dice until a sum of either 5 or 7 appears.
What is the probability that a sum of 5 occurs first?




Try:



Let $A$ be the event that a sum of $5$ occurs on the ith roll and $B$ that the sum of 7 occurs on the ith roll. We are interested on the event $A | A^c cup B^c $. We have



$$ P(A | A^c cup B^c) = dfrac P(A cap (A^c cup B^c))P(A^c cup B^c) = fracP(A cap B^c)1 - P(A cap B) = fracP(A cap B^c)1-0 = P(A cap B^c)$$



Is this approach correct so far?










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  • Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
    – Rob
    47 mins ago










  • Possible duplicate of Probability sum of 5 before sum of 7
    – Key Flex
    3 mins ago














up vote
1
down vote

favorite













Suppose we roll pair of dice until a sum of either 5 or 7 appears.
What is the probability that a sum of 5 occurs first?




Try:



Let $A$ be the event that a sum of $5$ occurs on the ith roll and $B$ that the sum of 7 occurs on the ith roll. We are interested on the event $A | A^c cup B^c $. We have



$$ P(A | A^c cup B^c) = dfrac P(A cap (A^c cup B^c))P(A^c cup B^c) = fracP(A cap B^c)1 - P(A cap B) = fracP(A cap B^c)1-0 = P(A cap B^c)$$



Is this approach correct so far?










share|cite|improve this question





















  • Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
    – Rob
    47 mins ago










  • Possible duplicate of Probability sum of 5 before sum of 7
    – Key Flex
    3 mins ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Suppose we roll pair of dice until a sum of either 5 or 7 appears.
What is the probability that a sum of 5 occurs first?




Try:



Let $A$ be the event that a sum of $5$ occurs on the ith roll and $B$ that the sum of 7 occurs on the ith roll. We are interested on the event $A | A^c cup B^c $. We have



$$ P(A | A^c cup B^c) = dfrac P(A cap (A^c cup B^c))P(A^c cup B^c) = fracP(A cap B^c)1 - P(A cap B) = fracP(A cap B^c)1-0 = P(A cap B^c)$$



Is this approach correct so far?










share|cite|improve this question














Suppose we roll pair of dice until a sum of either 5 or 7 appears.
What is the probability that a sum of 5 occurs first?




Try:



Let $A$ be the event that a sum of $5$ occurs on the ith roll and $B$ that the sum of 7 occurs on the ith roll. We are interested on the event $A | A^c cup B^c $. We have



$$ P(A | A^c cup B^c) = dfrac P(A cap (A^c cup B^c))P(A^c cup B^c) = fracP(A cap B^c)1 - P(A cap B) = fracP(A cap B^c)1-0 = P(A cap B^c)$$



Is this approach correct so far?







probability






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asked 53 mins ago









Jimmy Sabater

1,468115




1,468115











  • Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
    – Rob
    47 mins ago










  • Possible duplicate of Probability sum of 5 before sum of 7
    – Key Flex
    3 mins ago
















  • Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
    – Rob
    47 mins ago










  • Possible duplicate of Probability sum of 5 before sum of 7
    – Key Flex
    3 mins ago















Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
– Rob
47 mins ago




Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
– Rob
47 mins ago












Possible duplicate of Probability sum of 5 before sum of 7
– Key Flex
3 mins ago




Possible duplicate of Probability sum of 5 before sum of 7
– Key Flex
3 mins ago










3 Answers
3






active

oldest

votes

















up vote
1
down vote



accepted










Guide:



  • Compute the probability of $7$ appears, call it $q$.

  • Compute the probability of $5$ appears, call it $p$.
    We are interested in

$$sum_i=1^infty (1-p-q)^i-1p$$



Remark about your attempt:



Are you intending to use some indices to denote $i$-th throw?






share|cite|improve this answer



























    up vote
    2
    down vote













    A cleaner approach would be to consider the experiment until either a sum of $5$ or $7$ is rolled at some step, say $n$. (Note that $n<infty$ almost surely.) Then, the event $A$ is that $5$ was rolled and $B$ is that $7$ was rolled, and you need
    $$
    mathbbP[A|A cup B]
    $$

    where $A$ and $B$ are disjoint...






    share|cite|improve this answer



























      up vote
      2
      down vote













      All events are ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space.



      $$S=(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$$



      Now it's just the law of classical probability.






      share|cite|improve this answer




















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        1
        down vote



        accepted










        Guide:



        • Compute the probability of $7$ appears, call it $q$.

        • Compute the probability of $5$ appears, call it $p$.
          We are interested in

        $$sum_i=1^infty (1-p-q)^i-1p$$



        Remark about your attempt:



        Are you intending to use some indices to denote $i$-th throw?






        share|cite|improve this answer
























          up vote
          1
          down vote



          accepted










          Guide:



          • Compute the probability of $7$ appears, call it $q$.

          • Compute the probability of $5$ appears, call it $p$.
            We are interested in

          $$sum_i=1^infty (1-p-q)^i-1p$$



          Remark about your attempt:



          Are you intending to use some indices to denote $i$-th throw?






          share|cite|improve this answer






















            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            Guide:



            • Compute the probability of $7$ appears, call it $q$.

            • Compute the probability of $5$ appears, call it $p$.
              We are interested in

            $$sum_i=1^infty (1-p-q)^i-1p$$



            Remark about your attempt:



            Are you intending to use some indices to denote $i$-th throw?






            share|cite|improve this answer












            Guide:



            • Compute the probability of $7$ appears, call it $q$.

            • Compute the probability of $5$ appears, call it $p$.
              We are interested in

            $$sum_i=1^infty (1-p-q)^i-1p$$



            Remark about your attempt:



            Are you intending to use some indices to denote $i$-th throw?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 47 mins ago









            Siong Thye Goh

            85.3k1457107




            85.3k1457107




















                up vote
                2
                down vote













                A cleaner approach would be to consider the experiment until either a sum of $5$ or $7$ is rolled at some step, say $n$. (Note that $n<infty$ almost surely.) Then, the event $A$ is that $5$ was rolled and $B$ is that $7$ was rolled, and you need
                $$
                mathbbP[A|A cup B]
                $$

                where $A$ and $B$ are disjoint...






                share|cite|improve this answer
























                  up vote
                  2
                  down vote













                  A cleaner approach would be to consider the experiment until either a sum of $5$ or $7$ is rolled at some step, say $n$. (Note that $n<infty$ almost surely.) Then, the event $A$ is that $5$ was rolled and $B$ is that $7$ was rolled, and you need
                  $$
                  mathbbP[A|A cup B]
                  $$

                  where $A$ and $B$ are disjoint...






                  share|cite|improve this answer






















                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    A cleaner approach would be to consider the experiment until either a sum of $5$ or $7$ is rolled at some step, say $n$. (Note that $n<infty$ almost surely.) Then, the event $A$ is that $5$ was rolled and $B$ is that $7$ was rolled, and you need
                    $$
                    mathbbP[A|A cup B]
                    $$

                    where $A$ and $B$ are disjoint...






                    share|cite|improve this answer












                    A cleaner approach would be to consider the experiment until either a sum of $5$ or $7$ is rolled at some step, say $n$. (Note that $n<infty$ almost surely.) Then, the event $A$ is that $5$ was rolled and $B$ is that $7$ was rolled, and you need
                    $$
                    mathbbP[A|A cup B]
                    $$

                    where $A$ and $B$ are disjoint...







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 49 mins ago









                    gt6989b

                    31.2k22349




                    31.2k22349




















                        up vote
                        2
                        down vote













                        All events are ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space.



                        $$S=(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$$



                        Now it's just the law of classical probability.






                        share|cite|improve this answer
























                          up vote
                          2
                          down vote













                          All events are ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space.



                          $$S=(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$$



                          Now it's just the law of classical probability.






                          share|cite|improve this answer






















                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            All events are ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space.



                            $$S=(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$$



                            Now it's just the law of classical probability.






                            share|cite|improve this answer












                            All events are ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space.



                            $$S=(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$$



                            Now it's just the law of classical probability.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 42 mins ago









                            David Peterson

                            8,30121834




                            8,30121834



























                                 

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