A coin is thrown until the tail is above. Which is the probability of making at most 3 throws.












2












$begingroup$


A coin is thrown until the tail is above. Which is the probability of making at most 3 throws.



The answer is :$frac{7}{8}$.



I don't understand why. What are the possible cases and what are the favorable ones?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    It may be useful to know that this is a geometrically distributed random variable.
    $endgroup$
    – Peter Foreman
    4 hours ago
















2












$begingroup$


A coin is thrown until the tail is above. Which is the probability of making at most 3 throws.



The answer is :$frac{7}{8}$.



I don't understand why. What are the possible cases and what are the favorable ones?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    It may be useful to know that this is a geometrically distributed random variable.
    $endgroup$
    – Peter Foreman
    4 hours ago














2












2








2


0



$begingroup$


A coin is thrown until the tail is above. Which is the probability of making at most 3 throws.



The answer is :$frac{7}{8}$.



I don't understand why. What are the possible cases and what are the favorable ones?






probability







share|cite|improve this question











$endgroup$




A coin is thrown until the tail is above. Which is the probability of making at most 3 throws.



The answer is :$frac{7}{8}$.



I don't understand why. What are the possible cases and what are the favorable ones?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 22 mins ago









YuiTo Cheng

2,94741139




2,94741139










asked 4 hours ago









Ica SanduIca Sandu

1499




1499












  • $begingroup$
    It may be useful to know that this is a geometrically distributed random variable.
    $endgroup$
    – Peter Foreman
    4 hours ago


















  • $begingroup$
    It may be useful to know that this is a geometrically distributed random variable.
    $endgroup$
    – Peter Foreman
    4 hours ago
















$begingroup$
It may be useful to know that this is a geometrically distributed random variable.
$endgroup$
– Peter Foreman
4 hours ago




$begingroup$
It may be useful to know that this is a geometrically distributed random variable.
$endgroup$
– Peter Foreman
4 hours ago










4 Answers
4






active

oldest

votes


















4












$begingroup$

There are three possibilities: T, HT, HHT with probabilities $1/2$, $1/2cdot1/2$, and $1/2cdot1/2cdot1/2$, resp. They sum up to $7/8$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    You will need more than three throws only when head comes out thrice.



    Hence



    $$p=1-frac1{2^3}.$$



    More generally,



    $$p=1-frac1{2^n}$$ for the probability of making at most $n$ throws.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      Let $T$ be the event that a tail is obtained within 3 flips. We want to find $pr(T)$. It is a bit difficult to compute this because we have to consider obtaining tails within 1 flip, 2 flips, AND 3 flips, to which there is overlap. It is a bit easier to compute $pr(T')$ or the probability $T$ does not happen.



      For event $T'$ to occur, we would need to not obtain tails within 3 flips, i.e. get heads three times in a row. Since the probability of throwing heads is $frac{1}{2}$, getting three heads in a row is $frac{1}{2^3}=frac{1}{8}$. Hence, $pr(T')=frac{1}{8}$. Since any given event either happens or doesn't, we have in general for event $A$, $$pr(A')=1-pr(A).$$ Therefore,
      $$frac{1}{8}=pr(T')=1-pr(T)implies pr(T)=1-frac{1}{8}=frac{7}{8}$$.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        A really simple way to see this is just to list out all the possible outcomes of flipping a coin three times. There are only $8$ of them:
        $$ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$$



        Because the coin flips are independent of each other, each of these combinations are equally likely. So the probability of each is $frac 18$. As you can see, $7$ of them include at least one tail, so the probability of getting a tails in three coin tosses is $frac 78$.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          ok but I flipping until the tail is above not 3 times. Then one possible case is, for example HHHHHT or T or HT...
          $endgroup$
          – Ica Sandu
          41 mins ago












        • $begingroup$
          That doesn't matter. What you do after finding your tail isn't going to change the probability that you found it in the first place. So, even though there is no particular reason for you to keep flipping after you get a tail, pretend that you do so anyway. You get this scenario, and find that your probability was $7/8$. That holds even though some of the sequences contain unnecessary flips.
          $endgroup$
          – Paul Sinclair
          36 mins ago










        • $begingroup$
          As for 3 flips, that is because your question is about 3 flips. It also doesn't matter to that question how many more times you flip the coin after those three flips are over. Their outcome is already decided.
          $endgroup$
          – Paul Sinclair
          34 mins ago










        • $begingroup$
          I will think a time. I must undestand ,and i will accept your answer.
          $endgroup$
          – Ica Sandu
          22 mins ago












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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        There are three possibilities: T, HT, HHT with probabilities $1/2$, $1/2cdot1/2$, and $1/2cdot1/2cdot1/2$, resp. They sum up to $7/8$.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          There are three possibilities: T, HT, HHT with probabilities $1/2$, $1/2cdot1/2$, and $1/2cdot1/2cdot1/2$, resp. They sum up to $7/8$.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            There are three possibilities: T, HT, HHT with probabilities $1/2$, $1/2cdot1/2$, and $1/2cdot1/2cdot1/2$, resp. They sum up to $7/8$.






            share|cite|improve this answer









            $endgroup$



            There are three possibilities: T, HT, HHT with probabilities $1/2$, $1/2cdot1/2$, and $1/2cdot1/2cdot1/2$, resp. They sum up to $7/8$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            Michael HoppeMichael Hoppe

            11.3k31837




            11.3k31837























                4












                $begingroup$

                You will need more than three throws only when head comes out thrice.



                Hence



                $$p=1-frac1{2^3}.$$



                More generally,



                $$p=1-frac1{2^n}$$ for the probability of making at most $n$ throws.






                share|cite|improve this answer











                $endgroup$


















                  4












                  $begingroup$

                  You will need more than three throws only when head comes out thrice.



                  Hence



                  $$p=1-frac1{2^3}.$$



                  More generally,



                  $$p=1-frac1{2^n}$$ for the probability of making at most $n$ throws.






                  share|cite|improve this answer











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    You will need more than three throws only when head comes out thrice.



                    Hence



                    $$p=1-frac1{2^3}.$$



                    More generally,



                    $$p=1-frac1{2^n}$$ for the probability of making at most $n$ throws.






                    share|cite|improve this answer











                    $endgroup$



                    You will need more than three throws only when head comes out thrice.



                    Hence



                    $$p=1-frac1{2^3}.$$



                    More generally,



                    $$p=1-frac1{2^n}$$ for the probability of making at most $n$ throws.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 4 hours ago

























                    answered 4 hours ago









                    Yves DaoustYves Daoust

                    134k676232




                    134k676232























                        1












                        $begingroup$

                        Let $T$ be the event that a tail is obtained within 3 flips. We want to find $pr(T)$. It is a bit difficult to compute this because we have to consider obtaining tails within 1 flip, 2 flips, AND 3 flips, to which there is overlap. It is a bit easier to compute $pr(T')$ or the probability $T$ does not happen.



                        For event $T'$ to occur, we would need to not obtain tails within 3 flips, i.e. get heads three times in a row. Since the probability of throwing heads is $frac{1}{2}$, getting three heads in a row is $frac{1}{2^3}=frac{1}{8}$. Hence, $pr(T')=frac{1}{8}$. Since any given event either happens or doesn't, we have in general for event $A$, $$pr(A')=1-pr(A).$$ Therefore,
                        $$frac{1}{8}=pr(T')=1-pr(T)implies pr(T)=1-frac{1}{8}=frac{7}{8}$$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Let $T$ be the event that a tail is obtained within 3 flips. We want to find $pr(T)$. It is a bit difficult to compute this because we have to consider obtaining tails within 1 flip, 2 flips, AND 3 flips, to which there is overlap. It is a bit easier to compute $pr(T')$ or the probability $T$ does not happen.



                          For event $T'$ to occur, we would need to not obtain tails within 3 flips, i.e. get heads three times in a row. Since the probability of throwing heads is $frac{1}{2}$, getting three heads in a row is $frac{1}{2^3}=frac{1}{8}$. Hence, $pr(T')=frac{1}{8}$. Since any given event either happens or doesn't, we have in general for event $A$, $$pr(A')=1-pr(A).$$ Therefore,
                          $$frac{1}{8}=pr(T')=1-pr(T)implies pr(T)=1-frac{1}{8}=frac{7}{8}$$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Let $T$ be the event that a tail is obtained within 3 flips. We want to find $pr(T)$. It is a bit difficult to compute this because we have to consider obtaining tails within 1 flip, 2 flips, AND 3 flips, to which there is overlap. It is a bit easier to compute $pr(T')$ or the probability $T$ does not happen.



                            For event $T'$ to occur, we would need to not obtain tails within 3 flips, i.e. get heads three times in a row. Since the probability of throwing heads is $frac{1}{2}$, getting three heads in a row is $frac{1}{2^3}=frac{1}{8}$. Hence, $pr(T')=frac{1}{8}$. Since any given event either happens or doesn't, we have in general for event $A$, $$pr(A')=1-pr(A).$$ Therefore,
                            $$frac{1}{8}=pr(T')=1-pr(T)implies pr(T)=1-frac{1}{8}=frac{7}{8}$$.






                            share|cite|improve this answer









                            $endgroup$



                            Let $T$ be the event that a tail is obtained within 3 flips. We want to find $pr(T)$. It is a bit difficult to compute this because we have to consider obtaining tails within 1 flip, 2 flips, AND 3 flips, to which there is overlap. It is a bit easier to compute $pr(T')$ or the probability $T$ does not happen.



                            For event $T'$ to occur, we would need to not obtain tails within 3 flips, i.e. get heads three times in a row. Since the probability of throwing heads is $frac{1}{2}$, getting three heads in a row is $frac{1}{2^3}=frac{1}{8}$. Hence, $pr(T')=frac{1}{8}$. Since any given event either happens or doesn't, we have in general for event $A$, $$pr(A')=1-pr(A).$$ Therefore,
                            $$frac{1}{8}=pr(T')=1-pr(T)implies pr(T)=1-frac{1}{8}=frac{7}{8}$$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 4 hours ago









                            Jane CooperJane Cooper

                            536




                            536























                                1












                                $begingroup$

                                A really simple way to see this is just to list out all the possible outcomes of flipping a coin three times. There are only $8$ of them:
                                $$ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$$



                                Because the coin flips are independent of each other, each of these combinations are equally likely. So the probability of each is $frac 18$. As you can see, $7$ of them include at least one tail, so the probability of getting a tails in three coin tosses is $frac 78$.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  ok but I flipping until the tail is above not 3 times. Then one possible case is, for example HHHHHT or T or HT...
                                  $endgroup$
                                  – Ica Sandu
                                  41 mins ago












                                • $begingroup$
                                  That doesn't matter. What you do after finding your tail isn't going to change the probability that you found it in the first place. So, even though there is no particular reason for you to keep flipping after you get a tail, pretend that you do so anyway. You get this scenario, and find that your probability was $7/8$. That holds even though some of the sequences contain unnecessary flips.
                                  $endgroup$
                                  – Paul Sinclair
                                  36 mins ago










                                • $begingroup$
                                  As for 3 flips, that is because your question is about 3 flips. It also doesn't matter to that question how many more times you flip the coin after those three flips are over. Their outcome is already decided.
                                  $endgroup$
                                  – Paul Sinclair
                                  34 mins ago










                                • $begingroup$
                                  I will think a time. I must undestand ,and i will accept your answer.
                                  $endgroup$
                                  – Ica Sandu
                                  22 mins ago
















                                1












                                $begingroup$

                                A really simple way to see this is just to list out all the possible outcomes of flipping a coin three times. There are only $8$ of them:
                                $$ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$$



                                Because the coin flips are independent of each other, each of these combinations are equally likely. So the probability of each is $frac 18$. As you can see, $7$ of them include at least one tail, so the probability of getting a tails in three coin tosses is $frac 78$.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  ok but I flipping until the tail is above not 3 times. Then one possible case is, for example HHHHHT or T or HT...
                                  $endgroup$
                                  – Ica Sandu
                                  41 mins ago












                                • $begingroup$
                                  That doesn't matter. What you do after finding your tail isn't going to change the probability that you found it in the first place. So, even though there is no particular reason for you to keep flipping after you get a tail, pretend that you do so anyway. You get this scenario, and find that your probability was $7/8$. That holds even though some of the sequences contain unnecessary flips.
                                  $endgroup$
                                  – Paul Sinclair
                                  36 mins ago










                                • $begingroup$
                                  As for 3 flips, that is because your question is about 3 flips. It also doesn't matter to that question how many more times you flip the coin after those three flips are over. Their outcome is already decided.
                                  $endgroup$
                                  – Paul Sinclair
                                  34 mins ago










                                • $begingroup$
                                  I will think a time. I must undestand ,and i will accept your answer.
                                  $endgroup$
                                  – Ica Sandu
                                  22 mins ago














                                1












                                1








                                1





                                $begingroup$

                                A really simple way to see this is just to list out all the possible outcomes of flipping a coin three times. There are only $8$ of them:
                                $$ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$$



                                Because the coin flips are independent of each other, each of these combinations are equally likely. So the probability of each is $frac 18$. As you can see, $7$ of them include at least one tail, so the probability of getting a tails in three coin tosses is $frac 78$.






                                share|cite|improve this answer









                                $endgroup$



                                A really simple way to see this is just to list out all the possible outcomes of flipping a coin three times. There are only $8$ of them:
                                $$ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$$



                                Because the coin flips are independent of each other, each of these combinations are equally likely. So the probability of each is $frac 18$. As you can see, $7$ of them include at least one tail, so the probability of getting a tails in three coin tosses is $frac 78$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 44 mins ago









                                Paul SinclairPaul Sinclair

                                21.1k21543




                                21.1k21543












                                • $begingroup$
                                  ok but I flipping until the tail is above not 3 times. Then one possible case is, for example HHHHHT or T or HT...
                                  $endgroup$
                                  – Ica Sandu
                                  41 mins ago












                                • $begingroup$
                                  That doesn't matter. What you do after finding your tail isn't going to change the probability that you found it in the first place. So, even though there is no particular reason for you to keep flipping after you get a tail, pretend that you do so anyway. You get this scenario, and find that your probability was $7/8$. That holds even though some of the sequences contain unnecessary flips.
                                  $endgroup$
                                  – Paul Sinclair
                                  36 mins ago










                                • $begingroup$
                                  As for 3 flips, that is because your question is about 3 flips. It also doesn't matter to that question how many more times you flip the coin after those three flips are over. Their outcome is already decided.
                                  $endgroup$
                                  – Paul Sinclair
                                  34 mins ago










                                • $begingroup$
                                  I will think a time. I must undestand ,and i will accept your answer.
                                  $endgroup$
                                  – Ica Sandu
                                  22 mins ago


















                                • $begingroup$
                                  ok but I flipping until the tail is above not 3 times. Then one possible case is, for example HHHHHT or T or HT...
                                  $endgroup$
                                  – Ica Sandu
                                  41 mins ago












                                • $begingroup$
                                  That doesn't matter. What you do after finding your tail isn't going to change the probability that you found it in the first place. So, even though there is no particular reason for you to keep flipping after you get a tail, pretend that you do so anyway. You get this scenario, and find that your probability was $7/8$. That holds even though some of the sequences contain unnecessary flips.
                                  $endgroup$
                                  – Paul Sinclair
                                  36 mins ago










                                • $begingroup$
                                  As for 3 flips, that is because your question is about 3 flips. It also doesn't matter to that question how many more times you flip the coin after those three flips are over. Their outcome is already decided.
                                  $endgroup$
                                  – Paul Sinclair
                                  34 mins ago










                                • $begingroup$
                                  I will think a time. I must undestand ,and i will accept your answer.
                                  $endgroup$
                                  – Ica Sandu
                                  22 mins ago
















                                $begingroup$
                                ok but I flipping until the tail is above not 3 times. Then one possible case is, for example HHHHHT or T or HT...
                                $endgroup$
                                – Ica Sandu
                                41 mins ago






                                $begingroup$
                                ok but I flipping until the tail is above not 3 times. Then one possible case is, for example HHHHHT or T or HT...
                                $endgroup$
                                – Ica Sandu
                                41 mins ago














                                $begingroup$
                                That doesn't matter. What you do after finding your tail isn't going to change the probability that you found it in the first place. So, even though there is no particular reason for you to keep flipping after you get a tail, pretend that you do so anyway. You get this scenario, and find that your probability was $7/8$. That holds even though some of the sequences contain unnecessary flips.
                                $endgroup$
                                – Paul Sinclair
                                36 mins ago




                                $begingroup$
                                That doesn't matter. What you do after finding your tail isn't going to change the probability that you found it in the first place. So, even though there is no particular reason for you to keep flipping after you get a tail, pretend that you do so anyway. You get this scenario, and find that your probability was $7/8$. That holds even though some of the sequences contain unnecessary flips.
                                $endgroup$
                                – Paul Sinclair
                                36 mins ago












                                $begingroup$
                                As for 3 flips, that is because your question is about 3 flips. It also doesn't matter to that question how many more times you flip the coin after those three flips are over. Their outcome is already decided.
                                $endgroup$
                                – Paul Sinclair
                                34 mins ago




                                $begingroup$
                                As for 3 flips, that is because your question is about 3 flips. It also doesn't matter to that question how many more times you flip the coin after those three flips are over. Their outcome is already decided.
                                $endgroup$
                                – Paul Sinclair
                                34 mins ago












                                $begingroup$
                                I will think a time. I must undestand ,and i will accept your answer.
                                $endgroup$
                                – Ica Sandu
                                22 mins ago




                                $begingroup$
                                I will think a time. I must undestand ,and i will accept your answer.
                                $endgroup$
                                – Ica Sandu
                                22 mins ago


















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                                Ida-Boy-Ed-Garten

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                                Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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                                Verónica Boquete

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                                Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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