probability of choosing a biased coin C which has probability 3/15 of getting heads assuming we got head on...












2














Full question: there are 3 biased coins A, B, and C each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability 1/4 for A, 1/4 for B, and 1/2 for C of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin C?



My approach: Since the coin picked was C and the result was head we merely multiply the probability of both those things happening concerning C:
$$1/15 * 1/2 = 1/30$$






probability discrete-mathematics conditional-probability







share|cite|improve this question



























    2














    Full question: there are 3 biased coins A, B, and C each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability 1/4 for A, 1/4 for B, and 1/2 for C of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin C?



    My approach: Since the coin picked was C and the result was head we merely multiply the probability of both those things happening concerning C:
    $$1/15 * 1/2 = 1/30$$






    probability discrete-mathematics conditional-probability







    share|cite|improve this question

























      2












      2








      2







      Full question: there are 3 biased coins A, B, and C each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability 1/4 for A, 1/4 for B, and 1/2 for C of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin C?



      My approach: Since the coin picked was C and the result was head we merely multiply the probability of both those things happening concerning C:
      $$1/15 * 1/2 = 1/30$$






      probability discrete-mathematics conditional-probability







      share|cite|improve this question













      Full question: there are 3 biased coins A, B, and C each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability 1/4 for A, 1/4 for B, and 1/2 for C of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin C?



      My approach: Since the coin picked was C and the result was head we merely multiply the probability of both those things happening concerning C:
      $$1/15 * 1/2 = 1/30$$






      probability discrete-mathematics conditional-probability




      probability discrete-mathematics conditional-probability






      share|cite|improve this question













      share|cite|improve this question











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      asked 2 hours ago









      hussain sagar

      776




      776






















          2 Answers
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          An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:




          1. Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.

          2. Draw $B$: $1/4 times 3/15 = 1/20$.

          3. Draw $C$: $1/2 times 1/15 = 1/30$.


          Thus, the chance that $C$ was drawn is
          $$
          frac{1/30}{1/12 + 1/20 + 1/30}
          = frac{1}{5/2+3/2 + 1}
          = frac15.
          $$






          share|cite|improve this answer





















          • Thank you, this makes a lot of sense.
            – hussain sagar
            1 hour ago










          • @hussainsagar you are welcome
            – gt6989b
            1 hour ago



















          1














          What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.



          Guide:
          Use Bayes rule, that is



          $$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:




            1. Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.

            2. Draw $B$: $1/4 times 3/15 = 1/20$.

            3. Draw $C$: $1/2 times 1/15 = 1/30$.


            Thus, the chance that $C$ was drawn is
            $$
            frac{1/30}{1/12 + 1/20 + 1/30}
            = frac{1}{5/2+3/2 + 1}
            = frac15.
            $$






            share|cite|improve this answer





















            • Thank you, this makes a lot of sense.
              – hussain sagar
              1 hour ago










            • @hussainsagar you are welcome
              – gt6989b
              1 hour ago
















            3














            An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:




            1. Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.

            2. Draw $B$: $1/4 times 3/15 = 1/20$.

            3. Draw $C$: $1/2 times 1/15 = 1/30$.


            Thus, the chance that $C$ was drawn is
            $$
            frac{1/30}{1/12 + 1/20 + 1/30}
            = frac{1}{5/2+3/2 + 1}
            = frac15.
            $$






            share|cite|improve this answer





















            • Thank you, this makes a lot of sense.
              – hussain sagar
              1 hour ago










            • @hussainsagar you are welcome
              – gt6989b
              1 hour ago














            3












            3








            3






            An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:




            1. Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.

            2. Draw $B$: $1/4 times 3/15 = 1/20$.

            3. Draw $C$: $1/2 times 1/15 = 1/30$.


            Thus, the chance that $C$ was drawn is
            $$
            frac{1/30}{1/12 + 1/20 + 1/30}
            = frac{1}{5/2+3/2 + 1}
            = frac15.
            $$






            share|cite|improve this answer












            An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:




            1. Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.

            2. Draw $B$: $1/4 times 3/15 = 1/20$.

            3. Draw $C$: $1/2 times 1/15 = 1/30$.


            Thus, the chance that $C$ was drawn is
            $$
            frac{1/30}{1/12 + 1/20 + 1/30}
            = frac{1}{5/2+3/2 + 1}
            = frac15.
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            gt6989b

            32.9k22452




            32.9k22452












            • Thank you, this makes a lot of sense.
              – hussain sagar
              1 hour ago










            • @hussainsagar you are welcome
              – gt6989b
              1 hour ago


















            • Thank you, this makes a lot of sense.
              – hussain sagar
              1 hour ago










            • @hussainsagar you are welcome
              – gt6989b
              1 hour ago
















            Thank you, this makes a lot of sense.
            – hussain sagar
            1 hour ago




            Thank you, this makes a lot of sense.
            – hussain sagar
            1 hour ago












            @hussainsagar you are welcome
            – gt6989b
            1 hour ago




            @hussainsagar you are welcome
            – gt6989b
            1 hour ago











            1














            What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.



            Guide:
            Use Bayes rule, that is



            $$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$






            share|cite|improve this answer


























              1














              What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.



              Guide:
              Use Bayes rule, that is



              $$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$






              share|cite|improve this answer
























                1












                1








                1






                What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.



                Guide:
                Use Bayes rule, that is



                $$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$






                share|cite|improve this answer












                What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.



                Guide:
                Use Bayes rule, that is



                $$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Siong Thye Goh

                98.4k1463116




                98.4k1463116






























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