Probability of sum 15 when roll 3rd dice, and first roll of 2 dice at least 10












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enter image description hereTwo dice have been thown, giving a total of at least 10.
What is the probability that the throw of a third die
will bring the total of the three numbers shown to 15 or higher?
I work out is 2/27. but physicsandmathstutor.com give a solution is 4/9.
i wander if you can help and explain in details. thanks






probability dice







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  • 1




    For example, how did you get $frac 2{27}$? Could you sketch your method?
    – lulu
    Nov 19 '18 at 0:12










  • The answer must be at least $frac26$ since the third die being $5$ or $6$ would be sufficient in every case
    – Henry
    Nov 19 '18 at 0:13
















0














enter image description hereTwo dice have been thown, giving a total of at least 10.
What is the probability that the throw of a third die
will bring the total of the three numbers shown to 15 or higher?
I work out is 2/27. but physicsandmathstutor.com give a solution is 4/9.
i wander if you can help and explain in details. thanks






probability dice







share|cite|improve this question




















  • 1




    For example, how did you get $frac 2{27}$? Could you sketch your method?
    – lulu
    Nov 19 '18 at 0:12










  • The answer must be at least $frac26$ since the third die being $5$ or $6$ would be sufficient in every case
    – Henry
    Nov 19 '18 at 0:13














0












0








0







enter image description hereTwo dice have been thown, giving a total of at least 10.
What is the probability that the throw of a third die
will bring the total of the three numbers shown to 15 or higher?
I work out is 2/27. but physicsandmathstutor.com give a solution is 4/9.
i wander if you can help and explain in details. thanks






probability dice







share|cite|improve this question















enter image description hereTwo dice have been thown, giving a total of at least 10.
What is the probability that the throw of a third die
will bring the total of the three numbers shown to 15 or higher?
I work out is 2/27. but physicsandmathstutor.com give a solution is 4/9.
i wander if you can help and explain in details. thanks






probability dice




probability dice






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edited Nov 19 '18 at 20:20

























asked Nov 19 '18 at 0:04









Jan Jin

11




11








  • 1




    For example, how did you get $frac 2{27}$? Could you sketch your method?
    – lulu
    Nov 19 '18 at 0:12










  • The answer must be at least $frac26$ since the third die being $5$ or $6$ would be sufficient in every case
    – Henry
    Nov 19 '18 at 0:13














  • 1




    For example, how did you get $frac 2{27}$? Could you sketch your method?
    – lulu
    Nov 19 '18 at 0:12










  • The answer must be at least $frac26$ since the third die being $5$ or $6$ would be sufficient in every case
    – Henry
    Nov 19 '18 at 0:13








1




1




For example, how did you get $frac 2{27}$? Could you sketch your method?
– lulu
Nov 19 '18 at 0:12




For example, how did you get $frac 2{27}$? Could you sketch your method?
– lulu
Nov 19 '18 at 0:12












The answer must be at least $frac26$ since the third die being $5$ or $6$ would be sufficient in every case
– Henry
Nov 19 '18 at 0:13




The answer must be at least $frac26$ since the third die being $5$ or $6$ would be sufficient in every case
– Henry
Nov 19 '18 at 0:13










3 Answers
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Hint: I suggest a Probability Tree with total values then if it is 15 or more as the second branch. Calculate the probabilities individually. This should help clear any doubt you have with your answer.






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    0














    if you list all possible events is 16, over sample space 216 events. 16/216 = 2/27




    
    D1   D2   D3
    
1   4 -> 6 => 5
    
2   4 -> 6 => 6
    
3   5 -> 5 => 5
    
4   5 -> 5 => 6
    
5   6 -> 4 => 5
    
6   6 -> 4 => 6
    
7   5 -> 6 => 4
    
8   5 -> 6 => 5
    
9   5 -> 6 => 6
    
10  6 -> 5 => 4
    
11  6 -> 5 => 5
    
12  6 -> 5 => 6
    
13  6 -> 6 => 3
    
14  6 -> 6 => 4
    
15  6 -> 6 => 5
    
16  6 -> 6 => 6


    for people think use conditional probability, I think the wrong part is use probabilty 1/6 for two dice sum 10, applies to each event.






    share|cite|improve this answer





























      0














      Let $A$ be the event "total of the three numbers shown to 15 or higher", $B_1$ be the event "total of first two is 10", $B_2$ be the event "total of first two is 11", $B_3$ be the event "total of first two is 12". We have
      begin{align}
      P(B_1) = 3/6^2\
      P(B_2) = 2/6^2\
      P(B_3) = 1/6^2
      end{align}
      If $B$ is the event "total of first two is at least 10", then $P(B_1|B)=1/2$, $P(B_2|B)=1/3$ and $P(B_3|B)=1/6$. Finally
      begin{align}
      P(A|B)&=P(A|B_1,B)P(B_1|B)+P(A|B_2,B)P(B_2|B)+P(A|B_3,B)P(B_3|B)\
      &=1/3times1/2+1/2times1/3+2/3times1/6=4/9
      end{align}






      share|cite|improve this answer























      • I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
        – Jan Jin
        Nov 19 '18 at 20:43










      • yup that was a typo
        – dynamic89
        Nov 19 '18 at 21:13











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      3 Answers
      3






      active

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      3 Answers
      3






      active

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      active

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      active

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      0














      Hint: I suggest a Probability Tree with total values then if it is 15 or more as the second branch. Calculate the probabilities individually. This should help clear any doubt you have with your answer.






      share|cite|improve this answer


























        0














        Hint: I suggest a Probability Tree with total values then if it is 15 or more as the second branch. Calculate the probabilities individually. This should help clear any doubt you have with your answer.






        share|cite|improve this answer
























          0












          0








          0






          Hint: I suggest a Probability Tree with total values then if it is 15 or more as the second branch. Calculate the probabilities individually. This should help clear any doubt you have with your answer.






          share|cite|improve this answer












          Hint: I suggest a Probability Tree with total values then if it is 15 or more as the second branch. Calculate the probabilities individually. This should help clear any doubt you have with your answer.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 '18 at 0:20









          anuj1610

          12




          12























              0














              if you list all possible events is 16, over sample space 216 events. 16/216 = 2/27




              
    D1   D2   D3
              
1   4 -> 6 => 5
              
2   4 -> 6 => 6
              
3   5 -> 5 => 5
              
4   5 -> 5 => 6
              
5   6 -> 4 => 5
              
6   6 -> 4 => 6
              
7   5 -> 6 => 4
              
8   5 -> 6 => 5
              
9   5 -> 6 => 6
              
10  6 -> 5 => 4
              
11  6 -> 5 => 5
              
12  6 -> 5 => 6
              
13  6 -> 6 => 3
              
14  6 -> 6 => 4
              
15  6 -> 6 => 5
              
16  6 -> 6 => 6


              for people think use conditional probability, I think the wrong part is use probabilty 1/6 for two dice sum 10, applies to each event.






              share|cite|improve this answer


























                0














                if you list all possible events is 16, over sample space 216 events. 16/216 = 2/27




                
    D1   D2   D3
                
1   4 -> 6 => 5
                
2   4 -> 6 => 6
                
3   5 -> 5 => 5
                
4   5 -> 5 => 6
                
5   6 -> 4 => 5
                
6   6 -> 4 => 6
                
7   5 -> 6 => 4
                
8   5 -> 6 => 5
                
9   5 -> 6 => 6
                
10  6 -> 5 => 4
                
11  6 -> 5 => 5
                
12  6 -> 5 => 6
                
13  6 -> 6 => 3
                
14  6 -> 6 => 4
                
15  6 -> 6 => 5
                
16  6 -> 6 => 6


                for people think use conditional probability, I think the wrong part is use probabilty 1/6 for two dice sum 10, applies to each event.






                share|cite|improve this answer
























                  0












                  0








                  0






                  if you list all possible events is 16, over sample space 216 events. 16/216 = 2/27




                  
    D1   D2   D3
                  
1   4 -> 6 => 5
                  
2   4 -> 6 => 6
                  
3   5 -> 5 => 5
                  
4   5 -> 5 => 6
                  
5   6 -> 4 => 5
                  
6   6 -> 4 => 6
                  
7   5 -> 6 => 4
                  
8   5 -> 6 => 5
                  
9   5 -> 6 => 6
                  
10  6 -> 5 => 4
                  
11  6 -> 5 => 5
                  
12  6 -> 5 => 6
                  
13  6 -> 6 => 3
                  
14  6 -> 6 => 4
                  
15  6 -> 6 => 5
                  
16  6 -> 6 => 6


                  for people think use conditional probability, I think the wrong part is use probabilty 1/6 for two dice sum 10, applies to each event.






                  share|cite|improve this answer












                  if you list all possible events is 16, over sample space 216 events. 16/216 = 2/27




                  
    D1   D2   D3
                  
1   4 -> 6 => 5
                  
2   4 -> 6 => 6
                  
3   5 -> 5 => 5
                  
4   5 -> 5 => 6
                  
5   6 -> 4 => 5
                  
6   6 -> 4 => 6
                  
7   5 -> 6 => 4
                  
8   5 -> 6 => 5
                  
9   5 -> 6 => 6
                  
10  6 -> 5 => 4
                  
11  6 -> 5 => 5
                  
12  6 -> 5 => 6
                  
13  6 -> 6 => 3
                  
14  6 -> 6 => 4
                  
15  6 -> 6 => 5
                  
16  6 -> 6 => 6


                  for people think use conditional probability, I think the wrong part is use probabilty 1/6 for two dice sum 10, applies to each event.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 19 '18 at 20:33









                  Jan Jin

                  11




                  11























                      0














                      Let $A$ be the event "total of the three numbers shown to 15 or higher", $B_1$ be the event "total of first two is 10", $B_2$ be the event "total of first two is 11", $B_3$ be the event "total of first two is 12". We have
                      begin{align}
                      P(B_1) = 3/6^2\
                      P(B_2) = 2/6^2\
                      P(B_3) = 1/6^2
                      end{align}
                      If $B$ is the event "total of first two is at least 10", then $P(B_1|B)=1/2$, $P(B_2|B)=1/3$ and $P(B_3|B)=1/6$. Finally
                      begin{align}
                      P(A|B)&=P(A|B_1,B)P(B_1|B)+P(A|B_2,B)P(B_2|B)+P(A|B_3,B)P(B_3|B)\
                      &=1/3times1/2+1/2times1/3+2/3times1/6=4/9
                      end{align}






                      share|cite|improve this answer























                      • I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
                        – Jan Jin
                        Nov 19 '18 at 20:43










                      • yup that was a typo
                        – dynamic89
                        Nov 19 '18 at 21:13
















                      0














                      Let $A$ be the event "total of the three numbers shown to 15 or higher", $B_1$ be the event "total of first two is 10", $B_2$ be the event "total of first two is 11", $B_3$ be the event "total of first two is 12". We have
                      begin{align}
                      P(B_1) = 3/6^2\
                      P(B_2) = 2/6^2\
                      P(B_3) = 1/6^2
                      end{align}
                      If $B$ is the event "total of first two is at least 10", then $P(B_1|B)=1/2$, $P(B_2|B)=1/3$ and $P(B_3|B)=1/6$. Finally
                      begin{align}
                      P(A|B)&=P(A|B_1,B)P(B_1|B)+P(A|B_2,B)P(B_2|B)+P(A|B_3,B)P(B_3|B)\
                      &=1/3times1/2+1/2times1/3+2/3times1/6=4/9
                      end{align}






                      share|cite|improve this answer























                      • I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
                        – Jan Jin
                        Nov 19 '18 at 20:43










                      • yup that was a typo
                        – dynamic89
                        Nov 19 '18 at 21:13














                      0












                      0








                      0






                      Let $A$ be the event "total of the three numbers shown to 15 or higher", $B_1$ be the event "total of first two is 10", $B_2$ be the event "total of first two is 11", $B_3$ be the event "total of first two is 12". We have
                      begin{align}
                      P(B_1) = 3/6^2\
                      P(B_2) = 2/6^2\
                      P(B_3) = 1/6^2
                      end{align}
                      If $B$ is the event "total of first two is at least 10", then $P(B_1|B)=1/2$, $P(B_2|B)=1/3$ and $P(B_3|B)=1/6$. Finally
                      begin{align}
                      P(A|B)&=P(A|B_1,B)P(B_1|B)+P(A|B_2,B)P(B_2|B)+P(A|B_3,B)P(B_3|B)\
                      &=1/3times1/2+1/2times1/3+2/3times1/6=4/9
                      end{align}






                      share|cite|improve this answer














                      Let $A$ be the event "total of the three numbers shown to 15 or higher", $B_1$ be the event "total of first two is 10", $B_2$ be the event "total of first two is 11", $B_3$ be the event "total of first two is 12". We have
                      begin{align}
                      P(B_1) = 3/6^2\
                      P(B_2) = 2/6^2\
                      P(B_3) = 1/6^2
                      end{align}
                      If $B$ is the event "total of first two is at least 10", then $P(B_1|B)=1/2$, $P(B_2|B)=1/3$ and $P(B_3|B)=1/6$. Finally
                      begin{align}
                      P(A|B)&=P(A|B_1,B)P(B_1|B)+P(A|B_2,B)P(B_2|B)+P(A|B_3,B)P(B_3|B)\
                      &=1/3times1/2+1/2times1/3+2/3times1/6=4/9
                      end{align}







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 19 '18 at 21:13

























                      answered Nov 19 '18 at 1:58









                      dynamic89

                      38418




                      38418












                      • I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
                        – Jan Jin
                        Nov 19 '18 at 20:43










                      • yup that was a typo
                        – dynamic89
                        Nov 19 '18 at 21:13


















                      • I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
                        – Jan Jin
                        Nov 19 '18 at 20:43










                      • yup that was a typo
                        – dynamic89
                        Nov 19 '18 at 21:13
















                      I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
                      – Jan Jin
                      Nov 19 '18 at 20:43




                      I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
                      – Jan Jin
                      Nov 19 '18 at 20:43












                      yup that was a typo
                      – dynamic89
                      Nov 19 '18 at 21:13




                      yup that was a typo
                      – dynamic89
                      Nov 19 '18 at 21:13


















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