Rolling three dice…am I doing this correctly?












5












$begingroup$


Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?



Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first



For 1) The first die, you have 6/6. The second die needs to be equal to the first, so you have probability of 1/6. Then the third die can't be equal to the first and second dice, so it's 5/6.
All together you get $1*frac{1}{6}*frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3*(1*frac{1}{6}*frac{5}{6})=frac{5}{12}$$



Did I do this correctly?
Thank you.






probability







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  • 1




    $begingroup$
    This is the answer I get and your reasoning seems sound.
    $endgroup$
    – user45150
    Feb 12 '13 at 6:41
















5












$begingroup$


Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?



Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first



For 1) The first die, you have 6/6. The second die needs to be equal to the first, so you have probability of 1/6. Then the third die can't be equal to the first and second dice, so it's 5/6.
All together you get $1*frac{1}{6}*frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3*(1*frac{1}{6}*frac{5}{6})=frac{5}{12}$$



Did I do this correctly?
Thank you.






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This is the answer I get and your reasoning seems sound.
    $endgroup$
    – user45150
    Feb 12 '13 at 6:41














5












5








5





$begingroup$


Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?



Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first



For 1) The first die, you have 6/6. The second die needs to be equal to the first, so you have probability of 1/6. Then the third die can't be equal to the first and second dice, so it's 5/6.
All together you get $1*frac{1}{6}*frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3*(1*frac{1}{6}*frac{5}{6})=frac{5}{12}$$



Did I do this correctly?
Thank you.






probability







share|cite|improve this question









$endgroup$




Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?



Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first



For 1) The first die, you have 6/6. The second die needs to be equal to the first, so you have probability of 1/6. Then the third die can't be equal to the first and second dice, so it's 5/6.
All together you get $1*frac{1}{6}*frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3*(1*frac{1}{6}*frac{5}{6})=frac{5}{12}$$



Did I do this correctly?
Thank you.






probability




probability






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share|cite|improve this question











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asked Feb 12 '13 at 6:40









AltiAlti

1,18342246




1,18342246








  • 1




    $begingroup$
    This is the answer I get and your reasoning seems sound.
    $endgroup$
    – user45150
    Feb 12 '13 at 6:41














  • 1




    $begingroup$
    This is the answer I get and your reasoning seems sound.
    $endgroup$
    – user45150
    Feb 12 '13 at 6:41








1




1




$begingroup$
This is the answer I get and your reasoning seems sound.
$endgroup$
– user45150
Feb 12 '13 at 6:41




$begingroup$
This is the answer I get and your reasoning seems sound.
$endgroup$
– user45150
Feb 12 '13 at 6:41










3 Answers
3






active

oldest

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2












$begingroup$

Yes your answer is correct. $${3choose 2}cdotfrac{1}{6}cdotfrac{5}{6} = frac{5}{12}$$



Good job!






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It's $1/5$ or $1/6$?
    $endgroup$
    – MonsieurGalois
    Jan 17 '17 at 20:30





















3












$begingroup$

As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3cdot6cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $frac{90}{216}=frac{5}{12}$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    One more way you can solve this is using complimentary counting. There are 6*$dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $frac{90}{216}=frac{5}{12}$






    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









      2












      $begingroup$

      Yes your answer is correct. $${3choose 2}cdotfrac{1}{6}cdotfrac{5}{6} = frac{5}{12}$$



      Good job!






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        It's $1/5$ or $1/6$?
        $endgroup$
        – MonsieurGalois
        Jan 17 '17 at 20:30


















      2












      $begingroup$

      Yes your answer is correct. $${3choose 2}cdotfrac{1}{6}cdotfrac{5}{6} = frac{5}{12}$$



      Good job!






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        It's $1/5$ or $1/6$?
        $endgroup$
        – MonsieurGalois
        Jan 17 '17 at 20:30
















      2












      2








      2





      $begingroup$

      Yes your answer is correct. $${3choose 2}cdotfrac{1}{6}cdotfrac{5}{6} = frac{5}{12}$$



      Good job!






      share|cite|improve this answer











      $endgroup$



      Yes your answer is correct. $${3choose 2}cdotfrac{1}{6}cdotfrac{5}{6} = frac{5}{12}$$



      Good job!







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Oct 10 '17 at 17:27

























      answered Feb 12 '13 at 6:44









      MITjanitorMITjanitor

      1,9631543




      1,9631543








      • 1




        $begingroup$
        It's $1/5$ or $1/6$?
        $endgroup$
        – MonsieurGalois
        Jan 17 '17 at 20:30
















      • 1




        $begingroup$
        It's $1/5$ or $1/6$?
        $endgroup$
        – MonsieurGalois
        Jan 17 '17 at 20:30










      1




      1




      $begingroup$
      It's $1/5$ or $1/6$?
      $endgroup$
      – MonsieurGalois
      Jan 17 '17 at 20:30






      $begingroup$
      It's $1/5$ or $1/6$?
      $endgroup$
      – MonsieurGalois
      Jan 17 '17 at 20:30













      3












      $begingroup$

      As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3cdot6cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $frac{90}{216}=frac{5}{12}$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3cdot6cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $frac{90}{216}=frac{5}{12}$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3cdot6cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $frac{90}{216}=frac{5}{12}$.






          share|cite|improve this answer









          $endgroup$



          As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3cdot6cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $frac{90}{216}=frac{5}{12}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 12 '13 at 6:57









          robjohn♦robjohn

          271k27315643




          271k27315643























              0












              $begingroup$

              One more way you can solve this is using complimentary counting. There are 6*$dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $frac{90}{216}=frac{5}{12}$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                One more way you can solve this is using complimentary counting. There are 6*$dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $frac{90}{216}=frac{5}{12}$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  One more way you can solve this is using complimentary counting. There are 6*$dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $frac{90}{216}=frac{5}{12}$






                  share|cite|improve this answer









                  $endgroup$



                  One more way you can solve this is using complimentary counting. There are 6*$dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $frac{90}{216}=frac{5}{12}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 27 '18 at 16:15









                  BobBob

                  1




                  1






























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