Probability of rolling a pair of dice












4












$begingroup$


Suppose we have two fair dice and rolled them.



Let





  • $A$ be the event "the sum of the two dice is equal to $3$";


  • $B$ be the event "the sum of the two dice is equal to $7$";


  • $C$ be the event "at least one dice shows $1$".


How to calculate $P(A mid C)$?



In this case can we say that $A$ and $C$ are independent? Can we say that $B$ and $C$ are independent?






probability dice







share|cite|improve this question









New contributor




Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I don't see how $A cap C$ and $B cap C$ can be independent, since they're mutually exclusive and each has non-zero probability. Did you mean something else?
    $endgroup$
    – Brian Tung
    21 hours ago










  • $begingroup$
    Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
    $endgroup$
    – Davor
    18 hours ago
















4












$begingroup$


Suppose we have two fair dice and rolled them.



Let





  • $A$ be the event "the sum of the two dice is equal to $3$";


  • $B$ be the event "the sum of the two dice is equal to $7$";


  • $C$ be the event "at least one dice shows $1$".


How to calculate $P(A mid C)$?



In this case can we say that $A$ and $C$ are independent? Can we say that $B$ and $C$ are independent?






probability dice







share|cite|improve this question









New contributor




Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I don't see how $A cap C$ and $B cap C$ can be independent, since they're mutually exclusive and each has non-zero probability. Did you mean something else?
    $endgroup$
    – Brian Tung
    21 hours ago










  • $begingroup$
    Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
    $endgroup$
    – Davor
    18 hours ago














4












4








4


2



$begingroup$


Suppose we have two fair dice and rolled them.



Let





  • $A$ be the event "the sum of the two dice is equal to $3$";


  • $B$ be the event "the sum of the two dice is equal to $7$";


  • $C$ be the event "at least one dice shows $1$".


How to calculate $P(A mid C)$?



In this case can we say that $A$ and $C$ are independent? Can we say that $B$ and $C$ are independent?






probability dice







share|cite|improve this question









New contributor




Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Suppose we have two fair dice and rolled them.



Let





  • $A$ be the event "the sum of the two dice is equal to $3$";


  • $B$ be the event "the sum of the two dice is equal to $7$";


  • $C$ be the event "at least one dice shows $1$".


How to calculate $P(A mid C)$?



In this case can we say that $A$ and $C$ are independent? Can we say that $B$ and $C$ are independent?






probability dice




probability dice






share|cite|improve this question









New contributor




Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 16 hours ago









L. F.

18110




18110






New contributor




Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 23 hours ago









Ali J.Ali J.

464




464




New contributor




Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    I don't see how $A cap C$ and $B cap C$ can be independent, since they're mutually exclusive and each has non-zero probability. Did you mean something else?
    $endgroup$
    – Brian Tung
    21 hours ago










  • $begingroup$
    Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
    $endgroup$
    – Davor
    18 hours ago


















  • $begingroup$
    I don't see how $A cap C$ and $B cap C$ can be independent, since they're mutually exclusive and each has non-zero probability. Did you mean something else?
    $endgroup$
    – Brian Tung
    21 hours ago










  • $begingroup$
    Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
    $endgroup$
    – Davor
    18 hours ago
















$begingroup$
I don't see how $A cap C$ and $B cap C$ can be independent, since they're mutually exclusive and each has non-zero probability. Did you mean something else?
$endgroup$
– Brian Tung
21 hours ago




$begingroup$
I don't see how $A cap C$ and $B cap C$ can be independent, since they're mutually exclusive and each has non-zero probability. Did you mean something else?
$endgroup$
– Brian Tung
21 hours ago












$begingroup$
Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
$endgroup$
– Davor
18 hours ago




$begingroup$
Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
$endgroup$
– Davor
18 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

$P(C)$ is actually $frac{11}{36}$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(Acap C)=frac2{36}$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus
$$P(A|C)=frac{P(Acap C)}{P(C)}=frac{2/36}{11/36}=frac2{11}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. Can I say that $P(B|C)=frac{2}{11}$?
    $endgroup$
    – Ali J.
    22 hours ago












  • $begingroup$
    @AliJ Yes, since $P(Bcap C)=P(Acap C)$.
    $endgroup$
    – Parcly Taxel
    22 hours ago



















3












$begingroup$

If one of the dices shows $1$ then there are only two ways to get $3.$ If both the dices show $1$ then there is no chance of getting $3.$ So the required probability is $frac {2} {11}.$ Because when $C$ has already occurred then the reduced sample space has $11$ elements where at least one of the events is $1$ which are $${(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1) }$$ and only two of them suit your purpose which are $${(1,2),(2,1) }.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean that $P(A|C)=frac{1}{18}$?
    $endgroup$
    – Ali J.
    22 hours ago










  • $begingroup$
    May you please help about $P(B|C)$?
    $endgroup$
    – Ali J.
    22 hours ago










  • $begingroup$
    Same thing happens for $P(B mid C).$
    $endgroup$
    – Dbchatto67
    22 hours ago










  • $begingroup$
    Can't we use the rule $P(A|C)=frac{P(A.C)}{P(A)}$
    $endgroup$
    – Ali J.
    22 hours ago












  • $begingroup$
    Try to think it on your own. If the event $C$ has already occured then in how many ways can the event $B$ occur?
    $endgroup$
    – Dbchatto67
    22 hours ago













Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

$P(C)$ is actually $frac{11}{36}$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(Acap C)=frac2{36}$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus
$$P(A|C)=frac{P(Acap C)}{P(C)}=frac{2/36}{11/36}=frac2{11}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. Can I say that $P(B|C)=frac{2}{11}$?
    $endgroup$
    – Ali J.
    22 hours ago












  • $begingroup$
    @AliJ Yes, since $P(Bcap C)=P(Acap C)$.
    $endgroup$
    – Parcly Taxel
    22 hours ago
















3












$begingroup$

$P(C)$ is actually $frac{11}{36}$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(Acap C)=frac2{36}$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus
$$P(A|C)=frac{P(Acap C)}{P(C)}=frac{2/36}{11/36}=frac2{11}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. Can I say that $P(B|C)=frac{2}{11}$?
    $endgroup$
    – Ali J.
    22 hours ago












  • $begingroup$
    @AliJ Yes, since $P(Bcap C)=P(Acap C)$.
    $endgroup$
    – Parcly Taxel
    22 hours ago














3












3








3





$begingroup$

$P(C)$ is actually $frac{11}{36}$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(Acap C)=frac2{36}$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus
$$P(A|C)=frac{P(Acap C)}{P(C)}=frac{2/36}{11/36}=frac2{11}$$






share|cite|improve this answer









$endgroup$



$P(C)$ is actually $frac{11}{36}$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(Acap C)=frac2{36}$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus
$$P(A|C)=frac{P(Acap C)}{P(C)}=frac{2/36}{11/36}=frac2{11}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 22 hours ago









Parcly TaxelParcly Taxel

44.3k1375107




44.3k1375107












  • $begingroup$
    Thank you. Can I say that $P(B|C)=frac{2}{11}$?
    $endgroup$
    – Ali J.
    22 hours ago












  • $begingroup$
    @AliJ Yes, since $P(Bcap C)=P(Acap C)$.
    $endgroup$
    – Parcly Taxel
    22 hours ago


















  • $begingroup$
    Thank you. Can I say that $P(B|C)=frac{2}{11}$?
    $endgroup$
    – Ali J.
    22 hours ago












  • $begingroup$
    @AliJ Yes, since $P(Bcap C)=P(Acap C)$.
    $endgroup$
    – Parcly Taxel
    22 hours ago
















$begingroup$
Thank you. Can I say that $P(B|C)=frac{2}{11}$?
$endgroup$
– Ali J.
22 hours ago






$begingroup$
Thank you. Can I say that $P(B|C)=frac{2}{11}$?
$endgroup$
– Ali J.
22 hours ago














$begingroup$
@AliJ Yes, since $P(Bcap C)=P(Acap C)$.
$endgroup$
– Parcly Taxel
22 hours ago




$begingroup$
@AliJ Yes, since $P(Bcap C)=P(Acap C)$.
$endgroup$
– Parcly Taxel
22 hours ago











3












$begingroup$

If one of the dices shows $1$ then there are only two ways to get $3.$ If both the dices show $1$ then there is no chance of getting $3.$ So the required probability is $frac {2} {11}.$ Because when $C$ has already occurred then the reduced sample space has $11$ elements where at least one of the events is $1$ which are $${(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1) }$$ and only two of them suit your purpose which are $${(1,2),(2,1) }.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean that $P(A|C)=frac{1}{18}$?
    $endgroup$
    – Ali J.
    22 hours ago










  • $begingroup$
    May you please help about $P(B|C)$?
    $endgroup$
    – Ali J.
    22 hours ago










  • $begingroup$
    Same thing happens for $P(B mid C).$
    $endgroup$
    – Dbchatto67
    22 hours ago










  • $begingroup$
    Can't we use the rule $P(A|C)=frac{P(A.C)}{P(A)}$
    $endgroup$
    – Ali J.
    22 hours ago












  • $begingroup$
    Try to think it on your own. If the event $C$ has already occured then in how many ways can the event $B$ occur?
    $endgroup$
    – Dbchatto67
    22 hours ago


















3












$begingroup$

If one of the dices shows $1$ then there are only two ways to get $3.$ If both the dices show $1$ then there is no chance of getting $3.$ So the required probability is $frac {2} {11}.$ Because when $C$ has already occurred then the reduced sample space has $11$ elements where at least one of the events is $1$ which are $${(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1) }$$ and only two of them suit your purpose which are $${(1,2),(2,1) }.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean that $P(A|C)=frac{1}{18}$?
    $endgroup$
    – Ali J.
    22 hours ago










  • $begingroup$
    May you please help about $P(B|C)$?
    $endgroup$
    – Ali J.
    22 hours ago










  • $begingroup$
    Same thing happens for $P(B mid C).$
    $endgroup$
    – Dbchatto67
    22 hours ago










  • $begingroup$
    Can't we use the rule $P(A|C)=frac{P(A.C)}{P(A)}$
    $endgroup$
    – Ali J.
    22 hours ago












  • $begingroup$
    Try to think it on your own. If the event $C$ has already occured then in how many ways can the event $B$ occur?
    $endgroup$
    – Dbchatto67
    22 hours ago
















3












3








3





$begingroup$

If one of the dices shows $1$ then there are only two ways to get $3.$ If both the dices show $1$ then there is no chance of getting $3.$ So the required probability is $frac {2} {11}.$ Because when $C$ has already occurred then the reduced sample space has $11$ elements where at least one of the events is $1$ which are $${(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1) }$$ and only two of them suit your purpose which are $${(1,2),(2,1) }.$$






share|cite|improve this answer











$endgroup$



If one of the dices shows $1$ then there are only two ways to get $3.$ If both the dices show $1$ then there is no chance of getting $3.$ So the required probability is $frac {2} {11}.$ Because when $C$ has already occurred then the reduced sample space has $11$ elements where at least one of the events is $1$ which are $${(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1) }$$ and only two of them suit your purpose which are $${(1,2),(2,1) }.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 22 hours ago

























answered 23 hours ago









Dbchatto67Dbchatto67

1,789219




1,789219












  • $begingroup$
    You mean that $P(A|C)=frac{1}{18}$?
    $endgroup$
    – Ali J.
    22 hours ago










  • $begingroup$
    May you please help about $P(B|C)$?
    $endgroup$
    – Ali J.
    22 hours ago










  • $begingroup$
    Same thing happens for $P(B mid C).$
    $endgroup$
    – Dbchatto67
    22 hours ago










  • $begingroup$
    Can't we use the rule $P(A|C)=frac{P(A.C)}{P(A)}$
    $endgroup$
    – Ali J.
    22 hours ago












  • $begingroup$
    Try to think it on your own. If the event $C$ has already occured then in how many ways can the event $B$ occur?
    $endgroup$
    – Dbchatto67
    22 hours ago




















  • $begingroup$
    You mean that $P(A|C)=frac{1}{18}$?
    $endgroup$
    – Ali J.
    22 hours ago










  • $begingroup$
    May you please help about $P(B|C)$?
    $endgroup$
    – Ali J.
    22 hours ago










  • $begingroup$
    Same thing happens for $P(B mid C).$
    $endgroup$
    – Dbchatto67
    22 hours ago










  • $begingroup$
    Can't we use the rule $P(A|C)=frac{P(A.C)}{P(A)}$
    $endgroup$
    – Ali J.
    22 hours ago












  • $begingroup$
    Try to think it on your own. If the event $C$ has already occured then in how many ways can the event $B$ occur?
    $endgroup$
    – Dbchatto67
    22 hours ago


















$begingroup$
You mean that $P(A|C)=frac{1}{18}$?
$endgroup$
– Ali J.
22 hours ago




$begingroup$
You mean that $P(A|C)=frac{1}{18}$?
$endgroup$
– Ali J.
22 hours ago












$begingroup$
May you please help about $P(B|C)$?
$endgroup$
– Ali J.
22 hours ago




$begingroup$
May you please help about $P(B|C)$?
$endgroup$
– Ali J.
22 hours ago












$begingroup$
Same thing happens for $P(B mid C).$
$endgroup$
– Dbchatto67
22 hours ago




$begingroup$
Same thing happens for $P(B mid C).$
$endgroup$
– Dbchatto67
22 hours ago












$begingroup$
Can't we use the rule $P(A|C)=frac{P(A.C)}{P(A)}$
$endgroup$
– Ali J.
22 hours ago






$begingroup$
Can't we use the rule $P(A|C)=frac{P(A.C)}{P(A)}$
$endgroup$
– Ali J.
22 hours ago














$begingroup$
Try to think it on your own. If the event $C$ has already occured then in how many ways can the event $B$ occur?
$endgroup$
– Dbchatto67
22 hours ago






$begingroup$
Try to think it on your own. If the event $C$ has already occured then in how many ways can the event $B$ occur?
$endgroup$
– Dbchatto67
22 hours ago












Ali J. is a new contributor. Be nice, and check out our Code of Conduct.










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Ali J. is a new contributor. Be nice, and check out our Code of Conduct.













Ali J. is a new contributor. Be nice, and check out our Code of Conduct.












Ali J. is a new contributor. Be nice, and check out our Code of Conduct.
















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