Probability of rolling a pair of dice
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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
edited Mar 13 at 13:07
L. F.
20511
asked Mar 13 at 6:10
Ali J.Ali J.
484
$endgroup$
add a comment |
$begingroup$
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
edited Mar 13 at 13:07
L. F.
20511
asked Mar 13 at 6:10
Ali J.Ali J.
484
$endgroup$
$begingroup$
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
Mar 13 at 7:18
$begingroup$
Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
$endgroup$
– Davor
Mar 13 at 11:01
add a comment |
$begingroup$
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
edited Mar 13 at 13:07
L. F.
20511
asked Mar 13 at 6:10
Ali J.Ali J.
484
$endgroup$
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
edited Mar 13 at 13:07
L. F.
20511
asked Mar 13 at 6:10
Ali J.Ali J.
484
edited Mar 13 at 13:07
L. F.
20511
asked Mar 13 at 6:10
Ali J.Ali J.
484
edited Mar 13 at 13:07
L. F.
20511
edited Mar 13 at 13:07
L. F.
20511
edited Mar 13 at 13:07
L. F.
20511
20511
asked Mar 13 at 6:10
Ali J.Ali J.
484
asked Mar 13 at 6:10
Ali J.Ali J.
484
asked Mar 13 at 6:10
Ali J.Ali J.
484
484
$begingroup$
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
Mar 13 at 7:18
$begingroup$
Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
$endgroup$
– Davor
Mar 13 at 11:01
add a comment |
$begingroup$
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
Mar 13 at 7:18
$begingroup$
Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
$endgroup$
– Davor
Mar 13 at 11:01
$begingroup$
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
Mar 13 at 7:18
$begingroup$
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
Mar 13 at 7:18
$begingroup$
Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
$endgroup$
– Davor
Mar 13 at 11:01
$begingroup$
Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
$endgroup$
– Davor
Mar 13 at 11:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$P(C)$ is actually $frac1136$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(Acap C)=frac236$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus
$$P(A|C)=fracP(Acap C)P(C)=frac2/3611/36=frac211$$
answered Mar 13 at 6:19
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
$begingroup$
Thank you. Can I say that $P(B|C)=frac211$?
$endgroup$
– Ali J.
Mar 13 at 6:44
$begingroup$
@AliJ Yes, since $P(Bcap C)=P(Acap C)$.
$endgroup$
– Parcly Taxel
Mar 13 at 6:48
add a comment |
$begingroup$
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 $$(1,2),(2,1) .$$
answered Mar 13 at 6:13
Dbchatto67Dbchatto67
2,777622
$endgroup$
$begingroup$
You mean that $P(A|C)=frac118$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
May you please help about $P(B|C)$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
Same thing happens for $P(B mid C).$
$endgroup$
– Dbchatto67
Mar 13 at 6:18
$begingroup$
Can't we use the rule $P(A|C)=fracP(A.C)P(A)$
$endgroup$
– Ali J.
Mar 13 at 6:19
$begingroup$
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
Mar 13 at 6:19
|
show 5 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$P(C)$ is actually $frac1136$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(Acap C)=frac236$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus
$$P(A|C)=fracP(Acap C)P(C)=frac2/3611/36=frac211$$
answered Mar 13 at 6:19
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
$begingroup$
Thank you. Can I say that $P(B|C)=frac211$?
$endgroup$
– Ali J.
Mar 13 at 6:44
$begingroup$
@AliJ Yes, since $P(Bcap C)=P(Acap C)$.
$endgroup$
– Parcly Taxel
Mar 13 at 6:48
add a comment |
$begingroup$
$P(C)$ is actually $frac1136$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(Acap C)=frac236$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus
$$P(A|C)=fracP(Acap C)P(C)=frac2/3611/36=frac211$$
answered Mar 13 at 6:19
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
$begingroup$
Thank you. Can I say that $P(B|C)=frac211$?
$endgroup$
– Ali J.
Mar 13 at 6:44
$begingroup$
@AliJ Yes, since $P(Bcap C)=P(Acap C)$.
$endgroup$
– Parcly Taxel
Mar 13 at 6:48
add a comment |
$begingroup$
$P(C)$ is actually $frac1136$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(Acap C)=frac236$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus
$$P(A|C)=fracP(Acap C)P(C)=frac2/3611/36=frac211$$
answered Mar 13 at 6:19
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
$P(C)$ is actually $frac1136$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(Acap C)=frac236$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus
$$P(A|C)=fracP(Acap C)P(C)=frac2/3611/36=frac211$$
answered Mar 13 at 6:19
Parcly TaxelParcly Taxel
44.7k1376109
answered Mar 13 at 6:19
Parcly TaxelParcly Taxel
44.7k1376109
answered Mar 13 at 6:19
Parcly TaxelParcly Taxel
44.7k1376109
answered Mar 13 at 6:19
Parcly TaxelParcly Taxel
44.7k1376109
44.7k1376109
$begingroup$
Thank you. Can I say that $P(B|C)=frac211$?
$endgroup$
– Ali J.
Mar 13 at 6:44
$begingroup$
@AliJ Yes, since $P(Bcap C)=P(Acap C)$.
$endgroup$
– Parcly Taxel
Mar 13 at 6:48
add a comment |
$begingroup$
Thank you. Can I say that $P(B|C)=frac211$?
$endgroup$
– Ali J.
Mar 13 at 6:44
$begingroup$
@AliJ Yes, since $P(Bcap C)=P(Acap C)$.
$endgroup$
– Parcly Taxel
Mar 13 at 6:48
$begingroup$
Thank you. Can I say that $P(B|C)=frac211$?
$endgroup$
– Ali J.
Mar 13 at 6:44
$begingroup$
Thank you. Can I say that $P(B|C)=frac211$?
$endgroup$
– Ali J.
Mar 13 at 6:44
$begingroup$
@AliJ Yes, since $P(Bcap C)=P(Acap C)$.
$endgroup$
– Parcly Taxel
Mar 13 at 6:48
$begingroup$
@AliJ Yes, since $P(Bcap C)=P(Acap C)$.
$endgroup$
– Parcly Taxel
Mar 13 at 6:48
add a comment |
$begingroup$
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 $$(1,2),(2,1) .$$
answered Mar 13 at 6:13
Dbchatto67Dbchatto67
2,777622
$endgroup$
$begingroup$
You mean that $P(A|C)=frac118$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
May you please help about $P(B|C)$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
Same thing happens for $P(B mid C).$
$endgroup$
– Dbchatto67
Mar 13 at 6:18
$begingroup$
Can't we use the rule $P(A|C)=fracP(A.C)P(A)$
$endgroup$
– Ali J.
Mar 13 at 6:19
$begingroup$
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
Mar 13 at 6:19
|
show 5 more comments
$begingroup$
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 $$(1,2),(2,1) .$$
answered Mar 13 at 6:13
Dbchatto67Dbchatto67
2,777622
$endgroup$
$begingroup$
You mean that $P(A|C)=frac118$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
May you please help about $P(B|C)$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
Same thing happens for $P(B mid C).$
$endgroup$
– Dbchatto67
Mar 13 at 6:18
$begingroup$
Can't we use the rule $P(A|C)=fracP(A.C)P(A)$
$endgroup$
– Ali J.
Mar 13 at 6:19
$begingroup$
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
Mar 13 at 6:19
|
show 5 more comments
$begingroup$
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 $$(1,2),(2,1) .$$
answered Mar 13 at 6:13
Dbchatto67Dbchatto67
2,777622
$endgroup$
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 $$(1,2),(2,1) .$$
answered Mar 13 at 6:13
Dbchatto67Dbchatto67
2,777622
edited Mar 13 at 6:29
answered Mar 13 at 6:13
Dbchatto67Dbchatto67
2,777622
answered Mar 13 at 6:13
Dbchatto67Dbchatto67
2,777622
answered Mar 13 at 6:13
Dbchatto67Dbchatto67
2,777622
2,777622
$begingroup$
You mean that $P(A|C)=frac118$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
May you please help about $P(B|C)$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
Same thing happens for $P(B mid C).$
$endgroup$
– Dbchatto67
Mar 13 at 6:18
$begingroup$
Can't we use the rule $P(A|C)=fracP(A.C)P(A)$
$endgroup$
– Ali J.
Mar 13 at 6:19
$begingroup$
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
Mar 13 at 6:19
|
show 5 more comments
$begingroup$
You mean that $P(A|C)=frac118$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
May you please help about $P(B|C)$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
Same thing happens for $P(B mid C).$
$endgroup$
– Dbchatto67
Mar 13 at 6:18
$begingroup$
Can't we use the rule $P(A|C)=fracP(A.C)P(A)$
$endgroup$
– Ali J.
Mar 13 at 6:19
$begingroup$
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
Mar 13 at 6:19
$begingroup$
You mean that $P(A|C)=frac118$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
You mean that $P(A|C)=frac118$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
May you please help about $P(B|C)$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
May you please help about $P(B|C)$?
$endgroup$
– Ali J.
Mar 13 at 6:16
$begingroup$
Same thing happens for $P(B mid C).$
$endgroup$
– Dbchatto67
Mar 13 at 6:18
$begingroup$
Same thing happens for $P(B mid C).$
$endgroup$
– Dbchatto67
Mar 13 at 6:18
$begingroup$
Can't we use the rule $P(A|C)=fracP(A.C)P(A)$
$endgroup$
– Ali J.
Mar 13 at 6:19
$begingroup$
Can't we use the rule $P(A|C)=fracP(A.C)P(A)$
$endgroup$
– Ali J.
Mar 13 at 6:19
$begingroup$
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
Mar 13 at 6:19
$begingroup$
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
Mar 13 at 6:19
|
show 5 more comments
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$begingroup$
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
Mar 13 at 7:18
$begingroup$
Dice is already plural. The singular is die. So it's one die, 2 dice, 3 dice...
$endgroup$
– Davor
Mar 13 at 11:01