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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
asked Nov 19 at 0:04
Jan Jin
11
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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
asked Nov 19 at 0:04
Jan Jin
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 19 at 0:07
1
For example, how did you get $frac 2{27}$? Could you sketch your method?
– lulu
Nov 19 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 at 0:13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
asked Nov 19 at 0:04
Jan Jin
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
asked Nov 19 at 0:04
Jan Jin
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 19 at 0:04
Jan Jin
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
asked Nov 19 at 0:04
Jan Jin
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 19 at 0:04
Jan Jin
11
asked Nov 19 at 0:04
Jan Jin
11
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 19 at 0:07
1
For example, how did you get $frac 2{27}$? Could you sketch your method?
– lulu
Nov 19 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 at 0:13
add a comment |
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 19 at 0:07
1
For example, how did you get $frac 2{27}$? Could you sketch your method?
– lulu
Nov 19 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 at 0:13
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 19 at 0:07
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 19 at 0:07
1
1
For example, how did you get $frac 2{27}$? Could you sketch your method?
– lulu
Nov 19 at 0:12
For example, how did you get $frac 2{27}$? Could you sketch your method?
– lulu
Nov 19 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 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 at 0:13
add a comment |
3 Answers
3
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0
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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.
answered Nov 19 at 0:20
Anuj Doshi
11
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Anuj Doshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.
answered 2 days ago
Jan Jin
11
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Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
0
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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}
answered Nov 19 at 1:58
dynamic89
38418
I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
– Jan Jin
2 days ago
yup that was a typo
– dynamic89
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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.
answered Nov 19 at 0:20
Anuj Doshi
11
New contributor
Anuj Doshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
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.
answered Nov 19 at 0:20
Anuj Doshi
11
New contributor
Anuj Doshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
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.
answered Nov 19 at 0:20
Anuj Doshi
11
New contributor
Anuj Doshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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.
answered Nov 19 at 0:20
Anuj Doshi
11
New contributor
Anuj Doshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 19 at 0:20
Anuj Doshi
11
New contributor
Anuj Doshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 19 at 0:20
Anuj Doshi
11
answered Nov 19 at 0:20
Anuj Doshi
11
11
New contributor
Anuj Doshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Anuj Doshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Anuj Doshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
up vote
0
down vote
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.
answered 2 days ago
Jan Jin
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
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.
answered 2 days ago
Jan Jin
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
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.
answered 2 days ago
Jan Jin
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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.
answered 2 days ago
Jan Jin
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Jan Jin
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Jan Jin
11
answered 2 days ago
Jan Jin
11
11
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jan Jin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
up vote
0
down vote
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}
answered Nov 19 at 1:58
dynamic89
38418
I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
– Jan Jin
2 days ago
yup that was a typo
– dynamic89
2 days ago
add a comment |
up vote
0
down vote
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}
answered Nov 19 at 1:58
dynamic89
38418
I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
– Jan Jin
2 days ago
yup that was a typo
– dynamic89
2 days ago
add a comment |
up vote
0
down vote
up vote
0
down vote
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}
answered Nov 19 at 1:58
dynamic89
38418
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}
answered Nov 19 at 1:58
dynamic89
38418
edited 2 days ago
answered Nov 19 at 1:58
dynamic89
38418
answered Nov 19 at 1:58
dynamic89
38418
answered Nov 19 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
2 days ago
yup that was a typo
– dynamic89
2 days ago
add a comment |
I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
– Jan Jin
2 days ago
yup that was a typo
– dynamic89
2 days ago
I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
– Jan Jin
2 days ago
I think B1 be the event "total of first two is 10" P(B1) = 3/(6x6) = 3/36 ?
– Jan Jin
2 days ago
yup that was a typo
– dynamic89
2 days ago
yup that was a typo
– dynamic89
2 days ago
add a comment |
Jan Jin is a new contributor. Be nice, and check out our Code of Conduct.
Jan Jin is a new contributor. Be nice, and check out our Code of Conduct.
Jan Jin is a new contributor. Be nice, and check out our Code of Conduct.
Jan Jin is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 19 at 0:07
1
For example, how did you get $frac 2{27}$? Could you sketch your method?
– lulu
Nov 19 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 at 0:13