Mixing
Simple joint probability question of two dice throw
$begingroup$
I have a simple joint probability question I cannot understand. If I throw two dice and define the following events:
$X$ - number of times 4 was obtained.
$Y$ - number of even results obtained.
Now I need to find the joint probability function, so my problem is with $P(x=0, y=1) = P(text{obtaining zero times four and one time an even result})$, so it should be equal to probability of obtaining uneven results, and probability of obtaining an even result times $2/3$ (since 4 was not received). From this reasoning, I think the result should be: $frac{3}{6} cdot (frac{3}{6} cdot frac{2}{6})$, i.em $1/2$ chance to obtain uneven number times the probability to obtain an even number that is not ($3/6 cdot 2/6$).
However, the correct result should be: $frac{2 cdot 2 cdot3}{6 cdot 6}$, and I don't understand why. I would appreciate an explanation.
EDIT: similarly, why in $P(x=1, y=2) = P(text{two even results, of which only one is four})$ = $frac{2 cdot (1 cdot 2)}{6 cdot 6}$ and not: even ($3/6$) * even that is not four ($2/6$)?
probability dice
edited Feb 10 at 10:56
jvdhooft
5,67561641
asked Jan 28 at 12:22
q123q123
75
$endgroup$
add a comment |
$begingroup$
I have a simple joint probability question I cannot understand. If I throw two dice and define the following events:
$X$ - number of times 4 was obtained.
$Y$ - number of even results obtained.
Now I need to find the joint probability function, so my problem is with $P(x=0, y=1) = P(text{obtaining zero times four and one time an even result})$, so it should be equal to probability of obtaining uneven results, and probability of obtaining an even result times $2/3$ (since 4 was not received). From this reasoning, I think the result should be: $frac{3}{6} cdot (frac{3}{6} cdot frac{2}{6})$, i.em $1/2$ chance to obtain uneven number times the probability to obtain an even number that is not ($3/6 cdot 2/6$).
However, the correct result should be: $frac{2 cdot 2 cdot3}{6 cdot 6}$, and I don't understand why. I would appreciate an explanation.
EDIT: similarly, why in $P(x=1, y=2) = P(text{two even results, of which only one is four})$ = $frac{2 cdot (1 cdot 2)}{6 cdot 6}$ and not: even ($3/6$) * even that is not four ($2/6$)?
probability dice
edited Feb 10 at 10:56
jvdhooft
5,67561641
asked Jan 28 at 12:22
q123q123
75
$endgroup$
add a comment |
$begingroup$
I have a simple joint probability question I cannot understand. If I throw two dice and define the following events:
$X$ - number of times 4 was obtained.
$Y$ - number of even results obtained.
Now I need to find the joint probability function, so my problem is with $P(x=0, y=1) = P(text{obtaining zero times four and one time an even result})$, so it should be equal to probability of obtaining uneven results, and probability of obtaining an even result times $2/3$ (since 4 was not received). From this reasoning, I think the result should be: $frac{3}{6} cdot (frac{3}{6} cdot frac{2}{6})$, i.em $1/2$ chance to obtain uneven number times the probability to obtain an even number that is not ($3/6 cdot 2/6$).
However, the correct result should be: $frac{2 cdot 2 cdot3}{6 cdot 6}$, and I don't understand why. I would appreciate an explanation.
EDIT: similarly, why in $P(x=1, y=2) = P(text{two even results, of which only one is four})$ = $frac{2 cdot (1 cdot 2)}{6 cdot 6}$ and not: even ($3/6$) * even that is not four ($2/6$)?
probability dice
edited Feb 10 at 10:56
jvdhooft
5,67561641
asked Jan 28 at 12:22
q123q123
75
$endgroup$
I have a simple joint probability question I cannot understand. If I throw two dice and define the following events:
$X$ - number of times 4 was obtained.
$Y$ - number of even results obtained.
Now I need to find the joint probability function, so my problem is with $P(x=0, y=1) = P(text{obtaining zero times four and one time an even result})$, so it should be equal to probability of obtaining uneven results, and probability of obtaining an even result times $2/3$ (since 4 was not received). From this reasoning, I think the result should be: $frac{3}{6} cdot (frac{3}{6} cdot frac{2}{6})$, i.em $1/2$ chance to obtain uneven number times the probability to obtain an even number that is not ($3/6 cdot 2/6$).
However, the correct result should be: $frac{2 cdot 2 cdot3}{6 cdot 6}$, and I don't understand why. I would appreciate an explanation.
EDIT: similarly, why in $P(x=1, y=2) = P(text{two even results, of which only one is four})$ = $frac{2 cdot (1 cdot 2)}{6 cdot 6}$ and not: even ($3/6$) * even that is not four ($2/6$)?
probability dice
probability dice
edited Feb 10 at 10:56
jvdhooft
5,67561641
asked Jan 28 at 12:22
q123q123
75
edited Feb 10 at 10:56
jvdhooft
5,67561641
asked Jan 28 at 12:22
q123q123
75
edited Feb 10 at 10:56
jvdhooft
5,67561641
edited Feb 10 at 10:56
jvdhooft
5,67561641
edited Feb 10 at 10:56
jvdhooft
5,67561641
5,67561641
asked Jan 28 at 12:22
q123q123
75
asked Jan 28 at 12:22
q123q123
75
asked Jan 28 at 12:22
q123q123
75
75
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Give the dice a number.
Let $D_1$ denote the result of the first and let $D_2$ denote the result of the second die.
Then:
$$P(X=0,Y=1)=P(D_1in{2,6}, D_2in{1,3,5})+P(D_1in{1,3,5}, D_2in{2,6})=$$$$2P(D_1in{2,6})P(D_2in{1,3,5})=2frac26frac36$$
Edit (after edit of question):
$$P(X=1,Y=2)=P(D_1=4, D_2in{2,6})+P(D_1in{2,6}, D_2=4)=$$$$2P(D_1=4)P(D_2in{2,6})=2frac16frac26$$
answered Jan 28 at 12:37
drhabdrhab
104k545136
$endgroup$
$begingroup$
thank you very much, your explanation really helps me understand what i'm calculating and how
$endgroup$
– q123
Jan 28 at 12:43
$begingroup$
You are very welcome. In questions like this (you roll $n>1$ dice) it is often helpful to give the dice numbers.
$endgroup$
– drhab
Jan 28 at 12:45
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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$begingroup$
Give the dice a number.
Let $D_1$ denote the result of the first and let $D_2$ denote the result of the second die.
Then:
$$P(X=0,Y=1)=P(D_1in{2,6}, D_2in{1,3,5})+P(D_1in{1,3,5}, D_2in{2,6})=$$$$2P(D_1in{2,6})P(D_2in{1,3,5})=2frac26frac36$$
Edit (after edit of question):
$$P(X=1,Y=2)=P(D_1=4, D_2in{2,6})+P(D_1in{2,6}, D_2=4)=$$$$2P(D_1=4)P(D_2in{2,6})=2frac16frac26$$
answered Jan 28 at 12:37
drhabdrhab
104k545136
$endgroup$
$begingroup$
thank you very much, your explanation really helps me understand what i'm calculating and how
$endgroup$
– q123
Jan 28 at 12:43
$begingroup$
You are very welcome. In questions like this (you roll $n>1$ dice) it is often helpful to give the dice numbers.
$endgroup$
– drhab
Jan 28 at 12:45
add a comment |
$begingroup$
Give the dice a number.
Let $D_1$ denote the result of the first and let $D_2$ denote the result of the second die.
Then:
$$P(X=0,Y=1)=P(D_1in{2,6}, D_2in{1,3,5})+P(D_1in{1,3,5}, D_2in{2,6})=$$$$2P(D_1in{2,6})P(D_2in{1,3,5})=2frac26frac36$$
Edit (after edit of question):
$$P(X=1,Y=2)=P(D_1=4, D_2in{2,6})+P(D_1in{2,6}, D_2=4)=$$$$2P(D_1=4)P(D_2in{2,6})=2frac16frac26$$
answered Jan 28 at 12:37
drhabdrhab
104k545136
$endgroup$
$begingroup$
thank you very much, your explanation really helps me understand what i'm calculating and how
$endgroup$
– q123
Jan 28 at 12:43
$begingroup$
You are very welcome. In questions like this (you roll $n>1$ dice) it is often helpful to give the dice numbers.
$endgroup$
– drhab
Jan 28 at 12:45
add a comment |
$begingroup$
Give the dice a number.
Let $D_1$ denote the result of the first and let $D_2$ denote the result of the second die.
Then:
$$P(X=0,Y=1)=P(D_1in{2,6}, D_2in{1,3,5})+P(D_1in{1,3,5}, D_2in{2,6})=$$$$2P(D_1in{2,6})P(D_2in{1,3,5})=2frac26frac36$$
Edit (after edit of question):
$$P(X=1,Y=2)=P(D_1=4, D_2in{2,6})+P(D_1in{2,6}, D_2=4)=$$$$2P(D_1=4)P(D_2in{2,6})=2frac16frac26$$
answered Jan 28 at 12:37
drhabdrhab
104k545136
$endgroup$
Give the dice a number.
Let $D_1$ denote the result of the first and let $D_2$ denote the result of the second die.
Then:
$$P(X=0,Y=1)=P(D_1in{2,6}, D_2in{1,3,5})+P(D_1in{1,3,5}, D_2in{2,6})=$$$$2P(D_1in{2,6})P(D_2in{1,3,5})=2frac26frac36$$
Edit (after edit of question):
$$P(X=1,Y=2)=P(D_1=4, D_2in{2,6})+P(D_1in{2,6}, D_2=4)=$$$$2P(D_1=4)P(D_2in{2,6})=2frac16frac26$$
answered Jan 28 at 12:37
drhabdrhab
104k545136
answered Jan 28 at 12:37
drhabdrhab
104k545136
answered Jan 28 at 12:37
drhabdrhab
104k545136
answered Jan 28 at 12:37
drhabdrhab
104k545136
104k545136
$begingroup$
thank you very much, your explanation really helps me understand what i'm calculating and how
$endgroup$
– q123
Jan 28 at 12:43
$begingroup$
You are very welcome. In questions like this (you roll $n>1$ dice) it is often helpful to give the dice numbers.
$endgroup$
– drhab
Jan 28 at 12:45
add a comment |
$begingroup$
thank you very much, your explanation really helps me understand what i'm calculating and how
$endgroup$
– q123
Jan 28 at 12:43
$begingroup$
You are very welcome. In questions like this (you roll $n>1$ dice) it is often helpful to give the dice numbers.
$endgroup$
– drhab
Jan 28 at 12:45
$begingroup$
thank you very much, your explanation really helps me understand what i'm calculating and how
$endgroup$
– q123
Jan 28 at 12:43
$begingroup$
thank you very much, your explanation really helps me understand what i'm calculating and how
$endgroup$
– q123
Jan 28 at 12:43
$begingroup$
You are very welcome. In questions like this (you roll $n>1$ dice) it is often helpful to give the dice numbers.
$endgroup$
– drhab
Jan 28 at 12:45
$begingroup$
You are very welcome. In questions like this (you roll $n>1$ dice) it is often helpful to give the dice numbers.
$endgroup$
– drhab
Jan 28 at 12:45
add a comment |
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