Probability and permutations relationship
$begingroup$
A family with a mother, father, two daughters
and three sons lines up in a random order for a photo.
(a) Let D be the random variable denoting the number of daughters who are standing next to the mother and for i = 1,2 let Di be the indicator variable that is 1 if daughter i is next to the mother and 0 otherwise. What is the relationship between D, D1, and D2?
Solution: D = D1 + D2
Well, I don't really get the logic here...
discrete-mathematics
asked Dec 28 '18 at 20:02
Tom1999Tom1999
445
$endgroup$
add a comment |
$begingroup$
A family with a mother, father, two daughters
and three sons lines up in a random order for a photo.
(a) Let D be the random variable denoting the number of daughters who are standing next to the mother and for i = 1,2 let Di be the indicator variable that is 1 if daughter i is next to the mother and 0 otherwise. What is the relationship between D, D1, and D2?
Solution: D = D1 + D2
Well, I don't really get the logic here...
discrete-mathematics
asked Dec 28 '18 at 20:02
Tom1999Tom1999
445
$endgroup$
$begingroup$
What part don't you get? Do you know the definition of an indicator variable?
$endgroup$
– lulu
Dec 28 '18 at 20:15
add a comment |
$begingroup$
A family with a mother, father, two daughters
and three sons lines up in a random order for a photo.
(a) Let D be the random variable denoting the number of daughters who are standing next to the mother and for i = 1,2 let Di be the indicator variable that is 1 if daughter i is next to the mother and 0 otherwise. What is the relationship between D, D1, and D2?
Solution: D = D1 + D2
Well, I don't really get the logic here...
discrete-mathematics
asked Dec 28 '18 at 20:02
Tom1999Tom1999
445
$endgroup$
A family with a mother, father, two daughters
and three sons lines up in a random order for a photo.
(a) Let D be the random variable denoting the number of daughters who are standing next to the mother and for i = 1,2 let Di be the indicator variable that is 1 if daughter i is next to the mother and 0 otherwise. What is the relationship between D, D1, and D2?
Solution: D = D1 + D2
Well, I don't really get the logic here...
discrete-mathematics
discrete-mathematics
asked Dec 28 '18 at 20:02
Tom1999Tom1999
445
asked Dec 28 '18 at 20:02
Tom1999Tom1999
445
asked Dec 28 '18 at 20:02
Tom1999Tom1999
445
asked Dec 28 '18 at 20:02
Tom1999Tom1999
445
asked Dec 28 '18 at 20:02
Tom1999Tom1999
445
445
$begingroup$
What part don't you get? Do you know the definition of an indicator variable?
$endgroup$
– lulu
Dec 28 '18 at 20:15
add a comment |
$begingroup$
What part don't you get? Do you know the definition of an indicator variable?
$endgroup$
– lulu
Dec 28 '18 at 20:15
$begingroup$
What part don't you get? Do you know the definition of an indicator variable?
$endgroup$
– lulu
Dec 28 '18 at 20:15
$begingroup$
What part don't you get? Do you know the definition of an indicator variable?
$endgroup$
– lulu
Dec 28 '18 at 20:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The possible positions of the mother and two daughters are as follows: (where $x$ represents a person other than the daughters)
begin{align*}
ldots,x,m,d_i, x,x,ldots\
ldots,d_i,m,x,x,ldots\
ldots,d_i,m,d_j, x,x,ldots\
ldots,x,m,x,x,ldots\
end{align*}
In the first two cases, $D=1$, in the third case $D=2$, and in the last case $D=0$.
Likewise, in the first two cases, one of the indicator functions takes on the value $1$ and the other $0$. So $D=D_1+D_2$.
You can deal with the remaining cases similarly.
answered Dec 28 '18 at 20:20
Anurag AAnurag A
26.4k12351
$endgroup$
$begingroup$
Thank you Anurag :)
$endgroup$
– Tom1999
Dec 28 '18 at 21:43
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
The possible positions of the mother and two daughters are as follows: (where $x$ represents a person other than the daughters)
begin{align*}
ldots,x,m,d_i, x,x,ldots\
ldots,d_i,m,x,x,ldots\
ldots,d_i,m,d_j, x,x,ldots\
ldots,x,m,x,x,ldots\
end{align*}
In the first two cases, $D=1$, in the third case $D=2$, and in the last case $D=0$.
Likewise, in the first two cases, one of the indicator functions takes on the value $1$ and the other $0$. So $D=D_1+D_2$.
You can deal with the remaining cases similarly.
answered Dec 28 '18 at 20:20
Anurag AAnurag A
26.4k12351
$endgroup$
$begingroup$
Thank you Anurag :)
$endgroup$
– Tom1999
Dec 28 '18 at 21:43
add a comment |
$begingroup$
The possible positions of the mother and two daughters are as follows: (where $x$ represents a person other than the daughters)
begin{align*}
ldots,x,m,d_i, x,x,ldots\
ldots,d_i,m,x,x,ldots\
ldots,d_i,m,d_j, x,x,ldots\
ldots,x,m,x,x,ldots\
end{align*}
In the first two cases, $D=1$, in the third case $D=2$, and in the last case $D=0$.
Likewise, in the first two cases, one of the indicator functions takes on the value $1$ and the other $0$. So $D=D_1+D_2$.
You can deal with the remaining cases similarly.
answered Dec 28 '18 at 20:20
Anurag AAnurag A
26.4k12351
$endgroup$
$begingroup$
Thank you Anurag :)
$endgroup$
– Tom1999
Dec 28 '18 at 21:43
add a comment |
$begingroup$
The possible positions of the mother and two daughters are as follows: (where $x$ represents a person other than the daughters)
begin{align*}
ldots,x,m,d_i, x,x,ldots\
ldots,d_i,m,x,x,ldots\
ldots,d_i,m,d_j, x,x,ldots\
ldots,x,m,x,x,ldots\
end{align*}
In the first two cases, $D=1$, in the third case $D=2$, and in the last case $D=0$.
Likewise, in the first two cases, one of the indicator functions takes on the value $1$ and the other $0$. So $D=D_1+D_2$.
You can deal with the remaining cases similarly.
answered Dec 28 '18 at 20:20
Anurag AAnurag A
26.4k12351
$endgroup$
The possible positions of the mother and two daughters are as follows: (where $x$ represents a person other than the daughters)
begin{align*}
ldots,x,m,d_i, x,x,ldots\
ldots,d_i,m,x,x,ldots\
ldots,d_i,m,d_j, x,x,ldots\
ldots,x,m,x,x,ldots\
end{align*}
In the first two cases, $D=1$, in the third case $D=2$, and in the last case $D=0$.
Likewise, in the first two cases, one of the indicator functions takes on the value $1$ and the other $0$. So $D=D_1+D_2$.
You can deal with the remaining cases similarly.
answered Dec 28 '18 at 20:20
Anurag AAnurag A
26.4k12351
edited Dec 28 '18 at 20:45
answered Dec 28 '18 at 20:20
Anurag AAnurag A
26.4k12351
answered Dec 28 '18 at 20:20
Anurag AAnurag A
26.4k12351
answered Dec 28 '18 at 20:20
Anurag AAnurag A
26.4k12351
26.4k12351
$begingroup$
Thank you Anurag :)
$endgroup$
– Tom1999
Dec 28 '18 at 21:43
add a comment |
$begingroup$
Thank you Anurag :)
$endgroup$
– Tom1999
Dec 28 '18 at 21:43
$begingroup$
Thank you Anurag :)
$endgroup$
– Tom1999
Dec 28 '18 at 21:43
$begingroup$
Thank you Anurag :)
$endgroup$
– Tom1999
Dec 28 '18 at 21:43
add a comment |
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$begingroup$
What part don't you get? Do you know the definition of an indicator variable?
$endgroup$
– lulu
Dec 28 '18 at 20:15