Probability with Births and Discrete Math
$begingroup$
A couple is planning to have a family. Let us assume that the probability of having a girl is 0.48 and a boy is 0.52,and that the gender of this couple’s children are pairwise independent. They want to have at least one girl and at least one boy. At the same time, they know that raising too many kids is difficult. So here’s what they plan to do: they’ll keep trying to have children until they have at least one girl and at least one boy or until they have four kids. Once one of these two conditions are satisfied, they’ll stop. Our goal is to determine the expected number of children this couple will have.
Let X(s) be equal to the number of children with outcome s; e.g., X(GGB) = 3.
a. What are the possible values of X(s)?
b. For each such value i, list the outcomes in the event (X = i). For example, the outcome GGB is part of the event (X = 3).
c. For each such value i, what is P(X = i)? Keep in mind that P(G) = 0.48, P(B) = 0.52 and the gender of the couple’s children a independent of each other.
d. Finally, what isE[X]? That is, on average, how many kids will such a couple have?
I think I am very close to solving this, for A and B I have the possible outcomes of:
x = 1 -> G , B
x = 2 - > GG , GB , BB , BG
x = 3 - > GGG , GGB , BBB , BBG
x = 4 -> GGGG , GGGB , BBBB , BBBG
I have it like this because its impossible to have something like BBGG because once they would have a boy and a girl they would be done so they could never have 2 of each or more after the first boy and girl appear, or until 4.
For part C, im not sure if I am supposed to do this for each one or not. Here is what I was thinking I would do, e.g (x = 3) -> GGB would be (.48)(.48)(.52) = .12
But what I don't know is if GGG is even a possibility because technically its possible but we wouldn't be done at this point because that doesn't satisfy the condition of 4 children or at least 1 boy and 1 girl, so I don't know if I should find the probability to this or even include it .
Then for part D Im not sure of exactly how I would go about this, I was thinking something like $E[X] = E[X_1] + E[X_2] + E[X_3] + E[X_4] $ , but im not sure quite what to put in to these.
probability discrete-mathematics probability-distributions
asked Dec 9 '18 at 21:57
STPM222STPM222
1
$endgroup$
add a comment |
$begingroup$
A couple is planning to have a family. Let us assume that the probability of having a girl is 0.48 and a boy is 0.52,and that the gender of this couple’s children are pairwise independent. They want to have at least one girl and at least one boy. At the same time, they know that raising too many kids is difficult. So here’s what they plan to do: they’ll keep trying to have children until they have at least one girl and at least one boy or until they have four kids. Once one of these two conditions are satisfied, they’ll stop. Our goal is to determine the expected number of children this couple will have.
Let X(s) be equal to the number of children with outcome s; e.g., X(GGB) = 3.
a. What are the possible values of X(s)?
b. For each such value i, list the outcomes in the event (X = i). For example, the outcome GGB is part of the event (X = 3).
c. For each such value i, what is P(X = i)? Keep in mind that P(G) = 0.48, P(B) = 0.52 and the gender of the couple’s children a independent of each other.
d. Finally, what isE[X]? That is, on average, how many kids will such a couple have?
I think I am very close to solving this, for A and B I have the possible outcomes of:
x = 1 -> G , B
x = 2 - > GG , GB , BB , BG
x = 3 - > GGG , GGB , BBB , BBG
x = 4 -> GGGG , GGGB , BBBB , BBBG
I have it like this because its impossible to have something like BBGG because once they would have a boy and a girl they would be done so they could never have 2 of each or more after the first boy and girl appear, or until 4.
For part C, im not sure if I am supposed to do this for each one or not. Here is what I was thinking I would do, e.g (x = 3) -> GGB would be (.48)(.48)(.52) = .12
But what I don't know is if GGG is even a possibility because technically its possible but we wouldn't be done at this point because that doesn't satisfy the condition of 4 children or at least 1 boy and 1 girl, so I don't know if I should find the probability to this or even include it .
Then for part D Im not sure of exactly how I would go about this, I was thinking something like $E[X] = E[X_1] + E[X_2] + E[X_3] + E[X_4] $ , but im not sure quite what to put in to these.
probability discrete-mathematics probability-distributions
asked Dec 9 '18 at 21:57
STPM222STPM222
1
$endgroup$
$begingroup$
Notice that the event $X = 1$ (having a total of one child) cannot occur since the couple continues to have children until they have both a boy and a girl or have four children.
$endgroup$
– N. F. Taussig
Dec 9 '18 at 22:01
$begingroup$
There are of course some unjustified assumptions in the problem statement itself, e.g., that trying hard enough suffices to have four kids. I suppose the error introdiced by these assumptions outweighs the second decimal difference between boys and girls (and other phenotypes)
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 22:03
add a comment |
$begingroup$
A couple is planning to have a family. Let us assume that the probability of having a girl is 0.48 and a boy is 0.52,and that the gender of this couple’s children are pairwise independent. They want to have at least one girl and at least one boy. At the same time, they know that raising too many kids is difficult. So here’s what they plan to do: they’ll keep trying to have children until they have at least one girl and at least one boy or until they have four kids. Once one of these two conditions are satisfied, they’ll stop. Our goal is to determine the expected number of children this couple will have.
Let X(s) be equal to the number of children with outcome s; e.g., X(GGB) = 3.
a. What are the possible values of X(s)?
b. For each such value i, list the outcomes in the event (X = i). For example, the outcome GGB is part of the event (X = 3).
c. For each such value i, what is P(X = i)? Keep in mind that P(G) = 0.48, P(B) = 0.52 and the gender of the couple’s children a independent of each other.
d. Finally, what isE[X]? That is, on average, how many kids will such a couple have?
I think I am very close to solving this, for A and B I have the possible outcomes of:
x = 1 -> G , B
x = 2 - > GG , GB , BB , BG
x = 3 - > GGG , GGB , BBB , BBG
x = 4 -> GGGG , GGGB , BBBB , BBBG
I have it like this because its impossible to have something like BBGG because once they would have a boy and a girl they would be done so they could never have 2 of each or more after the first boy and girl appear, or until 4.
For part C, im not sure if I am supposed to do this for each one or not. Here is what I was thinking I would do, e.g (x = 3) -> GGB would be (.48)(.48)(.52) = .12
But what I don't know is if GGG is even a possibility because technically its possible but we wouldn't be done at this point because that doesn't satisfy the condition of 4 children or at least 1 boy and 1 girl, so I don't know if I should find the probability to this or even include it .
Then for part D Im not sure of exactly how I would go about this, I was thinking something like $E[X] = E[X_1] + E[X_2] + E[X_3] + E[X_4] $ , but im not sure quite what to put in to these.
probability discrete-mathematics probability-distributions
asked Dec 9 '18 at 21:57
STPM222STPM222
1
$endgroup$
A couple is planning to have a family. Let us assume that the probability of having a girl is 0.48 and a boy is 0.52,and that the gender of this couple’s children are pairwise independent. They want to have at least one girl and at least one boy. At the same time, they know that raising too many kids is difficult. So here’s what they plan to do: they’ll keep trying to have children until they have at least one girl and at least one boy or until they have four kids. Once one of these two conditions are satisfied, they’ll stop. Our goal is to determine the expected number of children this couple will have.
Let X(s) be equal to the number of children with outcome s; e.g., X(GGB) = 3.
a. What are the possible values of X(s)?
b. For each such value i, list the outcomes in the event (X = i). For example, the outcome GGB is part of the event (X = 3).
c. For each such value i, what is P(X = i)? Keep in mind that P(G) = 0.48, P(B) = 0.52 and the gender of the couple’s children a independent of each other.
d. Finally, what isE[X]? That is, on average, how many kids will such a couple have?
I think I am very close to solving this, for A and B I have the possible outcomes of:
x = 1 -> G , B
x = 2 - > GG , GB , BB , BG
x = 3 - > GGG , GGB , BBB , BBG
x = 4 -> GGGG , GGGB , BBBB , BBBG
I have it like this because its impossible to have something like BBGG because once they would have a boy and a girl they would be done so they could never have 2 of each or more after the first boy and girl appear, or until 4.
For part C, im not sure if I am supposed to do this for each one or not. Here is what I was thinking I would do, e.g (x = 3) -> GGB would be (.48)(.48)(.52) = .12
But what I don't know is if GGG is even a possibility because technically its possible but we wouldn't be done at this point because that doesn't satisfy the condition of 4 children or at least 1 boy and 1 girl, so I don't know if I should find the probability to this or even include it .
Then for part D Im not sure of exactly how I would go about this, I was thinking something like $E[X] = E[X_1] + E[X_2] + E[X_3] + E[X_4] $ , but im not sure quite what to put in to these.
probability discrete-mathematics probability-distributions
probability discrete-mathematics probability-distributions
asked Dec 9 '18 at 21:57
STPM222STPM222
1
asked Dec 9 '18 at 21:57
STPM222STPM222
1
asked Dec 9 '18 at 21:57
STPM222STPM222
1
asked Dec 9 '18 at 21:57
STPM222STPM222
1
asked Dec 9 '18 at 21:57
STPM222STPM222
1
1
$begingroup$
Notice that the event $X = 1$ (having a total of one child) cannot occur since the couple continues to have children until they have both a boy and a girl or have four children.
$endgroup$
– N. F. Taussig
Dec 9 '18 at 22:01
$begingroup$
There are of course some unjustified assumptions in the problem statement itself, e.g., that trying hard enough suffices to have four kids. I suppose the error introdiced by these assumptions outweighs the second decimal difference between boys and girls (and other phenotypes)
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 22:03
add a comment |
$begingroup$
Notice that the event $X = 1$ (having a total of one child) cannot occur since the couple continues to have children until they have both a boy and a girl or have four children.
$endgroup$
– N. F. Taussig
Dec 9 '18 at 22:01
$begingroup$
There are of course some unjustified assumptions in the problem statement itself, e.g., that trying hard enough suffices to have four kids. I suppose the error introdiced by these assumptions outweighs the second decimal difference between boys and girls (and other phenotypes)
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 22:03
$begingroup$
Notice that the event $X = 1$ (having a total of one child) cannot occur since the couple continues to have children until they have both a boy and a girl or have four children.
$endgroup$
– N. F. Taussig
Dec 9 '18 at 22:01
$begingroup$
Notice that the event $X = 1$ (having a total of one child) cannot occur since the couple continues to have children until they have both a boy and a girl or have four children.
$endgroup$
– N. F. Taussig
Dec 9 '18 at 22:01
$begingroup$
There are of course some unjustified assumptions in the problem statement itself, e.g., that trying hard enough suffices to have four kids. I suppose the error introdiced by these assumptions outweighs the second decimal difference between boys and girls (and other phenotypes)
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 22:03
$begingroup$
There are of course some unjustified assumptions in the problem statement itself, e.g., that trying hard enough suffices to have four kids. I suppose the error introdiced by these assumptions outweighs the second decimal difference between boys and girls (and other phenotypes)
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 22:03
add a comment |
1 Answer
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$begingroup$
For A, note that $X(S)$ cannot be $1$: they can't stop with just one child. So, you can only have 2,3, or 4.
For B, note that $BBB$ and $GGG$ are not possible outcomes: the couple would go for a fourth child.
For part C, im not sure if I am supposed to do this for each one or not. Here is what I was thinking I would do, e.g (x = 3) -> GGB would be (.48)(.48)(.52) = .12
You're right that $P(GGB)=(.48)(.48)(.52)$
But you also need to compute the probabilities of the other outcomes leading to $3$ children
But what I don't know is if GGG is even a possibility because technically its possible but we wouldn't be done at this point because that doesn't satisfy the condition of 4 children or at least 1 boy and 1 girl, so I don't know if I should find the probability to this or even include it .
That's right: $GGG$ should not be included (nor $BBB$). So the only other one to consider is $BBG$.
So: $P(X=3)=P(GGB)+P(BBG)=(.48)(.48)(.52)+(.52)(.52)(.48)$
Then for part D Im not sure of exactly how I would go about this, I was thinking something like $E[X] = E[X_1] + E[X_2] + E[X_3] + E[X_4] $ , but im not sure quite what to put in to these.
You need to multiply the number of children times the probability of that number. So, you get:
$E[X] = 2*P[X=2] + 3*P[X=3] + 4*P[X=4] $
answered Dec 9 '18 at 22:05
Bram28Bram28
61.8k44793
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1 Answer
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$begingroup$
For A, note that $X(S)$ cannot be $1$: they can't stop with just one child. So, you can only have 2,3, or 4.
For B, note that $BBB$ and $GGG$ are not possible outcomes: the couple would go for a fourth child.
For part C, im not sure if I am supposed to do this for each one or not. Here is what I was thinking I would do, e.g (x = 3) -> GGB would be (.48)(.48)(.52) = .12
You're right that $P(GGB)=(.48)(.48)(.52)$
But you also need to compute the probabilities of the other outcomes leading to $3$ children
But what I don't know is if GGG is even a possibility because technically its possible but we wouldn't be done at this point because that doesn't satisfy the condition of 4 children or at least 1 boy and 1 girl, so I don't know if I should find the probability to this or even include it .
That's right: $GGG$ should not be included (nor $BBB$). So the only other one to consider is $BBG$.
So: $P(X=3)=P(GGB)+P(BBG)=(.48)(.48)(.52)+(.52)(.52)(.48)$
Then for part D Im not sure of exactly how I would go about this, I was thinking something like $E[X] = E[X_1] + E[X_2] + E[X_3] + E[X_4] $ , but im not sure quite what to put in to these.
You need to multiply the number of children times the probability of that number. So, you get:
$E[X] = 2*P[X=2] + 3*P[X=3] + 4*P[X=4] $
answered Dec 9 '18 at 22:05
Bram28Bram28
61.8k44793
$endgroup$
add a comment |
$begingroup$
For A, note that $X(S)$ cannot be $1$: they can't stop with just one child. So, you can only have 2,3, or 4.
For B, note that $BBB$ and $GGG$ are not possible outcomes: the couple would go for a fourth child.
For part C, im not sure if I am supposed to do this for each one or not. Here is what I was thinking I would do, e.g (x = 3) -> GGB would be (.48)(.48)(.52) = .12
You're right that $P(GGB)=(.48)(.48)(.52)$
But you also need to compute the probabilities of the other outcomes leading to $3$ children
But what I don't know is if GGG is even a possibility because technically its possible but we wouldn't be done at this point because that doesn't satisfy the condition of 4 children or at least 1 boy and 1 girl, so I don't know if I should find the probability to this or even include it .
That's right: $GGG$ should not be included (nor $BBB$). So the only other one to consider is $BBG$.
So: $P(X=3)=P(GGB)+P(BBG)=(.48)(.48)(.52)+(.52)(.52)(.48)$
Then for part D Im not sure of exactly how I would go about this, I was thinking something like $E[X] = E[X_1] + E[X_2] + E[X_3] + E[X_4] $ , but im not sure quite what to put in to these.
You need to multiply the number of children times the probability of that number. So, you get:
$E[X] = 2*P[X=2] + 3*P[X=3] + 4*P[X=4] $
answered Dec 9 '18 at 22:05
Bram28Bram28
61.8k44793
$endgroup$
add a comment |
$begingroup$
For A, note that $X(S)$ cannot be $1$: they can't stop with just one child. So, you can only have 2,3, or 4.
For B, note that $BBB$ and $GGG$ are not possible outcomes: the couple would go for a fourth child.
For part C, im not sure if I am supposed to do this for each one or not. Here is what I was thinking I would do, e.g (x = 3) -> GGB would be (.48)(.48)(.52) = .12
You're right that $P(GGB)=(.48)(.48)(.52)$
But you also need to compute the probabilities of the other outcomes leading to $3$ children
But what I don't know is if GGG is even a possibility because technically its possible but we wouldn't be done at this point because that doesn't satisfy the condition of 4 children or at least 1 boy and 1 girl, so I don't know if I should find the probability to this or even include it .
That's right: $GGG$ should not be included (nor $BBB$). So the only other one to consider is $BBG$.
So: $P(X=3)=P(GGB)+P(BBG)=(.48)(.48)(.52)+(.52)(.52)(.48)$
Then for part D Im not sure of exactly how I would go about this, I was thinking something like $E[X] = E[X_1] + E[X_2] + E[X_3] + E[X_4] $ , but im not sure quite what to put in to these.
You need to multiply the number of children times the probability of that number. So, you get:
$E[X] = 2*P[X=2] + 3*P[X=3] + 4*P[X=4] $
answered Dec 9 '18 at 22:05
Bram28Bram28
61.8k44793
$endgroup$
For A, note that $X(S)$ cannot be $1$: they can't stop with just one child. So, you can only have 2,3, or 4.
For B, note that $BBB$ and $GGG$ are not possible outcomes: the couple would go for a fourth child.
For part C, im not sure if I am supposed to do this for each one or not. Here is what I was thinking I would do, e.g (x = 3) -> GGB would be (.48)(.48)(.52) = .12
You're right that $P(GGB)=(.48)(.48)(.52)$
But you also need to compute the probabilities of the other outcomes leading to $3$ children
But what I don't know is if GGG is even a possibility because technically its possible but we wouldn't be done at this point because that doesn't satisfy the condition of 4 children or at least 1 boy and 1 girl, so I don't know if I should find the probability to this or even include it .
That's right: $GGG$ should not be included (nor $BBB$). So the only other one to consider is $BBG$.
So: $P(X=3)=P(GGB)+P(BBG)=(.48)(.48)(.52)+(.52)(.52)(.48)$
Then for part D Im not sure of exactly how I would go about this, I was thinking something like $E[X] = E[X_1] + E[X_2] + E[X_3] + E[X_4] $ , but im not sure quite what to put in to these.
You need to multiply the number of children times the probability of that number. So, you get:
$E[X] = 2*P[X=2] + 3*P[X=3] + 4*P[X=4] $
answered Dec 9 '18 at 22:05
Bram28Bram28
61.8k44793
edited Dec 9 '18 at 22:12
answered Dec 9 '18 at 22:05
Bram28Bram28
61.8k44793
answered Dec 9 '18 at 22:05
Bram28Bram28
61.8k44793
answered Dec 9 '18 at 22:05
Bram28Bram28
61.8k44793
61.8k44793
add a comment |
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$begingroup$
Notice that the event $X = 1$ (having a total of one child) cannot occur since the couple continues to have children until they have both a boy and a girl or have four children.
$endgroup$
– N. F. Taussig
Dec 9 '18 at 22:01
$begingroup$
There are of course some unjustified assumptions in the problem statement itself, e.g., that trying hard enough suffices to have four kids. I suppose the error introdiced by these assumptions outweighs the second decimal difference between boys and girls (and other phenotypes)
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 22:03