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I am having trouble with a probability question regarding to G-distribution.
Background: Let's assume I have a disc that has two sides, one is black and one is white. Note: the disc has biased probability: the probability to get black is $p$ and the probability to get white is $1-p$. when u toss the disc time after time, the number of tosses ($Z$) until u get black at the first time distribute geometrically to: $Z$~$G(p)$ .
Note: formula for G-distribution: $P(Z=k)=p(1-p)^k-1$.
Questions:
1.let's toss the disc time after time until we get two black color in a row (for the first time).
1.a Find the probability function of the random variable: $T =$number of tosses until getting 2 black colors in a row for the first time. Make sure there normalization properties exists.
this is what i have achieved so far:
i know that i have to shoe that the sum of the probabilities of the experiment should equals to $1$.
$P(T=k)=((k-1)*(1-p)^k-2*p)*p$
the arguments in the brackets demonstrating one success of getting black color in $k-1$ tries.
the otter $p$ present success in the $k$-ith. time.
let's show normalization:
$fracp^2(1-p)^2sum_k=1^infty(k-1)(1-p)^k-2$
we know that $frac11-xsum_n=0^inftyx^n=> frac ddx=frac1(1-x)^2=sum_n=1^inftyn*x^n-1$
we multiply both sides with $x^2$ and then we will get :$frac (1-p)^2(1-(1-p)^2fracp^2(1-p)^2=1$. this is my answer but my professor told me its wrong and i don't understand why. if anyone can please help me understand what's wrong or give me a hint or a solution to this problem will be grateful.
probability probability-theory statistics probability-distributions geometric-probability
asked Aug 15 at 12:09
Tomer Levy
356
add a comment |Â
up vote
0
down vote
favorite
I am having trouble with a probability question regarding to G-distribution.
Background: Let's assume I have a disc that has two sides, one is black and one is white. Note: the disc has biased probability: the probability to get black is $p$ and the probability to get white is $1-p$. when u toss the disc time after time, the number of tosses ($Z$) until u get black at the first time distribute geometrically to: $Z$~$G(p)$ .
Note: formula for G-distribution: $P(Z=k)=p(1-p)^k-1$.
Questions:
1.let's toss the disc time after time until we get two black color in a row (for the first time).
1.a Find the probability function of the random variable: $T =$number of tosses until getting 2 black colors in a row for the first time. Make sure there normalization properties exists.
this is what i have achieved so far:
i know that i have to shoe that the sum of the probabilities of the experiment should equals to $1$.
$P(T=k)=((k-1)*(1-p)^k-2*p)*p$
the arguments in the brackets demonstrating one success of getting black color in $k-1$ tries.
the otter $p$ present success in the $k$-ith. time.
let's show normalization:
$fracp^2(1-p)^2sum_k=1^infty(k-1)(1-p)^k-2$
we know that $frac11-xsum_n=0^inftyx^n=> frac ddx=frac1(1-x)^2=sum_n=1^inftyn*x^n-1$
we multiply both sides with $x^2$ and then we will get :$frac (1-p)^2(1-(1-p)^2fracp^2(1-p)^2=1$. this is my answer but my professor told me its wrong and i don't understand why. if anyone can please help me understand what's wrong or give me a hint or a solution to this problem will be grateful.
probability probability-theory statistics probability-distributions geometric-probability
asked Aug 15 at 12:09
Tomer Levy
356
You have computed the probabilities for a different random variable, $T’$, which is the number of tosses it takes to get two black colors (not necessarily in a row).
– Mike Earnest
Aug 15 at 12:26
I just edited my post. if u can please help.
– Tomer Levy
Aug 15 at 13:23
Your expression for $P(T=k)$ is incorrect. $(k-1)(1-p)^k-1p^2$ is the probability that the first $k$ flips will contain exactly two blacks, with the second occurring at flip $k$. What you instead want to calculate is the probability that the first $k$ flips contain any number of black flips, as long as the only place where two consecutive black flips occur is in the last two flips. This is harder to write down a formula for.
– Mike Earnest
Aug 15 at 13:28
Am I understanding the problem correctly? In the below sequence of flips, is $T=10$? $$ W,B,W,W,B,W,B,W,B,B $$
– Mike Earnest
Aug 15 at 14:09
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am having trouble with a probability question regarding to G-distribution.
Background: Let's assume I have a disc that has two sides, one is black and one is white. Note: the disc has biased probability: the probability to get black is $p$ and the probability to get white is $1-p$. when u toss the disc time after time, the number of tosses ($Z$) until u get black at the first time distribute geometrically to: $Z$~$G(p)$ .
Note: formula for G-distribution: $P(Z=k)=p(1-p)^k-1$.
Questions:
1.let's toss the disc time after time until we get two black color in a row (for the first time).
1.a Find the probability function of the random variable: $T =$number of tosses until getting 2 black colors in a row for the first time. Make sure there normalization properties exists.
this is what i have achieved so far:
i know that i have to shoe that the sum of the probabilities of the experiment should equals to $1$.
$P(T=k)=((k-1)*(1-p)^k-2*p)*p$
the arguments in the brackets demonstrating one success of getting black color in $k-1$ tries.
the otter $p$ present success in the $k$-ith. time.
let's show normalization:
$fracp^2(1-p)^2sum_k=1^infty(k-1)(1-p)^k-2$
we know that $frac11-xsum_n=0^inftyx^n=> frac ddx=frac1(1-x)^2=sum_n=1^inftyn*x^n-1$
we multiply both sides with $x^2$ and then we will get :$frac (1-p)^2(1-(1-p)^2fracp^2(1-p)^2=1$. this is my answer but my professor told me its wrong and i don't understand why. if anyone can please help me understand what's wrong or give me a hint or a solution to this problem will be grateful.
probability probability-theory statistics probability-distributions geometric-probability
asked Aug 15 at 12:09
Tomer Levy
356
I am having trouble with a probability question regarding to G-distribution.
Background: Let's assume I have a disc that has two sides, one is black and one is white. Note: the disc has biased probability: the probability to get black is $p$ and the probability to get white is $1-p$. when u toss the disc time after time, the number of tosses ($Z$) until u get black at the first time distribute geometrically to: $Z$~$G(p)$ .
Note: formula for G-distribution: $P(Z=k)=p(1-p)^k-1$.
Questions:
1.let's toss the disc time after time until we get two black color in a row (for the first time).
1.a Find the probability function of the random variable: $T =$number of tosses until getting 2 black colors in a row for the first time. Make sure there normalization properties exists.
this is what i have achieved so far:
i know that i have to shoe that the sum of the probabilities of the experiment should equals to $1$.
$P(T=k)=((k-1)*(1-p)^k-2*p)*p$
the arguments in the brackets demonstrating one success of getting black color in $k-1$ tries.
the otter $p$ present success in the $k$-ith. time.
let's show normalization:
$fracp^2(1-p)^2sum_k=1^infty(k-1)(1-p)^k-2$
we know that $frac11-xsum_n=0^inftyx^n=> frac ddx=frac1(1-x)^2=sum_n=1^inftyn*x^n-1$
we multiply both sides with $x^2$ and then we will get :$frac (1-p)^2(1-(1-p)^2fracp^2(1-p)^2=1$. this is my answer but my professor told me its wrong and i don't understand why. if anyone can please help me understand what's wrong or give me a hint or a solution to this problem will be grateful.
probability probability-theory statistics probability-distributions geometric-probability
asked Aug 15 at 12:09
Tomer Levy
356
edited Aug 15 at 13:21
asked Aug 15 at 12:09
Tomer Levy
356
asked Aug 15 at 12:09
Tomer Levy
356
asked Aug 15 at 12:09
Tomer Levy
356
356
You have computed the probabilities for a different random variable, $T’$, which is the number of tosses it takes to get two black colors (not necessarily in a row).
– Mike Earnest
Aug 15 at 12:26
I just edited my post. if u can please help.
– Tomer Levy
Aug 15 at 13:23
Your expression for $P(T=k)$ is incorrect. $(k-1)(1-p)^k-1p^2$ is the probability that the first $k$ flips will contain exactly two blacks, with the second occurring at flip $k$. What you instead want to calculate is the probability that the first $k$ flips contain any number of black flips, as long as the only place where two consecutive black flips occur is in the last two flips. This is harder to write down a formula for.
– Mike Earnest
Aug 15 at 13:28
Am I understanding the problem correctly? In the below sequence of flips, is $T=10$? $$ W,B,W,W,B,W,B,W,B,B $$
– Mike Earnest
Aug 15 at 14:09
add a comment |Â
You have computed the probabilities for a different random variable, $T’$, which is the number of tosses it takes to get two black colors (not necessarily in a row).
– Mike Earnest
Aug 15 at 12:26
I just edited my post. if u can please help.
– Tomer Levy
Aug 15 at 13:23
Your expression for $P(T=k)$ is incorrect. $(k-1)(1-p)^k-1p^2$ is the probability that the first $k$ flips will contain exactly two blacks, with the second occurring at flip $k$. What you instead want to calculate is the probability that the first $k$ flips contain any number of black flips, as long as the only place where two consecutive black flips occur is in the last two flips. This is harder to write down a formula for.
– Mike Earnest
Aug 15 at 13:28
Am I understanding the problem correctly? In the below sequence of flips, is $T=10$? $$ W,B,W,W,B,W,B,W,B,B $$
– Mike Earnest
Aug 15 at 14:09
You have computed the probabilities for a different random variable, $T’$, which is the number of tosses it takes to get two black colors (not necessarily in a row).
– Mike Earnest
Aug 15 at 12:26
You have computed the probabilities for a different random variable, $T’$, which is the number of tosses it takes to get two black colors (not necessarily in a row).
– Mike Earnest
Aug 15 at 12:26
I just edited my post. if u can please help.
– Tomer Levy
Aug 15 at 13:23
I just edited my post. if u can please help.
– Tomer Levy
Aug 15 at 13:23
Your expression for $P(T=k)$ is incorrect. $(k-1)(1-p)^k-1p^2$ is the probability that the first $k$ flips will contain exactly two blacks, with the second occurring at flip $k$. What you instead want to calculate is the probability that the first $k$ flips contain any number of black flips, as long as the only place where two consecutive black flips occur is in the last two flips. This is harder to write down a formula for.
– Mike Earnest
Aug 15 at 13:28
Your expression for $P(T=k)$ is incorrect. $(k-1)(1-p)^k-1p^2$ is the probability that the first $k$ flips will contain exactly two blacks, with the second occurring at flip $k$. What you instead want to calculate is the probability that the first $k$ flips contain any number of black flips, as long as the only place where two consecutive black flips occur is in the last two flips. This is harder to write down a formula for.
– Mike Earnest
Aug 15 at 13:28
Am I understanding the problem correctly? In the below sequence of flips, is $T=10$? $$ W,B,W,W,B,W,B,W,B,B $$
– Mike Earnest
Aug 15 at 14:09
Am I understanding the problem correctly? In the below sequence of flips, is $T=10$? $$ W,B,W,W,B,W,B,W,B,B $$
– Mike Earnest
Aug 15 at 14:09
add a comment |Â
1 Answer
1
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up vote
0
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Let $a_k=P(T=k)$. You can show that
$$
a_k =(1-p)a_k-1+p(1-p)a_k-2,qquad k>2
$$
Why? There are two ways to have a sequence of $k>2$ coin flips whose first occurrence of $BB$ is at flips $k-1$ and $k$. Either you start with a white flip $W$, and then follow it up with a sequence which gets its first $BB$ at flips $k-2$ and $k-1$, or you start with $BW$ and get a sequence of $k-2$ flips ending in $BB$. You cannot start with $BB$, since that would imply $T=2$, and we are assuming $k>2$.
Combined with the base cases $a_1=0$, $a_2=p^2$, the above is a linear recurrence which can be solved using the methods in this article.
answered Aug 15 at 18:04
Mike Earnest
16.3k11648
so i tried to solve this linear recurrence, but i am getting solutions with roots in the process to solve this recurrence ( i also tried to use a variable to avoid roots, but no luck) and other stuff that are very not easy to solve. i really don't think my professor aimed for this solution. maybe there is another way you have in mind?
– Tomer Levy
Aug 16 at 7:17
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $a_k=P(T=k)$. You can show that
$$
a_k =(1-p)a_k-1+p(1-p)a_k-2,qquad k>2
$$
Why? There are two ways to have a sequence of $k>2$ coin flips whose first occurrence of $BB$ is at flips $k-1$ and $k$. Either you start with a white flip $W$, and then follow it up with a sequence which gets its first $BB$ at flips $k-2$ and $k-1$, or you start with $BW$ and get a sequence of $k-2$ flips ending in $BB$. You cannot start with $BB$, since that would imply $T=2$, and we are assuming $k>2$.
Combined with the base cases $a_1=0$, $a_2=p^2$, the above is a linear recurrence which can be solved using the methods in this article.
answered Aug 15 at 18:04
Mike Earnest
16.3k11648
so i tried to solve this linear recurrence, but i am getting solutions with roots in the process to solve this recurrence ( i also tried to use a variable to avoid roots, but no luck) and other stuff that are very not easy to solve. i really don't think my professor aimed for this solution. maybe there is another way you have in mind?
– Tomer Levy
Aug 16 at 7:17
add a comment |Â
up vote
0
down vote
Let $a_k=P(T=k)$. You can show that
$$
a_k =(1-p)a_k-1+p(1-p)a_k-2,qquad k>2
$$
Why? There are two ways to have a sequence of $k>2$ coin flips whose first occurrence of $BB$ is at flips $k-1$ and $k$. Either you start with a white flip $W$, and then follow it up with a sequence which gets its first $BB$ at flips $k-2$ and $k-1$, or you start with $BW$ and get a sequence of $k-2$ flips ending in $BB$. You cannot start with $BB$, since that would imply $T=2$, and we are assuming $k>2$.
Combined with the base cases $a_1=0$, $a_2=p^2$, the above is a linear recurrence which can be solved using the methods in this article.
answered Aug 15 at 18:04
Mike Earnest
16.3k11648
so i tried to solve this linear recurrence, but i am getting solutions with roots in the process to solve this recurrence ( i also tried to use a variable to avoid roots, but no luck) and other stuff that are very not easy to solve. i really don't think my professor aimed for this solution. maybe there is another way you have in mind?
– Tomer Levy
Aug 16 at 7:17
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $a_k=P(T=k)$. You can show that
$$
a_k =(1-p)a_k-1+p(1-p)a_k-2,qquad k>2
$$
Why? There are two ways to have a sequence of $k>2$ coin flips whose first occurrence of $BB$ is at flips $k-1$ and $k$. Either you start with a white flip $W$, and then follow it up with a sequence which gets its first $BB$ at flips $k-2$ and $k-1$, or you start with $BW$ and get a sequence of $k-2$ flips ending in $BB$. You cannot start with $BB$, since that would imply $T=2$, and we are assuming $k>2$.
Combined with the base cases $a_1=0$, $a_2=p^2$, the above is a linear recurrence which can be solved using the methods in this article.
answered Aug 15 at 18:04
Mike Earnest
16.3k11648
Let $a_k=P(T=k)$. You can show that
$$
a_k =(1-p)a_k-1+p(1-p)a_k-2,qquad k>2
$$
Why? There are two ways to have a sequence of $k>2$ coin flips whose first occurrence of $BB$ is at flips $k-1$ and $k$. Either you start with a white flip $W$, and then follow it up with a sequence which gets its first $BB$ at flips $k-2$ and $k-1$, or you start with $BW$ and get a sequence of $k-2$ flips ending in $BB$. You cannot start with $BB$, since that would imply $T=2$, and we are assuming $k>2$.
Combined with the base cases $a_1=0$, $a_2=p^2$, the above is a linear recurrence which can be solved using the methods in this article.
answered Aug 15 at 18:04
Mike Earnest
16.3k11648
answered Aug 15 at 18:04
Mike Earnest
16.3k11648
answered Aug 15 at 18:04
Mike Earnest
16.3k11648
answered Aug 15 at 18:04
Mike Earnest
16.3k11648
16.3k11648
so i tried to solve this linear recurrence, but i am getting solutions with roots in the process to solve this recurrence ( i also tried to use a variable to avoid roots, but no luck) and other stuff that are very not easy to solve. i really don't think my professor aimed for this solution. maybe there is another way you have in mind?
– Tomer Levy
Aug 16 at 7:17
add a comment |Â
so i tried to solve this linear recurrence, but i am getting solutions with roots in the process to solve this recurrence ( i also tried to use a variable to avoid roots, but no luck) and other stuff that are very not easy to solve. i really don't think my professor aimed for this solution. maybe there is another way you have in mind?
– Tomer Levy
Aug 16 at 7:17
so i tried to solve this linear recurrence, but i am getting solutions with roots in the process to solve this recurrence ( i also tried to use a variable to avoid roots, but no luck) and other stuff that are very not easy to solve. i really don't think my professor aimed for this solution. maybe there is another way you have in mind?
– Tomer Levy
Aug 16 at 7:17
so i tried to solve this linear recurrence, but i am getting solutions with roots in the process to solve this recurrence ( i also tried to use a variable to avoid roots, but no luck) and other stuff that are very not easy to solve. i really don't think my professor aimed for this solution. maybe there is another way you have in mind?
– Tomer Levy
Aug 16 at 7:17
add a comment |Â
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You have computed the probabilities for a different random variable, $T’$, which is the number of tosses it takes to get two black colors (not necessarily in a row).
– Mike Earnest
Aug 15 at 12:26
I just edited my post. if u can please help.
– Tomer Levy
Aug 15 at 13:23
Your expression for $P(T=k)$ is incorrect. $(k-1)(1-p)^k-1p^2$ is the probability that the first $k$ flips will contain exactly two blacks, with the second occurring at flip $k$. What you instead want to calculate is the probability that the first $k$ flips contain any number of black flips, as long as the only place where two consecutive black flips occur is in the last two flips. This is harder to write down a formula for.
– Mike Earnest
Aug 15 at 13:28
Am I understanding the problem correctly? In the below sequence of flips, is $T=10$? $$ W,B,W,W,B,W,B,W,B,B $$
– Mike Earnest
Aug 15 at 14:09