Largest possible value of a probability
$begingroup$
Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.
Try
Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,
$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$
notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,
$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$
$$ P(X=3) = 1.8-p-2p^2 $$
take derivative with respect to $p$ gives
$$ (P(X=3))' = -1 - 4p $$
which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?
probability
edited Nov 28 '18 at 19:51
LuxGiammi
17410
asked Nov 28 '18 at 18:37
NeymarNeymar
354114
$endgroup$
add a comment |
$begingroup$
Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.
Try
Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,
$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$
notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,
$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$
$$ P(X=3) = 1.8-p-2p^2 $$
take derivative with respect to $p$ gives
$$ (P(X=3))' = -1 - 4p $$
which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?
probability
edited Nov 28 '18 at 19:51
LuxGiammi
17410
asked Nov 28 '18 at 18:37
NeymarNeymar
354114
$endgroup$
4
$begingroup$
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:32
add a comment |
$begingroup$
Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.
Try
Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,
$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$
notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,
$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$
$$ P(X=3) = 1.8-p-2p^2 $$
take derivative with respect to $p$ gives
$$ (P(X=3))' = -1 - 4p $$
which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?
probability
edited Nov 28 '18 at 19:51
LuxGiammi
17410
asked Nov 28 '18 at 18:37
NeymarNeymar
354114
$endgroup$
Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.
Try
Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,
$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$
notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,
$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$
$$ P(X=3) = 1.8-p-2p^2 $$
take derivative with respect to $p$ gives
$$ (P(X=3))' = -1 - 4p $$
which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?
probability
probability
edited Nov 28 '18 at 19:51
LuxGiammi
17410
asked Nov 28 '18 at 18:37
NeymarNeymar
354114
edited Nov 28 '18 at 19:51
LuxGiammi
17410
asked Nov 28 '18 at 18:37
NeymarNeymar
354114
edited Nov 28 '18 at 19:51
LuxGiammi
17410
edited Nov 28 '18 at 19:51
LuxGiammi
17410
edited Nov 28 '18 at 19:51
LuxGiammi
17410
17410
asked Nov 28 '18 at 18:37
NeymarNeymar
354114
asked Nov 28 '18 at 18:37
NeymarNeymar
354114
asked Nov 28 '18 at 18:37
NeymarNeymar
354114
354114
4
$begingroup$
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:32
add a comment |
4
$begingroup$
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:32
4
4
$begingroup$
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:32
$begingroup$
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If your trials are independent
$P(X=3)= 0.6^3$
But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.
All of the trials have equal likelihood of showing disease, but they are not independent.
If $q$ is big... i.e. let $q = 1$
Either all three samples show disease or none do.
$P(X=3) = 0.6$
At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.
$P(X=3) = 0$
answered Nov 28 '18 at 19:38
Doug MDoug M
44.4k31854
$endgroup$
add a comment |
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$begingroup$
If your trials are independent
$P(X=3)= 0.6^3$
But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.
All of the trials have equal likelihood of showing disease, but they are not independent.
If $q$ is big... i.e. let $q = 1$
Either all three samples show disease or none do.
$P(X=3) = 0.6$
At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.
$P(X=3) = 0$
answered Nov 28 '18 at 19:38
Doug MDoug M
44.4k31854
$endgroup$
add a comment |
$begingroup$
If your trials are independent
$P(X=3)= 0.6^3$
But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.
All of the trials have equal likelihood of showing disease, but they are not independent.
If $q$ is big... i.e. let $q = 1$
Either all three samples show disease or none do.
$P(X=3) = 0.6$
At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.
$P(X=3) = 0$
answered Nov 28 '18 at 19:38
Doug MDoug M
44.4k31854
$endgroup$
add a comment |
$begingroup$
If your trials are independent
$P(X=3)= 0.6^3$
But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.
All of the trials have equal likelihood of showing disease, but they are not independent.
If $q$ is big... i.e. let $q = 1$
Either all three samples show disease or none do.
$P(X=3) = 0.6$
At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.
$P(X=3) = 0$
answered Nov 28 '18 at 19:38
Doug MDoug M
44.4k31854
$endgroup$
If your trials are independent
$P(X=3)= 0.6^3$
But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.
All of the trials have equal likelihood of showing disease, but they are not independent.
If $q$ is big... i.e. let $q = 1$
Either all three samples show disease or none do.
$P(X=3) = 0.6$
At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.
$P(X=3) = 0$
answered Nov 28 '18 at 19:38
Doug MDoug M
44.4k31854
edited Nov 28 '18 at 19:48
answered Nov 28 '18 at 19:38
Doug MDoug M
44.4k31854
answered Nov 28 '18 at 19:38
Doug MDoug M
44.4k31854
answered Nov 28 '18 at 19:38
Doug MDoug M
44.4k31854
44.4k31854
add a comment |
add a comment |
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$begingroup$
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:32