PMF of throwing a die 4 times
Clash Royale CLAN TAG#URR8PPP
$begingroup$
We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
What is the probability that $X ge 4$?
Compute the PMF of $X$.
Determine the mean and variance of $X$.
My attempt:
$(1/2)^4$ because that means each of the $4$ rolls, you either get a $4, 5$ or $6$.
for $X=1: (1/6)(6/6)(6/6)(6/6)$
for $X=2: (1/6)(5/6)(5/6)(5/6)$
for $X=3: (1/6)(4/6)(4/6)(4/6)$
for $X=4: (1/6)(3/6)(3/6)(3/6)$
for $X=5: (1/6)(2/6)(2/6)(2/6)$
for $X=6: (1/6)(1/6)(1/6)(1/6)$I can calculate this once I know I did the PMF correctly
Did I do (1) and (2) correctly?
probability probability-distributions dice
edited Feb 13 at 4:08
YuiTo Cheng
1,9452633
asked Feb 7 at 19:25
SraSra
324
$endgroup$
|
show 2 more comments
$begingroup$
We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
What is the probability that $X ge 4$?
Compute the PMF of $X$.
Determine the mean and variance of $X$.
My attempt:
$(1/2)^4$ because that means each of the $4$ rolls, you either get a $4, 5$ or $6$.
for $X=1: (1/6)(6/6)(6/6)(6/6)$
for $X=2: (1/6)(5/6)(5/6)(5/6)$
for $X=3: (1/6)(4/6)(4/6)(4/6)$
for $X=4: (1/6)(3/6)(3/6)(3/6)$
for $X=5: (1/6)(2/6)(2/6)(2/6)$
for $X=6: (1/6)(1/6)(1/6)(1/6)$I can calculate this once I know I did the PMF correctly
Did I do (1) and (2) correctly?
probability probability-distributions dice
edited Feb 13 at 4:08
YuiTo Cheng
1,9452633
asked Feb 7 at 19:25
SraSra
324
$endgroup$
$begingroup$
For (1) you need $left(frac12right)^4$
$endgroup$
– Daniel Mathias
Feb 7 at 19:29
$begingroup$
ad a) But you have to roll 4 times. So $P(X=4)=left(frac12right)^4$ And $P(X=5)=left(frac13right)^4$ And al last $P(X=6)=left(frac16right)^4$. To get $P(Xgeq 4)$ you have to sum up the probabilities.
$endgroup$
– callculus
Feb 7 at 19:31
$begingroup$
@calculus that should be $P(Xge4)$ and $P(Xge5)$
$endgroup$
– Daniel Mathias
Feb 7 at 19:34
$begingroup$
@DanielMathias No, since $X$ denote the minimal value rolled
$endgroup$
– callculus
Feb 7 at 19:35
$begingroup$
@callculus for $X=4$, one of the rolls must be 4. But $left(frac12right)^4$ includes the events that all rolls are greater than 4.
$endgroup$
– Daniel Mathias
Feb 7 at 19:38
|
show 2 more comments
$begingroup$
We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
What is the probability that $X ge 4$?
Compute the PMF of $X$.
Determine the mean and variance of $X$.
My attempt:
$(1/2)^4$ because that means each of the $4$ rolls, you either get a $4, 5$ or $6$.
for $X=1: (1/6)(6/6)(6/6)(6/6)$
for $X=2: (1/6)(5/6)(5/6)(5/6)$
for $X=3: (1/6)(4/6)(4/6)(4/6)$
for $X=4: (1/6)(3/6)(3/6)(3/6)$
for $X=5: (1/6)(2/6)(2/6)(2/6)$
for $X=6: (1/6)(1/6)(1/6)(1/6)$I can calculate this once I know I did the PMF correctly
Did I do (1) and (2) correctly?
probability probability-distributions dice
edited Feb 13 at 4:08
YuiTo Cheng
1,9452633
asked Feb 7 at 19:25
SraSra
324
$endgroup$
We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
What is the probability that $X ge 4$?
Compute the PMF of $X$.
Determine the mean and variance of $X$.
My attempt:
$(1/2)^4$ because that means each of the $4$ rolls, you either get a $4, 5$ or $6$.
for $X=1: (1/6)(6/6)(6/6)(6/6)$
for $X=2: (1/6)(5/6)(5/6)(5/6)$
for $X=3: (1/6)(4/6)(4/6)(4/6)$
for $X=4: (1/6)(3/6)(3/6)(3/6)$
for $X=5: (1/6)(2/6)(2/6)(2/6)$
for $X=6: (1/6)(1/6)(1/6)(1/6)$I can calculate this once I know I did the PMF correctly
Did I do (1) and (2) correctly?
probability probability-distributions dice
probability probability-distributions dice
edited Feb 13 at 4:08
YuiTo Cheng
1,9452633
asked Feb 7 at 19:25
SraSra
324
edited Feb 13 at 4:08
YuiTo Cheng
1,9452633
asked Feb 7 at 19:25
SraSra
324
edited Feb 13 at 4:08
YuiTo Cheng
1,9452633
edited Feb 13 at 4:08
YuiTo Cheng
1,9452633
edited Feb 13 at 4:08
YuiTo Cheng
1,9452633
1,9452633
asked Feb 7 at 19:25
SraSra
324
asked Feb 7 at 19:25
SraSra
324
asked Feb 7 at 19:25
SraSra
324
324
$begingroup$
For (1) you need $left(frac12right)^4$
$endgroup$
– Daniel Mathias
Feb 7 at 19:29
$begingroup$
ad a) But you have to roll 4 times. So $P(X=4)=left(frac12right)^4$ And $P(X=5)=left(frac13right)^4$ And al last $P(X=6)=left(frac16right)^4$. To get $P(Xgeq 4)$ you have to sum up the probabilities.
$endgroup$
– callculus
Feb 7 at 19:31
$begingroup$
@calculus that should be $P(Xge4)$ and $P(Xge5)$
$endgroup$
– Daniel Mathias
Feb 7 at 19:34
$begingroup$
@DanielMathias No, since $X$ denote the minimal value rolled
$endgroup$
– callculus
Feb 7 at 19:35
$begingroup$
@callculus for $X=4$, one of the rolls must be 4. But $left(frac12right)^4$ includes the events that all rolls are greater than 4.
$endgroup$
– Daniel Mathias
Feb 7 at 19:38
|
show 2 more comments
$begingroup$
For (1) you need $left(frac12right)^4$
$endgroup$
– Daniel Mathias
Feb 7 at 19:29
$begingroup$
ad a) But you have to roll 4 times. So $P(X=4)=left(frac12right)^4$ And $P(X=5)=left(frac13right)^4$ And al last $P(X=6)=left(frac16right)^4$. To get $P(Xgeq 4)$ you have to sum up the probabilities.
$endgroup$
– callculus
Feb 7 at 19:31
$begingroup$
@calculus that should be $P(Xge4)$ and $P(Xge5)$
$endgroup$
– Daniel Mathias
Feb 7 at 19:34
$begingroup$
@DanielMathias No, since $X$ denote the minimal value rolled
$endgroup$
– callculus
Feb 7 at 19:35
$begingroup$
@callculus for $X=4$, one of the rolls must be 4. But $left(frac12right)^4$ includes the events that all rolls are greater than 4.
$endgroup$
– Daniel Mathias
Feb 7 at 19:38
$begingroup$
For (1) you need $left(frac12right)^4$
$endgroup$
– Daniel Mathias
Feb 7 at 19:29
$begingroup$
For (1) you need $left(frac12right)^4$
$endgroup$
– Daniel Mathias
Feb 7 at 19:29
$begingroup$
ad a) But you have to roll 4 times. So $P(X=4)=left(frac12right)^4$ And $P(X=5)=left(frac13right)^4$ And al last $P(X=6)=left(frac16right)^4$. To get $P(Xgeq 4)$ you have to sum up the probabilities.
$endgroup$
– callculus
Feb 7 at 19:31
$begingroup$
ad a) But you have to roll 4 times. So $P(X=4)=left(frac12right)^4$ And $P(X=5)=left(frac13right)^4$ And al last $P(X=6)=left(frac16right)^4$. To get $P(Xgeq 4)$ you have to sum up the probabilities.
$endgroup$
– callculus
Feb 7 at 19:31
$begingroup$
@calculus that should be $P(Xge4)$ and $P(Xge5)$
$endgroup$
– Daniel Mathias
Feb 7 at 19:34
$begingroup$
@calculus that should be $P(Xge4)$ and $P(Xge5)$
$endgroup$
– Daniel Mathias
Feb 7 at 19:34
$begingroup$
@DanielMathias No, since $X$ denote the minimal value rolled
$endgroup$
– callculus
Feb 7 at 19:35
$begingroup$
@DanielMathias No, since $X$ denote the minimal value rolled
$endgroup$
– callculus
Feb 7 at 19:35
$begingroup$
@callculus for $X=4$, one of the rolls must be 4. But $left(frac12right)^4$ includes the events that all rolls are greater than 4.
$endgroup$
– Daniel Mathias
Feb 7 at 19:38
$begingroup$
@callculus for $X=4$, one of the rolls must be 4. But $left(frac12right)^4$ includes the events that all rolls are greater than 4.
$endgroup$
– Daniel Mathias
Feb 7 at 19:38
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
What is the probability that $X ge 4$?
For each roll, the value must be greater than three. This event has probability $frac36=frac12$. As this must occur on each of $4$ rolls, we have:
$$P(Xge4)=left(frac12right)^4=frac116$$
Compute the PMF of $X$.
$$P(X=1)=P(X>0)-P(X>1)=1-left(frac56right)^4=1-frac6251296=frac6711296$$
$$P(X=2)=P(X>1)-P(X>2)=left(frac56right)^4-left(frac46right)^4=frac6251296-frac2561296=frac3691296$$
$$P(X=3)=P(X>2)-P(X>3)=left(frac46right)^4-left(frac36right)^4=frac2561296-frac811296=frac1751296$$
$$P(X=4)=P(X>3)-P(X>4)=left(frac36right)^4-left(frac26right)^4=frac811296-frac161296=frac651296$$
$$P(X=5)=P(X>4)-P(X>5)=left(frac26right)^4-left(frac16right)^4=frac161296-frac11296=frac151296$$
$$P(X=6)=P(X>5)-P(X>6)=left(frac16right)^4-0=frac11296$$
answered Feb 7 at 21:09
Daniel MathiasDaniel Mathias
1,31518
$endgroup$
add a comment |
$begingroup$
1) is now correct after your edit based on Daniel's comment.
2) is incorrect since a) it doesn't sum to 1 and b) $X = 4, 5, 6$ doesn't sum to what you got in 1).
My approach to this question is to break it down into a bunch of simpler problems as such:
For $P(X=1)$, let's consider the $4$ dice as dice $1, 2, 3, 4$. There are $4$ possibilities, the result can have $1,2,3,$ or $4$ ones in it.
$1$ one: $4 choose 1 (frac16)^1(frac56)^3$
$2$ ones: $4 choose 2 (frac16)^2(frac56)^2$
$3$ ones: $4 choose 3 (frac16)^3(frac56)^1$
$4$ ones: $4 choose 4 (frac16)^4(frac56)^0 = (frac16)^4$
The sum of the above will give you the $P(X=1)$.
Similarly, you can solve for the rest. Make sure that the end results sums to $1$ and $P(X=4) + P(X=5) + P(X=6) = (frac12)^4$.
answered Feb 7 at 21:09
DubsDubs
55926
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
What is the probability that $X ge 4$?
For each roll, the value must be greater than three. This event has probability $frac36=frac12$. As this must occur on each of $4$ rolls, we have:
$$P(Xge4)=left(frac12right)^4=frac116$$
Compute the PMF of $X$.
$$P(X=1)=P(X>0)-P(X>1)=1-left(frac56right)^4=1-frac6251296=frac6711296$$
$$P(X=2)=P(X>1)-P(X>2)=left(frac56right)^4-left(frac46right)^4=frac6251296-frac2561296=frac3691296$$
$$P(X=3)=P(X>2)-P(X>3)=left(frac46right)^4-left(frac36right)^4=frac2561296-frac811296=frac1751296$$
$$P(X=4)=P(X>3)-P(X>4)=left(frac36right)^4-left(frac26right)^4=frac811296-frac161296=frac651296$$
$$P(X=5)=P(X>4)-P(X>5)=left(frac26right)^4-left(frac16right)^4=frac161296-frac11296=frac151296$$
$$P(X=6)=P(X>5)-P(X>6)=left(frac16right)^4-0=frac11296$$
answered Feb 7 at 21:09
Daniel MathiasDaniel Mathias
1,31518
$endgroup$
add a comment |
$begingroup$
We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
What is the probability that $X ge 4$?
For each roll, the value must be greater than three. This event has probability $frac36=frac12$. As this must occur on each of $4$ rolls, we have:
$$P(Xge4)=left(frac12right)^4=frac116$$
Compute the PMF of $X$.
$$P(X=1)=P(X>0)-P(X>1)=1-left(frac56right)^4=1-frac6251296=frac6711296$$
$$P(X=2)=P(X>1)-P(X>2)=left(frac56right)^4-left(frac46right)^4=frac6251296-frac2561296=frac3691296$$
$$P(X=3)=P(X>2)-P(X>3)=left(frac46right)^4-left(frac36right)^4=frac2561296-frac811296=frac1751296$$
$$P(X=4)=P(X>3)-P(X>4)=left(frac36right)^4-left(frac26right)^4=frac811296-frac161296=frac651296$$
$$P(X=5)=P(X>4)-P(X>5)=left(frac26right)^4-left(frac16right)^4=frac161296-frac11296=frac151296$$
$$P(X=6)=P(X>5)-P(X>6)=left(frac16right)^4-0=frac11296$$
answered Feb 7 at 21:09
Daniel MathiasDaniel Mathias
1,31518
$endgroup$
add a comment |
$begingroup$
We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
What is the probability that $X ge 4$?
For each roll, the value must be greater than three. This event has probability $frac36=frac12$. As this must occur on each of $4$ rolls, we have:
$$P(Xge4)=left(frac12right)^4=frac116$$
Compute the PMF of $X$.
$$P(X=1)=P(X>0)-P(X>1)=1-left(frac56right)^4=1-frac6251296=frac6711296$$
$$P(X=2)=P(X>1)-P(X>2)=left(frac56right)^4-left(frac46right)^4=frac6251296-frac2561296=frac3691296$$
$$P(X=3)=P(X>2)-P(X>3)=left(frac46right)^4-left(frac36right)^4=frac2561296-frac811296=frac1751296$$
$$P(X=4)=P(X>3)-P(X>4)=left(frac36right)^4-left(frac26right)^4=frac811296-frac161296=frac651296$$
$$P(X=5)=P(X>4)-P(X>5)=left(frac26right)^4-left(frac16right)^4=frac161296-frac11296=frac151296$$
$$P(X=6)=P(X>5)-P(X>6)=left(frac16right)^4-0=frac11296$$
answered Feb 7 at 21:09
Daniel MathiasDaniel Mathias
1,31518
$endgroup$
We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
What is the probability that $X ge 4$?
For each roll, the value must be greater than three. This event has probability $frac36=frac12$. As this must occur on each of $4$ rolls, we have:
$$P(Xge4)=left(frac12right)^4=frac116$$
Compute the PMF of $X$.
$$P(X=1)=P(X>0)-P(X>1)=1-left(frac56right)^4=1-frac6251296=frac6711296$$
$$P(X=2)=P(X>1)-P(X>2)=left(frac56right)^4-left(frac46right)^4=frac6251296-frac2561296=frac3691296$$
$$P(X=3)=P(X>2)-P(X>3)=left(frac46right)^4-left(frac36right)^4=frac2561296-frac811296=frac1751296$$
$$P(X=4)=P(X>3)-P(X>4)=left(frac36right)^4-left(frac26right)^4=frac811296-frac161296=frac651296$$
$$P(X=5)=P(X>4)-P(X>5)=left(frac26right)^4-left(frac16right)^4=frac161296-frac11296=frac151296$$
$$P(X=6)=P(X>5)-P(X>6)=left(frac16right)^4-0=frac11296$$
answered Feb 7 at 21:09
Daniel MathiasDaniel Mathias
1,31518
answered Feb 7 at 21:09
Daniel MathiasDaniel Mathias
1,31518
answered Feb 7 at 21:09
Daniel MathiasDaniel Mathias
1,31518
answered Feb 7 at 21:09
Daniel MathiasDaniel Mathias
1,31518
1,31518
add a comment |
add a comment |
$begingroup$
1) is now correct after your edit based on Daniel's comment.
2) is incorrect since a) it doesn't sum to 1 and b) $X = 4, 5, 6$ doesn't sum to what you got in 1).
My approach to this question is to break it down into a bunch of simpler problems as such:
For $P(X=1)$, let's consider the $4$ dice as dice $1, 2, 3, 4$. There are $4$ possibilities, the result can have $1,2,3,$ or $4$ ones in it.
$1$ one: $4 choose 1 (frac16)^1(frac56)^3$
$2$ ones: $4 choose 2 (frac16)^2(frac56)^2$
$3$ ones: $4 choose 3 (frac16)^3(frac56)^1$
$4$ ones: $4 choose 4 (frac16)^4(frac56)^0 = (frac16)^4$
The sum of the above will give you the $P(X=1)$.
Similarly, you can solve for the rest. Make sure that the end results sums to $1$ and $P(X=4) + P(X=5) + P(X=6) = (frac12)^4$.
answered Feb 7 at 21:09
DubsDubs
55926
$endgroup$
add a comment |
$begingroup$
1) is now correct after your edit based on Daniel's comment.
2) is incorrect since a) it doesn't sum to 1 and b) $X = 4, 5, 6$ doesn't sum to what you got in 1).
My approach to this question is to break it down into a bunch of simpler problems as such:
For $P(X=1)$, let's consider the $4$ dice as dice $1, 2, 3, 4$. There are $4$ possibilities, the result can have $1,2,3,$ or $4$ ones in it.
$1$ one: $4 choose 1 (frac16)^1(frac56)^3$
$2$ ones: $4 choose 2 (frac16)^2(frac56)^2$
$3$ ones: $4 choose 3 (frac16)^3(frac56)^1$
$4$ ones: $4 choose 4 (frac16)^4(frac56)^0 = (frac16)^4$
The sum of the above will give you the $P(X=1)$.
Similarly, you can solve for the rest. Make sure that the end results sums to $1$ and $P(X=4) + P(X=5) + P(X=6) = (frac12)^4$.
answered Feb 7 at 21:09
DubsDubs
55926
$endgroup$
add a comment |
$begingroup$
1) is now correct after your edit based on Daniel's comment.
2) is incorrect since a) it doesn't sum to 1 and b) $X = 4, 5, 6$ doesn't sum to what you got in 1).
My approach to this question is to break it down into a bunch of simpler problems as such:
For $P(X=1)$, let's consider the $4$ dice as dice $1, 2, 3, 4$. There are $4$ possibilities, the result can have $1,2,3,$ or $4$ ones in it.
$1$ one: $4 choose 1 (frac16)^1(frac56)^3$
$2$ ones: $4 choose 2 (frac16)^2(frac56)^2$
$3$ ones: $4 choose 3 (frac16)^3(frac56)^1$
$4$ ones: $4 choose 4 (frac16)^4(frac56)^0 = (frac16)^4$
The sum of the above will give you the $P(X=1)$.
Similarly, you can solve for the rest. Make sure that the end results sums to $1$ and $P(X=4) + P(X=5) + P(X=6) = (frac12)^4$.
answered Feb 7 at 21:09
DubsDubs
55926
$endgroup$
1) is now correct after your edit based on Daniel's comment.
2) is incorrect since a) it doesn't sum to 1 and b) $X = 4, 5, 6$ doesn't sum to what you got in 1).
My approach to this question is to break it down into a bunch of simpler problems as such:
For $P(X=1)$, let's consider the $4$ dice as dice $1, 2, 3, 4$. There are $4$ possibilities, the result can have $1,2,3,$ or $4$ ones in it.
$1$ one: $4 choose 1 (frac16)^1(frac56)^3$
$2$ ones: $4 choose 2 (frac16)^2(frac56)^2$
$3$ ones: $4 choose 3 (frac16)^3(frac56)^1$
$4$ ones: $4 choose 4 (frac16)^4(frac56)^0 = (frac16)^4$
The sum of the above will give you the $P(X=1)$.
Similarly, you can solve for the rest. Make sure that the end results sums to $1$ and $P(X=4) + P(X=5) + P(X=6) = (frac12)^4$.
answered Feb 7 at 21:09
DubsDubs
55926
answered Feb 7 at 21:09
DubsDubs
55926
answered Feb 7 at 21:09
DubsDubs
55926
answered Feb 7 at 21:09
DubsDubs
55926
55926
add a comment |
add a comment |
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For (1) you need $left(frac12right)^4$
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– Daniel Mathias
Feb 7 at 19:29
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ad a) But you have to roll 4 times. So $P(X=4)=left(frac12right)^4$ And $P(X=5)=left(frac13right)^4$ And al last $P(X=6)=left(frac16right)^4$. To get $P(Xgeq 4)$ you have to sum up the probabilities.
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– callculus
Feb 7 at 19:31
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@calculus that should be $P(Xge4)$ and $P(Xge5)$
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– Daniel Mathias
Feb 7 at 19:34
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@DanielMathias No, since $X$ denote the minimal value rolled
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– callculus
Feb 7 at 19:35
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@callculus for $X=4$, one of the rolls must be 4. But $left(frac12right)^4$ includes the events that all rolls are greater than 4.
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– Daniel Mathias
Feb 7 at 19:38