Probability that sheepdog performs at least one task successfuly

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What is the probability that a sheepdog performs at least $1$ of these tasks successfully?

My approach is to subtract the probability of performing at most $1$ of these tasks successfully from the probability of performing all $4$ tasks successfully.

$P(textfetch)=.9, P(textdrive)=.7, P(textherd)=.84, P(textseparate)=.75$.

The complement of these four probabilities is, $.1,.3,.16,$ and $,.25$, respectively.

So the probability that the sheepdog performs all four tasks successfully is simply, $(.9)(.7)(.84)(.75)$.

The probability that the sheepdog performs at most $1$ task successfully can be split into $4$ cases. Either the sheepdog performs the fetch task (and not the other 3) successfully, performs the drive task, performs the herd task, or performs the separate task.

This would look like:

$(.9)(.3)(.16)(.25)+(.1)(.7)(.16)(.25)+(.1)(.3)(.84)(.25)+(.1)(.3)(.16)(.75)$

Subtracting this from the case in which the sheepdog performs all four tasks would yield:

$(.9)(.7)(.84)(.75)-[(.9)(.3)(.16)(.25)+(.1)(.7)(.16)(.25)+(.1)(.3)(.84)(.25)+(.1)(.3)(.16)(.75)]$.

Is this correct?

edited 21 mins ago

N. F. Taussig

40.3k93253

asked 11 hours ago

rover2

515110

note: i would obviously simplify my answer at the end by multiplying all of the numbers out, i just want to see if this approach is correct...
â€“Â rover2
11 hours ago

add a commentÂ |Â

up vote
9
down vote

favorite

enter image description here

What is the probability that a sheepdog performs at least $1$ of these tasks successfully?

My approach is to subtract the probability of performing at most $1$ of these tasks successfully from the probability of performing all $4$ tasks successfully.

$P(textfetch)=.9, P(textdrive)=.7, P(textherd)=.84, P(textseparate)=.75$.

The complement of these four probabilities is, $.1,.3,.16,$ and $,.25$, respectively.

So the probability that the sheepdog performs all four tasks successfully is simply, $(.9)(.7)(.84)(.75)$.

This would look like:

$(.9)(.3)(.16)(.25)+(.1)(.7)(.16)(.25)+(.1)(.3)(.84)(.25)+(.1)(.3)(.16)(.75)$

Subtracting this from the case in which the sheepdog performs all four tasks would yield:

$(.9)(.7)(.84)(.75)-[(.9)(.3)(.16)(.25)+(.1)(.7)(.16)(.25)+(.1)(.3)(.84)(.25)+(.1)(.3)(.16)(.75)]$.

Is this correct?

edited 21 mins ago

N. F. Taussig

40.3k93253

asked 11 hours ago

rover2

515110

note: i would obviously simplify my answer at the end by multiplying all of the numbers out, i just want to see if this approach is correct...
â€“Â rover2
11 hours ago

add a commentÂ |Â

up vote
9
down vote

favorite

enter image description here

What is the probability that a sheepdog performs at least $1$ of these tasks successfully?

My approach is to subtract the probability of performing at most $1$ of these tasks successfully from the probability of performing all $4$ tasks successfully.

$P(textfetch)=.9, P(textdrive)=.7, P(textherd)=.84, P(textseparate)=.75$.

The complement of these four probabilities is, $.1,.3,.16,$ and $,.25$, respectively.

So the probability that the sheepdog performs all four tasks successfully is simply, $(.9)(.7)(.84)(.75)$.

This would look like:

$(.9)(.3)(.16)(.25)+(.1)(.7)(.16)(.25)+(.1)(.3)(.84)(.25)+(.1)(.3)(.16)(.75)$

Subtracting this from the case in which the sheepdog performs all four tasks would yield:

$(.9)(.7)(.84)(.75)-[(.9)(.3)(.16)(.25)+(.1)(.7)(.16)(.25)+(.1)(.3)(.84)(.25)+(.1)(.3)(.16)(.75)]$.

Is this correct?

edited 21 mins ago

N. F. Taussig

40.3k93253

asked 11 hours ago

rover2

515110

enter image description here

What is the probability that a sheepdog performs at least $1$ of these tasks successfully?

My approach is to subtract the probability of performing at most $1$ of these tasks successfully from the probability of performing all $4$ tasks successfully.

$P(textfetch)=.9, P(textdrive)=.7, P(textherd)=.84, P(textseparate)=.75$.

The complement of these four probabilities is, $.1,.3,.16,$ and $,.25$, respectively.

So the probability that the sheepdog performs all four tasks successfully is simply, $(.9)(.7)(.84)(.75)$.

This would look like:

$(.9)(.3)(.16)(.25)+(.1)(.7)(.16)(.25)+(.1)(.3)(.84)(.25)+(.1)(.3)(.16)(.75)$

Subtracting this from the case in which the sheepdog performs all four tasks would yield:

$(.9)(.7)(.84)(.75)-[(.9)(.3)(.16)(.25)+(.1)(.7)(.16)(.25)+(.1)(.3)(.84)(.25)+(.1)(.3)(.16)(.75)]$.

Is this correct?

probability combinatorics

edited 21 mins ago

N. F. Taussig

40.3k93253

asked 11 hours ago

rover2

515110

edited 21 mins ago

N. F. Taussig

40.3k93253

asked 11 hours ago

rover2

515110

edited 21 mins ago

N. F. Taussig

40.3k93253

edited 21 mins ago

N. F. Taussig

40.3k93253

edited 21 mins ago

N. F. Taussig

40.3k93253

asked 11 hours ago

rover2

515110

asked 11 hours ago

rover2

515110

asked 11 hours ago

rover2

515110

note: i would obviously simplify my answer at the end by multiplying all of the numbers out, i just want to see if this approach is correct...
â€“Â rover2
11 hours ago

add a commentÂ |Â

note: i would obviously simplify my answer at the end by multiplying all of the numbers out, i just want to see if this approach is correct...
â€“Â rover2
11 hours ago

note: i would obviously simplify my answer at the end by multiplying all of the numbers out, i just want to see if this approach is correct...
â€“Â rover2
11 hours ago

add a commentÂ |Â

3 Answers
3

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up vote
15
down vote

The complement of at least $1$ is not at most $1$.

The complement is if none of the task is perform.

Hence just compute $$1-prod_i=1^4 (1-p_i)$$

edited 11 hours ago

answered 11 hours ago

Siong Thye Goh

83.5k1456106

woowwwww...i cant believe i missed that. thank you ! i got it now
â€“Â rover2
11 hours ago

add a commentÂ |Â

up vote
2
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By sheepdog, I assume you mean 'good sheepdog' as defined in the question. Let $X$ denote the number of things in the list satisfied. If you want the probability the at least one of the 4things listed is satisfied $P(Xgeq 1)$, you work it out by solving $ 1-P(X=0)$ This is equal to $1-(0.1)(0.3)(0.16)(0.25) = 0.9988$

answered 11 hours ago

Displayname

1143

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up vote
0
down vote

Calculating the compliment of the chance of failing in all trials is undoubtedly the 'correct' approach to this problem, as per Siong Thye Goh's answer.

The same result can be calculated by accumulating the additional chance of success from each trial, given that the trial is needed, like this:


Chance this Chance of success Additional Cumulative chance
trial needed in this trial success from chance of success
 this trial in any trial
--------------------------------------------------------------------
A = 1.0 - D' B (input data) C = AB D = C + D'
--------------------------------------------------------------------
 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 
 1 0.9 0.9 0.9
 0.1 0.7 0.07 0.97
 0.03 0.84 0.0252 0.9952
 0.0048 0.75 0.0036 0.9988

answered 1 hour ago

Richard

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
15
down vote

The complement of at least $1$ is not at most $1$.

The complement is if none of the task is perform.

Hence just compute $$1-prod_i=1^4 (1-p_i)$$

edited 11 hours ago

answered 11 hours ago

Siong Thye Goh

83.5k1456106

woowwwww...i cant believe i missed that. thank you ! i got it now
â€“Â rover2
11 hours ago

add a commentÂ |Â

up vote
15
down vote

The complement of at least $1$ is not at most $1$.

The complement is if none of the task is perform.

Hence just compute $$1-prod_i=1^4 (1-p_i)$$

edited 11 hours ago

answered 11 hours ago

Siong Thye Goh

83.5k1456106

woowwwww...i cant believe i missed that. thank you ! i got it now
â€“Â rover2
11 hours ago

add a commentÂ |Â

up vote
15
down vote

The complement of at least $1$ is not at most $1$.

The complement is if none of the task is perform.

Hence just compute $$1-prod_i=1^4 (1-p_i)$$

edited 11 hours ago

answered 11 hours ago

Siong Thye Goh

83.5k1456106

The complement of at least $1$ is not at most $1$.

The complement is if none of the task is perform.

Hence just compute $$1-prod_i=1^4 (1-p_i)$$

edited 11 hours ago

answered 11 hours ago

Siong Thye Goh

83.5k1456106

edited 11 hours ago

answered 11 hours ago

Siong Thye Goh

83.5k1456106

answered 11 hours ago

Siong Thye Goh

83.5k1456106

answered 11 hours ago

Siong Thye Goh

83.5k1456106

woowwwww...i cant believe i missed that. thank you ! i got it now
â€“Â rover2
11 hours ago

add a commentÂ |Â

woowwwww...i cant believe i missed that. thank you ! i got it now
â€“Â rover2
11 hours ago

woowwwww...i cant believe i missed that. thank you ! i got it now
â€“Â rover2
11 hours ago

add a commentÂ |Â

up vote
2
down vote

answered 11 hours ago

Displayname

1143

New contributor

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up vote
2
down vote

answered 11 hours ago

Displayname

1143

New contributor

add a commentÂ |Â

up vote
2
down vote

answered 11 hours ago

Displayname

1143

New contributor

answered 11 hours ago

Displayname

1143

New contributor

answered 11 hours ago

Displayname

1143

New contributor

answered 11 hours ago

Displayname

1143

answered 11 hours ago

Displayname

1143

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Displayname is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote

Calculating the compliment of the chance of failing in all trials is undoubtedly the 'correct' approach to this problem, as per Siong Thye Goh's answer.

The same result can be calculated by accumulating the additional chance of success from each trial, given that the trial is needed, like this:


Chance this Chance of success Additional Cumulative chance
trial needed in this trial success from chance of success
 this trial in any trial
--------------------------------------------------------------------
A = 1.0 - D' B (input data) C = AB D = C + D'
--------------------------------------------------------------------
 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 
 1 0.9 0.9 0.9
 0.1 0.7 0.07 0.97
 0.03 0.84 0.0252 0.9952
 0.0048 0.75 0.0036 0.9988

answered 1 hour ago

Richard

New contributor

add a commentÂ |Â

up vote
0
down vote

Calculating the compliment of the chance of failing in all trials is undoubtedly the 'correct' approach to this problem, as per Siong Thye Goh's answer.

The same result can be calculated by accumulating the additional chance of success from each trial, given that the trial is needed, like this:


Chance this Chance of success Additional Cumulative chance
trial needed in this trial success from chance of success
 this trial in any trial
--------------------------------------------------------------------
A = 1.0 - D' B (input data) C = AB D = C + D'
--------------------------------------------------------------------
 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 
 1 0.9 0.9 0.9
 0.1 0.7 0.07 0.97
 0.03 0.84 0.0252 0.9952
 0.0048 0.75 0.0036 0.9988

answered 1 hour ago

Richard

New contributor

add a commentÂ |Â

up vote
0
down vote

Calculating the compliment of the chance of failing in all trials is undoubtedly the 'correct' approach to this problem, as per Siong Thye Goh's answer.

The same result can be calculated by accumulating the additional chance of success from each trial, given that the trial is needed, like this:


Chance this Chance of success Additional Cumulative chance
trial needed in this trial success from chance of success
 this trial in any trial
--------------------------------------------------------------------
A = 1.0 - D' B (input data) C = AB D = C + D'
--------------------------------------------------------------------
 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 
 1 0.9 0.9 0.9
 0.1 0.7 0.07 0.97
 0.03 0.84 0.0252 0.9952
 0.0048 0.75 0.0036 0.9988

answered 1 hour ago

Richard

New contributor

Calculating the compliment of the chance of failing in all trials is undoubtedly the 'correct' approach to this problem, as per Siong Thye Goh's answer.

The same result can be calculated by accumulating the additional chance of success from each trial, given that the trial is needed, like this:


Chance this Chance of success Additional Cumulative chance
trial needed in this trial success from chance of success
 this trial in any trial
--------------------------------------------------------------------
A = 1.0 - D' B (input data) C = AB D = C + D'
--------------------------------------------------------------------
 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 
 1 0.9 0.9 0.9
 0.1 0.7 0.07 0.97
 0.03 0.84 0.0252 0.9952
 0.0048 0.75 0.0036 0.9988

answered 1 hour ago

Richard

New contributor

answered 1 hour ago

Richard

New contributor

answered 1 hour ago

Richard

answered 1 hour ago

Richard

New contributor

Richard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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