Mixing
Probability Question - Am I doing this problem right?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
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?
probability combinatorics
asked 2 hours ago
rover2
480110
add a comment |Â
up vote
2
down vote
favorite
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?
probability combinatorics
asked 2 hours ago
rover2
480110
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...
– rover2
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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?
probability combinatorics
asked 2 hours ago
rover2
480110
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?
probability combinatorics
probability combinatorics
asked 2 hours ago
rover2
480110
asked 2 hours ago
rover2
480110
asked 2 hours ago
rover2
480110
asked 2 hours ago
rover2
480110
asked 2 hours ago
rover2
480110
480110
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...
– rover2
2 hours ago
add a comment |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...
– rover2
2 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...
– rover2
2 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...
– rover2
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
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 $$1-prod_i=1^4 (1-p_i)$$
answered 2 hours ago
Siong Thye Goh
83.4k1456104
woowwwww...i cant believe i missed that. thank you ! i got it now
– rover2
2 hours ago
add a comment |Â
up vote
1
down vote
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 2 hours ago
Displayname
963
New contributor
Displayname is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
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 $$1-prod_i=1^4 (1-p_i)$$
answered 2 hours ago
Siong Thye Goh
83.4k1456104
woowwwww...i cant believe i missed that. thank you ! i got it now
– rover2
2 hours ago
add a comment |Â
up vote
3
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 $$1-prod_i=1^4 (1-p_i)$$
answered 2 hours ago
Siong Thye Goh
83.4k1456104
woowwwww...i cant believe i missed that. thank you ! i got it now
– rover2
2 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
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 $$1-prod_i=1^4 (1-p_i)$$
answered 2 hours ago
Siong Thye Goh
83.4k1456104
The complement of at least $1$ is not at most $1$.
The complement is if none of the task is perform.
Hence just compute $$1-prod_i=1^4 (1-p_i)$$
answered 2 hours ago
Siong Thye Goh
83.4k1456104
edited 2 hours ago
answered 2 hours ago
Siong Thye Goh
83.4k1456104
answered 2 hours ago
Siong Thye Goh
83.4k1456104
answered 2 hours ago
Siong Thye Goh
83.4k1456104
83.4k1456104
woowwwww...i cant believe i missed that. thank you ! i got it now
– rover2
2 hours ago
add a comment |Â
woowwwww...i cant believe i missed that. thank you ! i got it now
– rover2
2 hours ago
woowwwww...i cant believe i missed that. thank you ! i got it now
– rover2
2 hours ago
woowwwww...i cant believe i missed that. thank you ! i got it now
– rover2
2 hours ago
add a comment |Â
up vote
1
down vote
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 2 hours ago
Displayname
963
New contributor
Displayname is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
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 2 hours ago
Displayname
963
New contributor
Displayname is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
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 2 hours ago
Displayname
963
New contributor
Displayname is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 hours ago
Displayname
963
New contributor
Displayname is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Displayname
963
New contributor
Displayname is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Displayname
963
answered 2 hours ago
Displayname
963
963
New contributor
Displayname is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Displayname is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Displayname is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2932482%2fprobability-question-am-i-doing-this-problem-right%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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...
– rover2
2 hours ago