Mixing
Conditional probability, In studying the causes of power failures, the following data have been gathered:
$begingroup$
In studying the causes of power failures, the following data have been gathered:
10% are due to a transformer damage, 75% are due to line damage, 5% involve both
problems. Based on these percentages, find the probability that a given power failure
involves:
a) line damage given that there is transformer damage
b) transformer damage given that there is line damage
c) transformer damage but not line damage
d) transformer damage given that there is no line damage
e) transformer damage or line damage.
Now, I am familiar with conditional probabilities ( to some degree at least ) and the first thing that came to my mind for point a) was Bayes
so
$ T $ - it involves transformer damage
$ L $ - it involves line damage
$ B $ - it involves both
$$ P(L/T)=frac{P(L)*P(T/L)}{P(T)} $$
but the problem is that I am stuck at $P(T/L)$ I have no idea from were to start and maybe this approach is not even the correct one so I would appreciate some help, maybe a hint on how to proceed...
probability conditional-probability
asked Feb 2 at 1:35
The VirtuosoThe Virtuoso
448
$endgroup$
add a comment |
$begingroup$
In studying the causes of power failures, the following data have been gathered:
10% are due to a transformer damage, 75% are due to line damage, 5% involve both
problems. Based on these percentages, find the probability that a given power failure
involves:
a) line damage given that there is transformer damage
b) transformer damage given that there is line damage
c) transformer damage but not line damage
d) transformer damage given that there is no line damage
e) transformer damage or line damage.
Now, I am familiar with conditional probabilities ( to some degree at least ) and the first thing that came to my mind for point a) was Bayes
so
$ T $ - it involves transformer damage
$ L $ - it involves line damage
$ B $ - it involves both
$$ P(L/T)=frac{P(L)*P(T/L)}{P(T)} $$
but the problem is that I am stuck at $P(T/L)$ I have no idea from were to start and maybe this approach is not even the correct one so I would appreciate some help, maybe a hint on how to proceed...
probability conditional-probability
asked Feb 2 at 1:35
The VirtuosoThe Virtuoso
448
$endgroup$
$begingroup$
Hint: use the definition of conditional probability
$endgroup$
– Alex
Feb 2 at 1:53
$begingroup$
Yes but first I need a complete system of events, and I just can't think of one
$endgroup$
– The Virtuoso
Feb 2 at 1:56
$begingroup$
Never mind, I'm such a douche we only worked with bayes and complete systems of events in my class that's why I couldn't think of a way to solve it, but I guess my answer could be found on wikipedia...
$endgroup$
– The Virtuoso
Feb 2 at 2:06
add a comment |
$begingroup$
In studying the causes of power failures, the following data have been gathered:
10% are due to a transformer damage, 75% are due to line damage, 5% involve both
problems. Based on these percentages, find the probability that a given power failure
involves:
a) line damage given that there is transformer damage
b) transformer damage given that there is line damage
c) transformer damage but not line damage
d) transformer damage given that there is no line damage
e) transformer damage or line damage.
Now, I am familiar with conditional probabilities ( to some degree at least ) and the first thing that came to my mind for point a) was Bayes
so
$ T $ - it involves transformer damage
$ L $ - it involves line damage
$ B $ - it involves both
$$ P(L/T)=frac{P(L)*P(T/L)}{P(T)} $$
but the problem is that I am stuck at $P(T/L)$ I have no idea from were to start and maybe this approach is not even the correct one so I would appreciate some help, maybe a hint on how to proceed...
probability conditional-probability
asked Feb 2 at 1:35
The VirtuosoThe Virtuoso
448
$endgroup$
In studying the causes of power failures, the following data have been gathered:
10% are due to a transformer damage, 75% are due to line damage, 5% involve both
problems. Based on these percentages, find the probability that a given power failure
involves:
a) line damage given that there is transformer damage
b) transformer damage given that there is line damage
c) transformer damage but not line damage
d) transformer damage given that there is no line damage
e) transformer damage or line damage.
Now, I am familiar with conditional probabilities ( to some degree at least ) and the first thing that came to my mind for point a) was Bayes
so
$ T $ - it involves transformer damage
$ L $ - it involves line damage
$ B $ - it involves both
$$ P(L/T)=frac{P(L)*P(T/L)}{P(T)} $$
but the problem is that I am stuck at $P(T/L)$ I have no idea from were to start and maybe this approach is not even the correct one so I would appreciate some help, maybe a hint on how to proceed...
probability conditional-probability
probability conditional-probability
asked Feb 2 at 1:35
The VirtuosoThe Virtuoso
448
asked Feb 2 at 1:35
The VirtuosoThe Virtuoso
448
asked Feb 2 at 1:35
The VirtuosoThe Virtuoso
448
asked Feb 2 at 1:35
The VirtuosoThe Virtuoso
448
asked Feb 2 at 1:35
The VirtuosoThe Virtuoso
448
448
$begingroup$
Hint: use the definition of conditional probability
$endgroup$
– Alex
Feb 2 at 1:53
$begingroup$
Yes but first I need a complete system of events, and I just can't think of one
$endgroup$
– The Virtuoso
Feb 2 at 1:56
$begingroup$
Never mind, I'm such a douche we only worked with bayes and complete systems of events in my class that's why I couldn't think of a way to solve it, but I guess my answer could be found on wikipedia...
$endgroup$
– The Virtuoso
Feb 2 at 2:06
add a comment |
$begingroup$
Hint: use the definition of conditional probability
$endgroup$
– Alex
Feb 2 at 1:53
$begingroup$
Yes but first I need a complete system of events, and I just can't think of one
$endgroup$
– The Virtuoso
Feb 2 at 1:56
$begingroup$
Never mind, I'm such a douche we only worked with bayes and complete systems of events in my class that's why I couldn't think of a way to solve it, but I guess my answer could be found on wikipedia...
$endgroup$
– The Virtuoso
Feb 2 at 2:06
$begingroup$
Hint: use the definition of conditional probability
$endgroup$
– Alex
Feb 2 at 1:53
$begingroup$
Hint: use the definition of conditional probability
$endgroup$
– Alex
Feb 2 at 1:53
$begingroup$
Yes but first I need a complete system of events, and I just can't think of one
$endgroup$
– The Virtuoso
Feb 2 at 1:56
$begingroup$
Yes but first I need a complete system of events, and I just can't think of one
$endgroup$
– The Virtuoso
Feb 2 at 1:56
$begingroup$
Never mind, I'm such a douche we only worked with bayes and complete systems of events in my class that's why I couldn't think of a way to solve it, but I guess my answer could be found on wikipedia...
$endgroup$
– The Virtuoso
Feb 2 at 2:06
$begingroup$
Never mind, I'm such a douche we only worked with bayes and complete systems of events in my class that's why I couldn't think of a way to solve it, but I guess my answer could be found on wikipedia...
$endgroup$
– The Virtuoso
Feb 2 at 2:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
According to your notation, we have $textbf{P}(T) = 0.1$, $textbf{P}(L) = 0.75$ and $textbf{P}(Tcap L) = 0.05$. Therefore:
(a) The sought probability is given by $textbf{P}(L|T)$, which can be rewritten as
begin{align*}
textbf{P}(L|T) = frac{textbf{P}(Lcap T)}{textbf{P}(T)} = frac{0.05}{0.1} = 0.5
end{align*}
(b) Analogously, we have
begin{align*}
textbf{P}(T|L) = frac{textbf{P}(Tcap L)}{textbf{P}(L)} = frac{0.05}{0.75} = frac{1}{15}
end{align*}
(c) Here, the event in which we are interested in is described by $textbf{P}(Tcap L^{c})$:
begin{align*}
textbf{P}(Tcap L^{c}) = textbf{P}(T) - textbf{P}(Tcap L) = 0.1 - 0.05 = 0.05
end{align*}
(d) In this case, the target event is described by $textbf{P}(T|L^{c})$:
begin{align*}
textbf{P}(T|L^{c}) = frac{textbf{P}(Tcap L^{c})}{textbf{P}(L^{c})} = frac{textbf{P}(Tcap L^{c})}{1 - textbf{P}(L)} = frac{0.05}{1 - 0.75} = frac{0.05}{0.25} = 0.2
end{align*}
(e) Finally, the last event is given by
begin{align*}
textbf{P}(Tcup L) = textbf{P}(T) + textbf{P}(L) - textbf{P}(Tcap L) = 0.1 + 0.75 - 0.05 = 0.8
end{align*}
Hope this helps.
answered Feb 2 at 1:59
APC89APC89
2,371720
$endgroup$
$begingroup$
Thank you very much, in my class we only worked with Bayes and complete system of events that's why when someone just mentioned me to use the definition of conditional probability I searched on wikipedia to make sure that I'm not missing something obvious and I just realized that I was...
$endgroup$
– The Virtuoso
Feb 2 at 2:04
$begingroup$
If you think my answer deserves it, give it a up vote.
$endgroup$
– APC89
Feb 2 at 2:52
add a comment |
$begingroup$
Going off my hint: recall that for two events $X$ and $Y$ (with $P(Y) neq 0$) we have
$$
P(X|Y) = frac{P(X cap Y)}{P(Y)}
$$
So in your case, you want
$$
P(L|T) = frac{P(L cap T)}{P(T)}
$$
Note that the event $L cap T$ is exactly $B$. So
$$P(L|T) = frac{P(B)}{P(T)}$$
answered Feb 2 at 1:59
AlexAlex
52538
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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active
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votes
$begingroup$
According to your notation, we have $textbf{P}(T) = 0.1$, $textbf{P}(L) = 0.75$ and $textbf{P}(Tcap L) = 0.05$. Therefore:
(a) The sought probability is given by $textbf{P}(L|T)$, which can be rewritten as
begin{align*}
textbf{P}(L|T) = frac{textbf{P}(Lcap T)}{textbf{P}(T)} = frac{0.05}{0.1} = 0.5
end{align*}
(b) Analogously, we have
begin{align*}
textbf{P}(T|L) = frac{textbf{P}(Tcap L)}{textbf{P}(L)} = frac{0.05}{0.75} = frac{1}{15}
end{align*}
(c) Here, the event in which we are interested in is described by $textbf{P}(Tcap L^{c})$:
begin{align*}
textbf{P}(Tcap L^{c}) = textbf{P}(T) - textbf{P}(Tcap L) = 0.1 - 0.05 = 0.05
end{align*}
(d) In this case, the target event is described by $textbf{P}(T|L^{c})$:
begin{align*}
textbf{P}(T|L^{c}) = frac{textbf{P}(Tcap L^{c})}{textbf{P}(L^{c})} = frac{textbf{P}(Tcap L^{c})}{1 - textbf{P}(L)} = frac{0.05}{1 - 0.75} = frac{0.05}{0.25} = 0.2
end{align*}
(e) Finally, the last event is given by
begin{align*}
textbf{P}(Tcup L) = textbf{P}(T) + textbf{P}(L) - textbf{P}(Tcap L) = 0.1 + 0.75 - 0.05 = 0.8
end{align*}
Hope this helps.
answered Feb 2 at 1:59
APC89APC89
2,371720
$endgroup$
$begingroup$
Thank you very much, in my class we only worked with Bayes and complete system of events that's why when someone just mentioned me to use the definition of conditional probability I searched on wikipedia to make sure that I'm not missing something obvious and I just realized that I was...
$endgroup$
– The Virtuoso
Feb 2 at 2:04
$begingroup$
If you think my answer deserves it, give it a up vote.
$endgroup$
– APC89
Feb 2 at 2:52
add a comment |
$begingroup$
According to your notation, we have $textbf{P}(T) = 0.1$, $textbf{P}(L) = 0.75$ and $textbf{P}(Tcap L) = 0.05$. Therefore:
(a) The sought probability is given by $textbf{P}(L|T)$, which can be rewritten as
begin{align*}
textbf{P}(L|T) = frac{textbf{P}(Lcap T)}{textbf{P}(T)} = frac{0.05}{0.1} = 0.5
end{align*}
(b) Analogously, we have
begin{align*}
textbf{P}(T|L) = frac{textbf{P}(Tcap L)}{textbf{P}(L)} = frac{0.05}{0.75} = frac{1}{15}
end{align*}
(c) Here, the event in which we are interested in is described by $textbf{P}(Tcap L^{c})$:
begin{align*}
textbf{P}(Tcap L^{c}) = textbf{P}(T) - textbf{P}(Tcap L) = 0.1 - 0.05 = 0.05
end{align*}
(d) In this case, the target event is described by $textbf{P}(T|L^{c})$:
begin{align*}
textbf{P}(T|L^{c}) = frac{textbf{P}(Tcap L^{c})}{textbf{P}(L^{c})} = frac{textbf{P}(Tcap L^{c})}{1 - textbf{P}(L)} = frac{0.05}{1 - 0.75} = frac{0.05}{0.25} = 0.2
end{align*}
(e) Finally, the last event is given by
begin{align*}
textbf{P}(Tcup L) = textbf{P}(T) + textbf{P}(L) - textbf{P}(Tcap L) = 0.1 + 0.75 - 0.05 = 0.8
end{align*}
Hope this helps.
answered Feb 2 at 1:59
APC89APC89
2,371720
$endgroup$
$begingroup$
Thank you very much, in my class we only worked with Bayes and complete system of events that's why when someone just mentioned me to use the definition of conditional probability I searched on wikipedia to make sure that I'm not missing something obvious and I just realized that I was...
$endgroup$
– The Virtuoso
Feb 2 at 2:04
$begingroup$
If you think my answer deserves it, give it a up vote.
$endgroup$
– APC89
Feb 2 at 2:52
add a comment |
$begingroup$
According to your notation, we have $textbf{P}(T) = 0.1$, $textbf{P}(L) = 0.75$ and $textbf{P}(Tcap L) = 0.05$. Therefore:
(a) The sought probability is given by $textbf{P}(L|T)$, which can be rewritten as
begin{align*}
textbf{P}(L|T) = frac{textbf{P}(Lcap T)}{textbf{P}(T)} = frac{0.05}{0.1} = 0.5
end{align*}
(b) Analogously, we have
begin{align*}
textbf{P}(T|L) = frac{textbf{P}(Tcap L)}{textbf{P}(L)} = frac{0.05}{0.75} = frac{1}{15}
end{align*}
(c) Here, the event in which we are interested in is described by $textbf{P}(Tcap L^{c})$:
begin{align*}
textbf{P}(Tcap L^{c}) = textbf{P}(T) - textbf{P}(Tcap L) = 0.1 - 0.05 = 0.05
end{align*}
(d) In this case, the target event is described by $textbf{P}(T|L^{c})$:
begin{align*}
textbf{P}(T|L^{c}) = frac{textbf{P}(Tcap L^{c})}{textbf{P}(L^{c})} = frac{textbf{P}(Tcap L^{c})}{1 - textbf{P}(L)} = frac{0.05}{1 - 0.75} = frac{0.05}{0.25} = 0.2
end{align*}
(e) Finally, the last event is given by
begin{align*}
textbf{P}(Tcup L) = textbf{P}(T) + textbf{P}(L) - textbf{P}(Tcap L) = 0.1 + 0.75 - 0.05 = 0.8
end{align*}
Hope this helps.
answered Feb 2 at 1:59
APC89APC89
2,371720
$endgroup$
According to your notation, we have $textbf{P}(T) = 0.1$, $textbf{P}(L) = 0.75$ and $textbf{P}(Tcap L) = 0.05$. Therefore:
(a) The sought probability is given by $textbf{P}(L|T)$, which can be rewritten as
begin{align*}
textbf{P}(L|T) = frac{textbf{P}(Lcap T)}{textbf{P}(T)} = frac{0.05}{0.1} = 0.5
end{align*}
(b) Analogously, we have
begin{align*}
textbf{P}(T|L) = frac{textbf{P}(Tcap L)}{textbf{P}(L)} = frac{0.05}{0.75} = frac{1}{15}
end{align*}
(c) Here, the event in which we are interested in is described by $textbf{P}(Tcap L^{c})$:
begin{align*}
textbf{P}(Tcap L^{c}) = textbf{P}(T) - textbf{P}(Tcap L) = 0.1 - 0.05 = 0.05
end{align*}
(d) In this case, the target event is described by $textbf{P}(T|L^{c})$:
begin{align*}
textbf{P}(T|L^{c}) = frac{textbf{P}(Tcap L^{c})}{textbf{P}(L^{c})} = frac{textbf{P}(Tcap L^{c})}{1 - textbf{P}(L)} = frac{0.05}{1 - 0.75} = frac{0.05}{0.25} = 0.2
end{align*}
(e) Finally, the last event is given by
begin{align*}
textbf{P}(Tcup L) = textbf{P}(T) + textbf{P}(L) - textbf{P}(Tcap L) = 0.1 + 0.75 - 0.05 = 0.8
end{align*}
Hope this helps.
answered Feb 2 at 1:59
APC89APC89
2,371720
edited Feb 2 at 2:55
answered Feb 2 at 1:59
APC89APC89
2,371720
answered Feb 2 at 1:59
APC89APC89
2,371720
answered Feb 2 at 1:59
APC89APC89
2,371720
2,371720
$begingroup$
Thank you very much, in my class we only worked with Bayes and complete system of events that's why when someone just mentioned me to use the definition of conditional probability I searched on wikipedia to make sure that I'm not missing something obvious and I just realized that I was...
$endgroup$
– The Virtuoso
Feb 2 at 2:04
$begingroup$
If you think my answer deserves it, give it a up vote.
$endgroup$
– APC89
Feb 2 at 2:52
add a comment |
$begingroup$
Thank you very much, in my class we only worked with Bayes and complete system of events that's why when someone just mentioned me to use the definition of conditional probability I searched on wikipedia to make sure that I'm not missing something obvious and I just realized that I was...
$endgroup$
– The Virtuoso
Feb 2 at 2:04
$begingroup$
If you think my answer deserves it, give it a up vote.
$endgroup$
– APC89
Feb 2 at 2:52
$begingroup$
Thank you very much, in my class we only worked with Bayes and complete system of events that's why when someone just mentioned me to use the definition of conditional probability I searched on wikipedia to make sure that I'm not missing something obvious and I just realized that I was...
$endgroup$
– The Virtuoso
Feb 2 at 2:04
$begingroup$
Thank you very much, in my class we only worked with Bayes and complete system of events that's why when someone just mentioned me to use the definition of conditional probability I searched on wikipedia to make sure that I'm not missing something obvious and I just realized that I was...
$endgroup$
– The Virtuoso
Feb 2 at 2:04
$begingroup$
If you think my answer deserves it, give it a up vote.
$endgroup$
– APC89
Feb 2 at 2:52
$begingroup$
If you think my answer deserves it, give it a up vote.
$endgroup$
– APC89
Feb 2 at 2:52
add a comment |
$begingroup$
Going off my hint: recall that for two events $X$ and $Y$ (with $P(Y) neq 0$) we have
$$
P(X|Y) = frac{P(X cap Y)}{P(Y)}
$$
So in your case, you want
$$
P(L|T) = frac{P(L cap T)}{P(T)}
$$
Note that the event $L cap T$ is exactly $B$. So
$$P(L|T) = frac{P(B)}{P(T)}$$
answered Feb 2 at 1:59
AlexAlex
52538
$endgroup$
add a comment |
$begingroup$
Going off my hint: recall that for two events $X$ and $Y$ (with $P(Y) neq 0$) we have
$$
P(X|Y) = frac{P(X cap Y)}{P(Y)}
$$
So in your case, you want
$$
P(L|T) = frac{P(L cap T)}{P(T)}
$$
Note that the event $L cap T$ is exactly $B$. So
$$P(L|T) = frac{P(B)}{P(T)}$$
answered Feb 2 at 1:59
AlexAlex
52538
$endgroup$
add a comment |
$begingroup$
Going off my hint: recall that for two events $X$ and $Y$ (with $P(Y) neq 0$) we have
$$
P(X|Y) = frac{P(X cap Y)}{P(Y)}
$$
So in your case, you want
$$
P(L|T) = frac{P(L cap T)}{P(T)}
$$
Note that the event $L cap T$ is exactly $B$. So
$$P(L|T) = frac{P(B)}{P(T)}$$
answered Feb 2 at 1:59
AlexAlex
52538
$endgroup$
Going off my hint: recall that for two events $X$ and $Y$ (with $P(Y) neq 0$) we have
$$
P(X|Y) = frac{P(X cap Y)}{P(Y)}
$$
So in your case, you want
$$
P(L|T) = frac{P(L cap T)}{P(T)}
$$
Note that the event $L cap T$ is exactly $B$. So
$$P(L|T) = frac{P(B)}{P(T)}$$
answered Feb 2 at 1:59
AlexAlex
52538
answered Feb 2 at 1:59
AlexAlex
52538
answered Feb 2 at 1:59
AlexAlex
52538
answered Feb 2 at 1:59
AlexAlex
52538
52538
add a comment |
add a comment |
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$begingroup$
Hint: use the definition of conditional probability
$endgroup$
– Alex
Feb 2 at 1:53
$begingroup$
Yes but first I need a complete system of events, and I just can't think of one
$endgroup$
– The Virtuoso
Feb 2 at 1:56
$begingroup$
Never mind, I'm such a douche we only worked with bayes and complete systems of events in my class that's why I couldn't think of a way to solve it, but I guess my answer could be found on wikipedia...
$endgroup$
– The Virtuoso
Feb 2 at 2:06