Probability of a device having anomaly type A
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$A$ = component has anomaly A
$B$ = component has anomaly B.
$P(B)=0.09$
$P(A|B) = 0.5$
$P(A|B^c) = 0.01$
I figured out that the probability of device having both anomalies is equal to $P(B cap A) = P(B)*P(A|B)=0.045$
The next question is to figure out what $P(A)$ is.
I tried $P(A cap B) = P(A)*P(B|A)$ but I only know $P(A cap B)$ in this equation. I dont see how to apply what I know about $P(A|B^c)$
probability statistics conditional-probability
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up vote
0
down vote
favorite
$A$ = component has anomaly A
$B$ = component has anomaly B.
$P(B)=0.09$
$P(A|B) = 0.5$
$P(A|B^c) = 0.01$
I figured out that the probability of device having both anomalies is equal to $P(B cap A) = P(B)*P(A|B)=0.045$
The next question is to figure out what $P(A)$ is.
I tried $P(A cap B) = P(A)*P(B|A)$ but I only know $P(A cap B)$ in this equation. I dont see how to apply what I know about $P(A|B^c)$
probability statistics conditional-probability
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$A$ = component has anomaly A
$B$ = component has anomaly B.
$P(B)=0.09$
$P(A|B) = 0.5$
$P(A|B^c) = 0.01$
I figured out that the probability of device having both anomalies is equal to $P(B cap A) = P(B)*P(A|B)=0.045$
The next question is to figure out what $P(A)$ is.
I tried $P(A cap B) = P(A)*P(B|A)$ but I only know $P(A cap B)$ in this equation. I dont see how to apply what I know about $P(A|B^c)$
probability statistics conditional-probability
$A$ = component has anomaly A
$B$ = component has anomaly B.
$P(B)=0.09$
$P(A|B) = 0.5$
$P(A|B^c) = 0.01$
I figured out that the probability of device having both anomalies is equal to $P(B cap A) = P(B)*P(A|B)=0.045$
The next question is to figure out what $P(A)$ is.
I tried $P(A cap B) = P(A)*P(B|A)$ but I only know $P(A cap B)$ in this equation. I dont see how to apply what I know about $P(A|B^c)$
probability statistics conditional-probability
probability statistics conditional-probability
asked Sep 2 at 10:46
novo
35118
35118
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1 Answer
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We have $mathbbP(B^c) = 1 - mathbbP(B) = 0.91$ , so you also know $mathbbP(A cap B^c) = 0.0091$ and $$ mathbbP(A) = mathbbP(A cap B) + mathbbP(A cap B^c) = 0.0541 $$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
We have $mathbbP(B^c) = 1 - mathbbP(B) = 0.91$ , so you also know $mathbbP(A cap B^c) = 0.0091$ and $$ mathbbP(A) = mathbbP(A cap B) + mathbbP(A cap B^c) = 0.0541 $$
add a comment |Â
up vote
0
down vote
accepted
We have $mathbbP(B^c) = 1 - mathbbP(B) = 0.91$ , so you also know $mathbbP(A cap B^c) = 0.0091$ and $$ mathbbP(A) = mathbbP(A cap B) + mathbbP(A cap B^c) = 0.0541 $$
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
We have $mathbbP(B^c) = 1 - mathbbP(B) = 0.91$ , so you also know $mathbbP(A cap B^c) = 0.0091$ and $$ mathbbP(A) = mathbbP(A cap B) + mathbbP(A cap B^c) = 0.0541 $$
We have $mathbbP(B^c) = 1 - mathbbP(B) = 0.91$ , so you also know $mathbbP(A cap B^c) = 0.0091$ and $$ mathbbP(A) = mathbbP(A cap B) + mathbbP(A cap B^c) = 0.0541 $$
answered Sep 2 at 10:49
LucaMac
1,44414
1,44414
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