Probability of Circuit Flowing
An electrical circuit can go from the starting point to the ending point through one of two paths: through relays A and C, or through relays B and D. The relays are independent and have a probability of $0.9$ of being conductive. It is considered to be flowing if either of these paths are conductive.
Find the probability that relays A and D are conductive if the circuit is flowing.
My solution:
Let $A, B, C, D$, be the events that the corresponding relay is conductive.
Since the relays are independent, the probability of any two relays being both conductive is $.9^2 = .81$.
$$
P(Flow) = P((A cap C) cup (B cap D))
$$
then
$$
P(overline{Flow}) = P(overline{(A cap C) cup (B cap D)}) = P(overline{(A cap C)} cap overline{(B cap D)}) = (1 - .81)^2 = 0.0361
$$
therefore
$$
P(Flow) = .9639
$$
I know everything up to here is right. The problem comes with the conditional probability aspect:
$$
P(A cap D | Flow) = frac{P((A cap D) cap Flow)}{P(Flow)} = frac{P(Flow|Acap D) P(A cap D)}{P(Flow)}
$$
My intuition here tells me that, in order for the circuit to be flowing given that A and D are both working, then either B or C must be working. As such:
$$
P(A cap D | Flow) = frac{P(B cup C) P(A cap D)}{P(Flow)} = frac{.99 cdot .81}{.9639} = .8319
$$
However this is the wrong answer (correct answer is $.916$). Where did my logic go wrong?
probability conditional-probability bayes-theorem
asked Dec 6 at 23:30
Bryden C
1858
add a comment |
An electrical circuit can go from the starting point to the ending point through one of two paths: through relays A and C, or through relays B and D. The relays are independent and have a probability of $0.9$ of being conductive. It is considered to be flowing if either of these paths are conductive.
Find the probability that relays A and D are conductive if the circuit is flowing.
My solution:
Let $A, B, C, D$, be the events that the corresponding relay is conductive.
Since the relays are independent, the probability of any two relays being both conductive is $.9^2 = .81$.
$$
P(Flow) = P((A cap C) cup (B cap D))
$$
then
$$
P(overline{Flow}) = P(overline{(A cap C) cup (B cap D)}) = P(overline{(A cap C)} cap overline{(B cap D)}) = (1 - .81)^2 = 0.0361
$$
therefore
$$
P(Flow) = .9639
$$
I know everything up to here is right. The problem comes with the conditional probability aspect:
$$
P(A cap D | Flow) = frac{P((A cap D) cap Flow)}{P(Flow)} = frac{P(Flow|Acap D) P(A cap D)}{P(Flow)}
$$
My intuition here tells me that, in order for the circuit to be flowing given that A and D are both working, then either B or C must be working. As such:
$$
P(A cap D | Flow) = frac{P(B cup C) P(A cap D)}{P(Flow)} = frac{.99 cdot .81}{.9639} = .8319
$$
However this is the wrong answer (correct answer is $.916$). Where did my logic go wrong?
probability conditional-probability bayes-theorem
asked Dec 6 at 23:30
Bryden C
1858
Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
– amd
Dec 6 at 23:51
add a comment |
An electrical circuit can go from the starting point to the ending point through one of two paths: through relays A and C, or through relays B and D. The relays are independent and have a probability of $0.9$ of being conductive. It is considered to be flowing if either of these paths are conductive.
Find the probability that relays A and D are conductive if the circuit is flowing.
My solution:
Let $A, B, C, D$, be the events that the corresponding relay is conductive.
Since the relays are independent, the probability of any two relays being both conductive is $.9^2 = .81$.
$$
P(Flow) = P((A cap C) cup (B cap D))
$$
then
$$
P(overline{Flow}) = P(overline{(A cap C) cup (B cap D)}) = P(overline{(A cap C)} cap overline{(B cap D)}) = (1 - .81)^2 = 0.0361
$$
therefore
$$
P(Flow) = .9639
$$
I know everything up to here is right. The problem comes with the conditional probability aspect:
$$
P(A cap D | Flow) = frac{P((A cap D) cap Flow)}{P(Flow)} = frac{P(Flow|Acap D) P(A cap D)}{P(Flow)}
$$
My intuition here tells me that, in order for the circuit to be flowing given that A and D are both working, then either B or C must be working. As such:
$$
P(A cap D | Flow) = frac{P(B cup C) P(A cap D)}{P(Flow)} = frac{.99 cdot .81}{.9639} = .8319
$$
However this is the wrong answer (correct answer is $.916$). Where did my logic go wrong?
probability conditional-probability bayes-theorem
asked Dec 6 at 23:30
Bryden C
1858
An electrical circuit can go from the starting point to the ending point through one of two paths: through relays A and C, or through relays B and D. The relays are independent and have a probability of $0.9$ of being conductive. It is considered to be flowing if either of these paths are conductive.
Find the probability that relays A and D are conductive if the circuit is flowing.
My solution:
Let $A, B, C, D$, be the events that the corresponding relay is conductive.
Since the relays are independent, the probability of any two relays being both conductive is $.9^2 = .81$.
$$
P(Flow) = P((A cap C) cup (B cap D))
$$
then
$$
P(overline{Flow}) = P(overline{(A cap C) cup (B cap D)}) = P(overline{(A cap C)} cap overline{(B cap D)}) = (1 - .81)^2 = 0.0361
$$
therefore
$$
P(Flow) = .9639
$$
I know everything up to here is right. The problem comes with the conditional probability aspect:
$$
P(A cap D | Flow) = frac{P((A cap D) cap Flow)}{P(Flow)} = frac{P(Flow|Acap D) P(A cap D)}{P(Flow)}
$$
My intuition here tells me that, in order for the circuit to be flowing given that A and D are both working, then either B or C must be working. As such:
$$
P(A cap D | Flow) = frac{P(B cup C) P(A cap D)}{P(Flow)} = frac{.99 cdot .81}{.9639} = .8319
$$
However this is the wrong answer (correct answer is $.916$). Where did my logic go wrong?
probability conditional-probability bayes-theorem
probability conditional-probability bayes-theorem
asked Dec 6 at 23:30
Bryden C
1858
asked Dec 6 at 23:30
Bryden C
1858
asked Dec 6 at 23:30
Bryden C
1858
asked Dec 6 at 23:30
Bryden C
1858
asked Dec 6 at 23:30
Bryden C
1858
1858
Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
– amd
Dec 6 at 23:51
add a comment |
Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
– amd
Dec 6 at 23:51
Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
– amd
Dec 6 at 23:51
Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
– amd
Dec 6 at 23:51
add a comment |
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Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
– amd
Dec 6 at 23:51