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I'm trying to understand the following question:

An engineer conducts tests to find out if circuits of a certain type are prone to overheating. 30% of all such circuits are prone to overheating. If the circuit is prone to overheating, the test will report it is not prone to overheating with probability 0.1, prone to overheating with probability 0.7, and produce an inconclusive result with probability 0.2. If it is not prone to overheating, the test will report it is not prone to overheating with probability 0.6, prone to overheating with probability 0.3, and produce an inconclusive result with probability 0.1. The experiment is performed twice on a particular circuit; the first time it produces an inconclusive result and the second time it reports that the circuit is prone to overheating. Assuming the results of the two tests are independent, what is the probability the circuit is prone to overheating, given the outcome of the tests?

This is how I tried to solve the question:

$$P(O) = 0.3 \ \quad P(O^c)= 0.7 \\ P(N|O) = 0.1 \quad P(P|O) = 0.7 \quad P(I|O) = 0.2 \\ P(N|O^c) = 0.6 \quad P(P|O^c) = 0.3 \quad P(I|O^c) = 0.1 \\ \\ P(O|I)= \frac {P(I|O)P(O)}{P(I|O)P(O) + P(I|O^c)P(O^c)} = \frac{0.2*0.3}{0.2*0.3 + 0.1*0.7} = \frac{6}{13}\\ P(O|P)= \frac {P(P|O)P(O)}{P(P|O)P(O) + P(P|O^c)P(O^c)} = \frac{0.7*0.3}{0.7*0.3 + 0.3*0.7} = \frac{1}{2}\\$$ So the probability that the circuit is prone to overheating is $\frac{6}{13}* \frac{1}{2} = \frac{3}{13} $

My answer was incorrect. The actual method is:

The probability that the circuit is prone to overheating and we observe the test results we have seen is: 0.3 × 0.2 × 0.7 = 0.042. The probability that the circuit is not prone to overheating and we observe the test results we have seen is: 0.7 × 0.1 × 0.3 = 0.021. The probability that we observe the test results we have seen is: 0.042 + 0.021 = 0.063. Therefore, the conditional probability that the circuit is prone to overheating given the outcomes of the tests is:$ \frac {0.042}{0.063} = \frac{2}{3}$

My understanding is clearly not correct. Could someone explain why my method doesn't work?

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  • $\begingroup$ While using mathematical notation for Bayes' Theorem, students frequently get lost in the maze of symbols. You haven't considered that two tests in series have been done. Study the explanation which is quite clear, and try to modify your formula accordingly. $\endgroup$ Apr 11 '16 at 14:25
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Your method doesn't work because you have to find:

P(O | I on $1^{st}$ test $\cap$ P on $2^{nd}$ test), but you have calculated

What you have calculated is $P(O | \text{I on a test}) * P (O | \text{P on a test})$ which isn't the probability of a specific event.

The first part:

$P(O | \text{I on a test})$

Includes cases where you get I on a test but not P on the other, and the second part:

$P (O | \text{P on a test})$

includes cases where you get P on a test but not I on the other

You need to find the conditional probability given both I and P happen. $I \cap P$


Calculation for completeness:

For simplicity I'll call the events I and P.

$P(O | I \cap P) = \frac{P(O \cap I \cap P)}{P(I \cap P)}$

$P(O | I \cap P) = \frac{P(O \cap I \cap P)}{P(O \cap I \cap P) + P(O^c \cap I \cap P)}$

$P(O \cap I \cap P) = P(O)*P(I \cap P | O) = 0.3 * 0.2 * 0.7 = 0.042 $

$P(O^c \cap I \cap P) = P(O^c)*P(I \cap P | O^c) = 0.7 * 0.1 * 0.3 = 0.021 $

$P(O | I \cap P) = \frac{0.042}{0.042 + 0.021} = \frac{0.042}{0.063} = \frac{2}{3} $

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