To compute $P(F \mid S, T, R)$, we use Bayes' Theorem in the form:

$P(F \mid S, T, R) = \frac{P(S, T, R \mid F) P(F)}{P(S, T, R)}$

The steps are as follows:

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Step 1: Compute $P(S, T, R \mid F)$

Using the assumption of independence among $S$, $T$, and $R$, given $F$, we can write: $P(S, T, R \mid F) = P(S \mid F) \cdot P(T \mid F) \cdot P(R \mid F)$ Substitute the given values: $P(S, T, R \mid F) = 0.2 \cdot 0.4 \cdot 0.3 = 0.024$

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Step 2: Compute $P(S, T, R \mid B)$

Similarly: $P(S, T, R \mid B) = P(S \mid B) \cdot P(T \mid B) \cdot P(R \mid B)$ Substitute the given values: $P(S, T, R \mid B) = 0.1 \cdot 0.2 \cdot 0.6 = 0.012$

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Step 3: Compute $P(S, T, R)$

Using the law of total probability: $P(S, T, R) = P(S, T, R \mid F) P(F) + P(S, T, R \mid B) P(B)$ Substitute the values: $P(S, T, R) = (0.024 \cdot 0.4) + (0.012 \cdot 0.6) = 0.0096 + 0.0072 = 0.0168$

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Step 4: Compute $P(F \mid S, T, R)$

Using Bayes' Theorem: $P(F \mid S, T, R) = \frac{P(S, T, R \mid F) P(F)}{P(S, T, R)}$ Substitute the values: $P(F \mid S, T, R) = \frac{0.024 \cdot 0.4}{0.0168} = \frac{0.0096}{0.0168} = 0.5714$

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Final Answer:

$P(F \mid S, T, R) = 0.5714 \, \text{or approximately } 57.14\%.$

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Would you like further clarifications? Here are related questions for deeper understanding:

1. What is the interpretation of conditional probability in real-world terms?
2. How does the law of total probability apply to this problem?
3. Can independence be assumed between $S, T, R$ in more complex cases?
4. How would the answer change if $P(F)$ and $P(B)$ were equal?
5. How does Bayes' Theorem generalize to continuous distributions?

Tip: Always verify assumptions like independence before applying formulas, as they can critically impact your solution.