Key Concepts and Formulas

1. Independent Events: Two events $A$ and $B$ are independent if the occurrence of one does not affect the probability of the other. The probability of both events occurring is the product of their individual probabilities: $P(A \cap B) = P(A) \cdot P(B)$
2. Complementary Events: The probability of an event $A$ not occurring is denoted as $P(A')$ or $P(\text{not } A)$, and it is calculated as: $P(A') = 1 - P(A)$
3. Conditional Probability: The probability of an event $A$ occurring given that another event $B$ has already occurred is denoted as $P(A|B)$ and is calculated using the formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ where $P(B) > 0$.

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Step-by-Step Solution

Step 1: Define Events and Initial Probabilities Let's clearly define the events related to the functioning of the two units:

• $U_1$: Event that the first unit functions.
• $U_2$: Event that the second unit functions.

From the problem statement, we are given the probabilities that these units function:

• $P(U_1) = 0.9$
• $P(U_2) = 0.8$

Since the units function independently, we can also find the probabilities of them failing, which are their complementary events:

• $U_1'$: Event that the first unit fails. $P(U_1') = 1 - P(U_1) = 1 - 0.9 = 0.1$
• $U_2'$: Event that the second unit fails. $P(U_2') = 1 - P(U_2) = 1 - 0.8 = 0.2$ Why this step? Clearly defining events and their respective probabilities is the foundational step. It translates the problem's verbal description into precise mathematical notation, which is essential for accurate calculations and understanding the problem structure.

Step 2: Calculate the Probability of the Instrument Operating and Failing The problem states that "Each unit must function independently for the instrument to operate." This means the instrument operates only if both unit 1 and unit 2 function.

• Let $I_O$ be the event that the instrument operates. $P(I_O) = P(U_1 \cap U_2)$ Since $U_1$ and $U_2$ are independent events: $P(I_O) = P(U_1) \cdot P(U_2) = 0.9 \cdot 0.8 = 0.72$

The problem then states, "The instrument is switched on and it fails to operate." Let $I_F$ be the event that the instrument fails to operate. This is the complementary event to the instrument operating:

• $P(I_F) = 1 - P(I_O) = 1 - 0.72 = 0.28$ Why this step? The problem specifies that the instrument "fails to operate" as a given condition for the conditional probability calculation. Therefore, calculating the probability of this event, $P(I_F)$, is crucial as it will form the denominator in our conditional probability formula. Using the complementary event approach is the most straightforward way to calculate this.

(Optional Alternative Calculation for $P(I_F)$ for deeper understanding): The instrument fails if at least one unit fails. This can occur in three mutually exclusive ways: a) Unit 1 fails and Unit 2 functions ($U_1' \cap U_2$) b) Unit 1 functions and Unit 2 fails ($U_1 \cap U_2'$) c) Both units fail ($U_1' \cap U_2'$) $P(U_1' \cap U_2) = P(U_1') \cdot P(U_2) = 0.1 \cdot 0.8 = 0.08$ $P(U_1 \cap U_2') = P(U_1) \cdot P(U_2') = 0.9 \cdot 0.2 = 0.18$ $P(U_1' \cap U_2') = P(U_1') \cdot P(U_2') = 0.1 \cdot 0.2 = 0.02$ Summing these mutually exclusive probabilities gives: $P(I_F) = 0.08 + 0.18 + 0.02 = 0.28$ This confirms our earlier calculation.

Step 3: Define and Calculate Probability of the Specific Failure Mode of Interest We are interested in the event "only the first unit failed and second unit is functioning." Let's call this event $A$.

• Event $A$: $U_1' \cap U_2$ Since the units operate independently, their failure/functioning are also independent events: $P(A) = P(U_1' \cap U_2) = P(U_1') \cdot P(U_2)$ $P(A) = 0.1 \cdot 0.8 = 0.08$ Why this step? This is the specific event whose probability we need to find given that the instrument has failed. Calculating $P(A)$ (which will also be $P(A \cap I_F)$ in this case, as explained in the next step) is necessary for the numerator of our conditional probability.

Step 4: Calculate the Conditional Probability $p = P(A|I_F)$ We need to find the probability $p$ that "only the first unit failed and second unit is functioning" (event $A$), given that "the instrument fails to operate" (event $I_F$). Using the conditional probability formula: $p = P(A|I_F) = \frac{P(A \cap I_F)}{P(I_F)}$

Now, let's analyze the intersection $A \cap I_F$: If event $A$ occurs (i.e., the first unit failed and the second unit is functioning), then by the definition of the instrument's operation (both units must function), the instrument must fail. This means that event $A$ is a specific way for the instrument to fail, making $A$ a subset of $I_F$ ($A \subseteq I_F$). Therefore, the intersection of $A$ and $I_F$ is simply $A$ itself: $A \cap I_F = A$. So, $P(A \cap I_F) = P(A) = 0.08$.

Now, substitute the calculated values into the conditional probability formula: $p = \frac{P(A)}{P(I_F)} = \frac{0.08}{0.28}$ To simplify the fraction, multiply the numerator and denominator by 100: $p = \frac{8}{28}$ Divide both by their greatest common divisor, 4: $p = \frac{8 \div 4}{28 \div 4} = \frac{2}{7}$ Why this step? This is the core calculation of the problem, directly applying the conditional probability formula. Recognizing the subset relationship ($A \subseteq I_F$) simplifies the calculation of the intersection probability $P(A \cap I_F)$ significantly.

Step 5: Calculate the Final Value $98p$ The problem asks for the value of $98p$. $98p = 98 \cdot \frac{2}{7}$ $98p = \left(\frac{98}{7}\right) \cdot 2$ $98p = 14 \cdot 2$ $98p = 28$ Why this step? This is the final step to provide the specific numerical answer requested by the problem statement.

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Common Mistakes & Tips

• Misinterpreting "Fails to Operate": Ensure you correctly identify all scenarios where the instrument fails. It's often easier to calculate the probability of operating and subtract from 1.
• Confusing $P(A \cap B)$ with $P(A|B)$: Remember that $P(A \cap B)$ is the probability of both events happening, while $P(A|B)$ is the probability of $A$ happening given that $B$ has already occurred.
• Incorrectly Handling Independence: Always check if events are truly independent. If they are, you can multiply their probabilities for intersections; if not, you'll need other methods (like total probability or Bayes' theorem).
• Understanding Subset Relationships: If event $A$ can only happen when event $B$ has already happened (i.e., $A$ is a subset of $B$), then $P(A \cap B) = P(A)$. This is a common simplification in conditional probability problems.

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Summary

This problem is a straightforward application of conditional probability involving independent events. The key steps involved defining the events, calculating the probability of the conditioning event (instrument failure), calculating the probability of the specific event of interest (only the first unit failed and the second unit functioned), and finally applying the conditional probability formula. Careful definition of events and understanding of independence are crucial for solving such problems accurately.

The final answer is $\boxed{28}$.