An electric instrument consists of two units. Each unit must function independently for the instrume

Q: Wait, why do we subtract from 1 to find the failure probability? Can't we just add the failure probabilities?

That is a classic trap! If you simply add $P(\bar{I_1}) + P(\bar{I_2})$, you are double-counting the case where *both* units fail. The complement rule $P(F) = 1 - P(Success)$ is the most robust way to capture the entire 'failure' region without any overlap. Always trust the complement rule when dealing with 'at least one' or 'failure' scenarios.

Q: Why is the intersection of $\bar{I_1}$ and $I_2$ just the product of their probabilities?

Because the problem explicitly states the units are independent. Independence is a powerful tool—it allows us to multiply probabilities directly. If they were dependent, we would need conditional probabilities, but here, the failure of one unit does not change the likelihood of the other unit working. It is a beautiful simplification!

Q: Why is the denominator $P(F)$ and not just 1?

Because we are given that the instrument *has already failed*. We are no longer looking at the entire sample space of all possibilities. We have 'zoomed in' to the subset of outcomes where the instrument is broken. This is the very essence of conditional probability: restricting the universe to the condition provided.

Question

An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is $p$, then $98p$ is equal to .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an engineer standing before a complex electric instrument. It is a dual-unit system, and for it to hum with life, both units must be in perfect harmony.

In the language of probability, we define $I_1$ as the event that the first unit functions, and $I_2$ as the event that the second unit functions. The instrument operates only if both occur—a beautiful intersection of events, $I_1 \cap I_2$.

But today, the instrument is silent. It has failed. This failure is not just a setback; it is a mathematical condition that changes everything.

### Defining the Universe

We are given the probabilities of success: $P(I_1) = 0.9$ and $P(I_2) = 0.8$. Because these units are independent, the probability of them both working is simply the product:

$$P(I_1 \cap I_2) = 0.9 	imes 0.8 = 0.72$$

This is our 'success' region. But our instrument has failed. Let $F$ be the event of failure.

The failure is the complement of success. Mathematically:

$$P(F) = 1 - P(I_1 \cap I_2) = 1 - 0.72 = 0.28$$

This $0.28$ represents the entire 'red zone' of failure—the scenarios where the instrument refuses to start.

### The Conditional Lens

Now, we are asked a specific question: given that the instrument has failed, what is the probability that *only* the first unit failed while the second unit is still functioning?

This is the heart of conditional probability. We are not looking at the whole world anymore; we are looking only at the $0.28$ probability space where the instrument is broken. We need to find $p = P(\bar{I_1} \cap I_2 \mid F)$.

Using the definition of conditional probability, we have:

$$p = \frac{P((\bar{I_1} \cap I_2) \cap F)}{P(F)}$$

Since the event 'only the first unit failed and the second functions' is a subset of the total failure event $F$, the intersection $(\bar{I_1} \cap I_2) \cap F$ is simply $(\bar{I_1} \cap I_2)$.

### The Final Calculation

First, let us calculate the probability of this specific failure mode: $P(\bar{I_1} \cap I_2)$. Since the units are independent, their complements are also independent.

Thus, $P(\bar{I_1} \cap I_2) = P(\bar{I_1}) 	imes P(I_2)$. We know $P(\bar{I_1}) = 1 - 0.9 = 0.1$.

So, $P(\bar{I_1} \cap I_2) = 0.1 	imes 0.8 = 0.08$. Now, we apply our conditional formula:

$$p = \frac{0.08}{0.28}$$

Simplifying this fraction is a joy. Multiplying by $100$ gives us $\frac{8}{28}$. Dividing both by $4$, we arrive at $p = \frac{2}{7}$.

The problem asks for $98p$. Substituting our value, we get:

$$98 	imes \frac{2}{7} = 14 	imes 2 = 28$$

And there it is—the elegance of the result. We navigated the failure, isolated the specific condition, and arrived at the final answer: **28**.

Enter Numerical Value:

Visualized Solution

Defining the Events $I_{1}$ and $I_{2}$

Given Probabilities

Probabilities of Failure

Instrument Failure Condition

Calculating $P (F)$

Defining the Target Event

Calculating $P (\overset{ˉ}{I_{1}} \cap I_{2})$

Applying Conditional Probability

Simplifying the Fraction

Final Calculation of $98 p$

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Defining the Universe

The Conditional Lens

The Final Calculation

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