There are four machines and it is known that exactly two of them are faulty. They are tested, one by

Q: Wait, why do we use combinations ($^nC_r$) instead of permutations? Doesn't the order of testing matter?

That is a fantastic question! In this specific problem, we are looking at the *set* of the first two machines tested. Whether we pick Machine A then Machine B, or Machine B then Machine A, the *pair* of machines tested is the same. Since we only care about the composition of the first two tests, combinations are the perfect tool to count the total possible pairs. If you used permutations, you would just be multiplying both the numerator and denominator by $2!$, and they would cancel out anyway!

Q: What if the question asked for the probability that exactly 3 tests are needed?

Great intuition! If we need exactly 3 tests, it means the first two tests must contain exactly one faulty machine and one good machine, and the third test must be the second faulty machine. You would calculate the number of ways to pick one faulty and one good machine in the first two tests, and then multiply by the probability that the third machine is the remaining faulty one. It's all about breaking the event down into its logical components!

Q: Is the conditional probability method always better than the combination method?

Neither is 'better'—they are just different lenses to view the same truth! The combination method is excellent for visualizing the entire sample space at once, while the conditional probability method (multiplying $P(F_1) \times P(F_2|F_1)$) is often faster and more intuitive for sequential events. As you progress in your JEE journey, mastering both will make you unstoppable because you can use one to verify the other!

Question

There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is

Options:

(A) $1/3$

(B) $1/6$

(C) $1/2$

(D) $1/4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a high-tech laboratory with four machines in a row. You know for an absolute fact that exactly two of them are faulty, and the other two are in perfect working condition.

Your mission is to test them one by one until you have identified both faulty machines. The moment you find the second faulty machine, you stop. We must determine the probability that you find both faulty machines in exactly two tests.

### Phase 1: The Sample Space

To solve this, we must first understand the "universe" of possibilities. We are picking two machines out of four to be our first two tests.

In the language of mathematics, we are looking for the number of ways to choose a subset of 2 items from a set of 4. This is a classic combination problem, denoted as $^4C_2$.

Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$, we calculate the total number of pairs:

$$^4C_2 = \frac{4 	imes 3}{2 	imes 1} = 6$$

There are exactly 6 different pairs of machines that could be our first two tests. This represents our total sample space.

### Phase 2: The Favorable Event

Now, let us look for the "Eureka" moment. We want the probability that exactly two tests are needed.

This means the very first machine we test must be faulty, AND the second machine we test must also be faulty. If either of these machines were good, we would be forced to continue testing.

Therefore, our favorable outcome is picking the specific pair consisting of the two faulty machines. Since there are only 2 faulty machines in total, there is only $^2C_2 = 1$ way to choose them. Our favorable outcome count is 1.

### Phase 3: The Elegant Alternative

Before we conclude, let us look at this through the lens of conditional probability. It is a beautiful way to think about sequential events.

What is the probability that the first machine you pick is faulty? Since there are 2 faulty machines out of 4, the probability is $\frac{2}{4}$.

Now, imagine you have successfully picked a faulty machine. You have one faulty machine left and three machines total remaining. The probability that your second pick is also faulty is $\frac{1}{3}$.

By multiplying these sequential steps, we arrive at the final probability:

$$P = \frac{2}{4} 	imes \frac{1}{3} = \frac{2}{12} = \frac{1}{6}$$

### Conclusion: The Beauty of Logic

Whether you use the combination method or the conditional probability method, the result is the same: **$\frac{1}{6}$**.

This is the beauty of mathematics—different paths, when walked with logical precision, always lead to the same summit. Probability is not merely a game of chance; it is the science of certainty.

Animated Solution for Mathematics - Probability: There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is

Select Answer:

Visualized Solution

Visualizing the Machines

Defining the Event

Total Possible Outcomes

Calculating Total Outcomes

Favorable Outcomes

Calculating Favorable Outcomes

Final Probability Calculation

Alternative Method

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Conclusion

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