A lot contains 50 defective and 50 non defective bulbs. Two bulbs are drawn at random, one at a time

Q: Wait, why do we reject the negative root here? Or rather, why is the intersection of all three events zero?

That is a fantastic question! It is not about a negative root, but about logical consistency. Event $A$ says the first is $D$. Event $B$ says the second is $ND$. Event $C$ says both are the same (either both $D$ or both $ND$). If $A$ and $B$ occur, we have $(D, ND)$. But if $C$ occurs, we must have $(D, D)$ or $(ND, ND)$. There is no overlap between the set of outcomes satisfying $(A \cap B)$ and the set of outcomes satisfying $C$. Therefore, the intersection is the empty set, and its probability is $0$.

Q: Does 'with replacement' really matter? What if it were without replacement?

It matters immensely! 'With replacement' ensures that the probability of drawing a defective bulb remains $\frac{1}{2}$ for the second draw, regardless of the first. If it were 'without replacement', the events would be dependent because the outcome of the first draw would change the composition of the lot for the second draw. Always check the sampling method first!

Q: I am confused. If $A, B,$ and $C$ are pairwise independent, why aren't they mutually independent?

Think of pairwise independence as a local property—it only looks at how two events interact. Mutual independence is a global property—it looks at the entire system. It is entirely possible for two events to be 'blind' to each other's influence while still being constrained by a third event. This is a classic JEE trap, so remember: pairwise independence is a necessary condition for mutual independence, but it is not sufficient.

Question

A lot contains 50 defective and 50 non defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events $A, B, C$ are defined as $A = (	ext{the first bulb is defective})$, $B = (	ext{the second bulb is non-defective})$, $C = (	ext{the two bulbs are both defective or both non defective})$. Determine whether (i) $A, B, C$ are pairwise independent (ii) $A, B, C$ are independent.

Sigma Percentile · Accepted Answer

### The Illusion of Independence: A Probability Masterclass

Welcome, future engineers! Today, we are diving into one of the most subtle and beautiful topics in probability theory: the distinction between pairwise and mutual independence.

Many students walk into the exam hall thinking that if events are independent in pairs, they must be independent as a group. Today, we are going to shatter that misconception using a simple lot of bulbs.

### Analyzing the Setup

Imagine you are standing in front of a bin containing $100$ bulbs. $50$ are defective, and $50$ are perfectly fine. We are drawing two bulbs, one after the other, with replacement.

Because we replace the bulb, the universe resets after the first draw. The probability of drawing a defective bulb ($D$) is:

$$P(D) = \frac{50}{100} = \frac{1}{2}$$

The probability of a non-defective bulb ($ND$) is also $P(ND) = \frac{1}{2}$.

To visualize this, let us construct a sample space grid. Since each draw has two outcomes, we have $2 	imes 2 = 4$ total outcomes: $(D, D), (D, ND), (ND, D),$ and $(ND, ND)$.

Because the draws are independent, each outcome has a probability of:

$$P(	ext{outcome}) = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$$

This grid is our map. If you can see the grid, you can solve the problem.

### Defining the Events

Let us define our players. Event $A$ is 'the first bulb is defective'. Looking at our grid, this corresponds to the first row: $(D, D)$ and $(D, ND)$.

Summing their probabilities:

$$P(A) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

Event $B$ is 'the second bulb is non-defective'. This corresponds to the second column: $(D, ND)$ and $(ND, ND)$.

Again, we find:

$$P(B) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

Finally, Event $C$ is 'the two bulbs are both defective or both non-defective'. This is the diagonal of our grid: $(D, D)$ and $(ND, ND)$.

Summing these, we get:

$$P(C) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

### The Pairwise Test

Now, let us test for pairwise independence. We check if $P(X \cap Y) = P(X)P(Y)$ for every pair.

For $A$ and $B$: The intersection $A \cap B$ is the outcome $(D, ND)$. Its probability is $\frac{1}{4}$.

Since $P(A)P(B) = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$, they are **independent**.

For $B$ and $C$: The intersection $B \cap C$ is the outcome $(ND, ND)$. Its probability is $\frac{1}{4}$.

Since $P(B)P(C) = \frac{1}{4}$, they are **independent**.

For $C$ and $A$: The intersection $C \cap A$ is the outcome $(D, D)$. Its probability is $\frac{1}{4}$.

Since $P(C)P(A) = \frac{1}{4}$, they are **independent**.

We have proven that $A, B,$ and $C$ are pairwise independent. It feels like we are done, right? Not so fast.

### The Mutual Independence Trap

This is where the JEE examiners love to catch you. For mutual independence, we need the following condition to hold:

$$P(A \cap B \cap C) = P(A)P(B)P(C)$$

Let us look at the intersection of all three: $A \cap B \cap C$. We need the first to be $D$, the second to be $ND$, AND both to be the same.

But if the first is $D$ and the second is $ND$, they are clearly different! It is a logical contradiction. The intersection is the empty set, so:

$$P(A \cap B \cap C) = 0$$

However, the product of the individual probabilities is:

$$P(A)P(B)P(C) = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$$

Since $0 
eq \frac{1}{8}$, the events are **NOT mutually independent**.

We have just witnessed a beautiful mathematical truth: pairwise independence is a local property that does not guarantee global independence. Keep this in your toolkit, and you will never fall for this trap again!

Visualized Solution

Understanding the Setup

Visualizing the Sample Space

Defining Event $A$

Defining Event $B$

Defining Event $C$

Pairwise Independence: $A$ and $B$

Pairwise Independence: $B$ and $C$

Pairwise Independence: $C$ and $A$

Mutual Independence Check

Final Conclusion

The Sigma Insight: Addition and Multiplication Theorems

The Illusion of Independence

Analyzing the Setup

Defining the Events

The Pairwise Test

The Mutual Independence Trap

Similar Questions

Let $A, B, C$ be three mutually independent events. Consider the two statements $S_{1}$ and $S_{2}$ \\ $S_{1} : A$ and $B \cup C$ are independent \\ $S_{2} : A$ and $B \cap C$ are independent \\ Then,

$A$ and $B$ are two independent events. $C$ is event in which exactly one of $A$ or $B$ occurs. Prove that $P (C) \geq P (A \cup B) P (\overset{ˉ}{A} \cap \overset{ˉ}{B})$ .

Let $A$ and $B$ be two events such that $P (A \cup B) = 5/6, P (A \cap B) = 1/4$ and $P (\overset{ˉ}{A}) = 1/4$ , where $\overset{ˉ}{A}$ stands for the complement of the event $A$ . Then the events $A$ and $B$ are

Let two fair six-faced dice $A$ and $B$ be thrown simultaneously. If $E_{1}$ is the event that die $A$ shows up four, $E_{2}$ is the event that die $B$ shows up two and $E_{3}$ is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true?

Two events $A$ and $B$ have probabilities $0.25$ and $0.50$ respectively. The probability that both $A$ and $B$ occur simultaneously is $0.14$ . Then the probability that neither $A$ nor $B$ occurs is

An experiment has 10 equally likely outcomes. Let $A$ and $B$ be non-empty events of the experiment. If $A$ consists of 4 outcomes, the number of outcomes that $B$ must have so that $A$ and $B$ are independent, is

Two fair dice are tossed. Let $x$ be the event that the first die shows an even number and $y$ be the event that the second die shows an odd number. The two events $x$ and $y$ are :

Let $A$ and $B$ be two events such that the probability that exactly one of them occurs is $2/5$ and the probability that $A$ or $B$ occurs is $1/2$ , then the probability of both of them occur together is

Visualized Solution

Understanding the Setup

Visualizing the Sample Space

Defining Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Defining Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Defining Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Pairwise Independence: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Pairwise Independence: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Pairwise Independence: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A