Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up

Q: Wait, if $E_1$ and $E_2$ are independent, and $E_2$ and $E_3$ are independent, why aren't they all independent?

This is the classic 'Pairwise vs. Mutual' trap! Pairwise independence only tells us about the relationships between two events at a time. It says nothing about the collective behavior of all three. Think of it like a three-way handshake—if two people are friends, and the second and third are friends, it doesn't automatically mean all three are in the same group. In probability, mutual independence is a much stronger condition that requires the product of all three probabilities to equal the probability of their intersection.

Q: How do I visualize the intersection of three events?

Think of the sample space as a $6 \times 6$ grid. $E_1$ is a vertical line (where die A is 4), $E_2$ is a horizontal line (where die B is 2), and $E_3$ is a checkerboard pattern (where the sum is odd). The intersection of all three is the specific point where all these conditions overlap. If that point doesn't exist or doesn't satisfy the condition of $E_3$, the intersection is empty, and its probability is zero.

Q: Why do we reject the negative root here? Wait, there are no roots here, but why is the answer 0?

You are right, there are no roots! The probability of an impossible event is always zero. Since $E_1$ forces the outcome $(4,2)$, and the sum $4+2=6$ is even, it is impossible for $E_3$ (which requires an odd sum) to occur simultaneously with $E_1$ and $E_2$. Thus, $P(E_1 \cap E_2 \cap E_3) = 0$.

Question

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?

Options:

(A) $E_1$ and $E_3$ are independent

(B) $E_1, E_2$ and $E_3$ are independent

(C) $E_1$ and $E_2$ are independent

(D) $E_2$ and $E_3$ are independent

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To understand the concept of independence, we consider two fair, six-faced dice, $A$ and $B$. The total number of outcomes in our sample space is:

$$6 	imes 6 = 36$$

We visualize this as a $6 	imes 6$ grid where the $x$-axis represents the outcome of die $A$ and the $y$-axis represents the outcome of die $B$. Every point on this grid is an equally likely outcome.

### Defining the Events

We define three specific events within this sample space:

1. **Event $E_1$**: Die $A$ shows a $4$. This corresponds to the vertical column where $x=4$. There are $6$ such points, so:
$$P(E_1) = \frac{6}{36} = \frac{1}{6}$$

2. **Event $E_2$**: Die $B$ shows a $2$. This corresponds to the horizontal row where $y=2$. There are $6$ such points, so:
$$P(E_2) = \frac{6}{36} = \frac{1}{6}$$

3. **Event $E_3$**: The sum of the numbers is odd. For a sum to be odd, we must have one odd number and one even number. Counting the points on our grid, we find $18$ such outcomes, so:
$$P(E_3) = \frac{18}{36} = \frac{1}{2}$$

### Testing Pairwise Independence

For any two events $X$ and $Y$, they are independent if $P(X \cap Y) = P(X)P(Y)$. Let us verify this for our pairs.

For $E_1$ and $E_2$, the intersection $E_1 \cap E_2$ is the single point $(4,2)$. The probability is $P(E_1 \cap E_2) = \frac{1}{36}$, which matches the product $P(E_1)P(E_2) = \frac{1}{6} 	imes \frac{1}{6} = \frac{1}{36}$.

For $E_1$ and $E_3$, the intersection $E_1 \cap E_3$ requires die $A=4$ and the sum to be odd. Since $4$ is even, die $B$ must be odd ($1, 3, 5$), yielding $3$ outcomes: $(4,1), (4,3), (4,5)$. Thus:
$$P(E_1 \cap E_3) = \frac{3}{36} = \frac{1}{12}$$

This matches the product $P(E_1)P(E_3) = \frac{1}{6} 	imes \frac{1}{2} = \frac{1}{12}$. By symmetry, $E_2$ and $E_3$ are also pairwise independent.

### The Trap: Mutual Independence

The events $E_1, E_2,$ and $E_3$ are mutually independent only if $P(E_1 \cap E_2 \cap E_3) = P(E_1)P(E_2)P(E_3)$.

Let us calculate the triple intersection. $E_1$ forces die $A=4$ and $E_2$ forces die $B=2$. The only outcome in $E_1 \cap E_2$ is $(4,2)$.

However, the sum of $(4,2)$ is $6$, which is even. Since $E_3$ requires an odd sum, the intersection $E_1 \cap E_2 \cap E_3$ is the empty set $\emptyset$, and its probability is $0$.

Comparing this to the product of the individual probabilities:
$$P(E_1)P(E_2)P(E_3) = \frac{1}{6} 	imes \frac{1}{6} 	imes \frac{1}{2} = \frac{1}{72}$$

Since $0 
eq \frac{1}{72}$, the events are **NOT mutually independent**. This proves that pairwise independence does not imply mutual independence, a critical distinction for JEE Advanced mathematics.

Select Answer:

Visualized Solution

Visualizing the Sample Space

Analyzing Event $E_{1}$

Analyzing Event $E_{2}$

Analyzing Event $E_{3}$

Pairwise Independence: $E_{1}$ and $E_{2}$

Pairwise Independence: $E_{1}$ and $E_{3}$

Pairwise Independence: $E_{2}$ and $E_{3}$

The Trap: Mutual Independence

Conclusion

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

Defining the Events

Testing Pairwise Independence

The Trap

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Select Answer:

Visualized Solution

Visualizing the Sample Space

Analyzing Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​

Analyzing Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Analyzing Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E3​

Pairwise Independence: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Pairwise Independence: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E3​

Pairwise Independence: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E3​