Two dice are thrown independently. Let A be the event that the number appeared on the 1st die is les

Q: Wait, why is $B \cap C$ empty? Can't a die roll be both?

Think about the parity. Event $B$ requires the first die to be even, while Event $C$ requires it to be odd. A single number cannot be both even and odd simultaneously. They are mutually exclusive, so their intersection is the empty set $\phi$. It is a logical impossibility, not a calculation error!

Q: Why did we use the Distributive Law instead of just drawing it?

Drawing is excellent for intuition, but the Distributive Law is your algebraic safety net. It guarantees you don't miss any overlaps or double-count cases. It turns a visual problem into a rigorous logical proof, which is exactly what JEE Advanced examiners are looking for.

Q: Is there a shortcut for $n(A \cap C)$?

While you could try to derive a complex formula, for a small sample space like $6 \times 6$, systematic counting is actually faster and less prone to error. JEE Advanced rewards the ability to choose the most efficient path—sometimes that's algebra, and sometimes it's careful, disciplined enumeration.

Question

Two dice are thrown independently. Let $A$ be the event that the number appeared on the $1^{st}$ die is less than the number appeared on the $2^{nd}$ die, $B$ be the event that the number appeared on the $1^{st}$ die is even and that on the second die is odd, and $C$ be the event that the number appeared on the $1^{st}$ die is odd and that on the $2^{nd}$ is even. Then

Options:

(A) the number of favourable cases of the event $(A \cup B) \cap C$ is 6

(B) A and B are mutually exclusive

(C) The number of favourable cases of the events A, B and C are 15, 6 and 6 respectively

(D) B and C are independent

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a vast, perfectly ordered grid. This is the universe of our problem: the sample space of two dice. When we throw two dice, we navigate a $6 	imes 6$ coordinate system where every point $(x, y)$ represents a unique reality.

With $x$ being the outcome of the first die and $y$ the second, we have a total of $6 	imes 6 = 36$ possible outcomes. This grid is your map.

The diagonal where $x = y$ splits the grid perfectly. Event $A$, defined by $x < y$, is the entire region strictly above this diagonal. Because the grid is symmetric and there are $6$ points on the diagonal, we are left with $30$ points. Half of these satisfy $x < y$, giving us $n(A) = 15$.

### The Surgical Precision of Set Theory

The problem asks us to evaluate the number of favorable cases for $(A \cup B) \cap C$. At first glance, this expression looks like a tangled knot of logic.

We apply the **Distributive Law of sets** to expand this expression into:
$$(A \cap C) \cup (B \cap C)$$

By doing this, we have transformed one complex problem into two simpler, independent missions. We do not need to hold the whole structure in our heads at once; we simply solve these two pieces and bring them together.

### The Clash of Parity

Let us examine the second piece: $B \cap C$. Event $B$ demands that the first die be even ($x \in \{2, 4, 6\}$) and the second be odd ($y \in \{1, 3, 5\}$).

Event $C$ demands the exact opposite: the first die must be odd ($x \in \{1, 3, 5\}$) and the second even ($y \in \{2, 4, 6\}$).

Ask yourself: can the first die be both even and odd at the same time? This is a logical contradiction. Therefore, $B \cap C = \phi$, the empty set. The number of favorable cases here is **zero**.

### The Final Count: $A \cap C$

We are left with only $A \cap C$. We need to find outcomes where $x < y$, $x$ is odd, and $y$ is even. We fix $x$ to its odd possibilities: $1, 3,$ and $5$.

If $x = 1$, then $y$ can be any even number greater than $1$, which means $y \in \{2, 4, 6\}$. This yields $3$ cases.

If $x = 3$, then $y$ must be an even number greater than $3$, so $y \in \{4, 6\}$. This yields $2$ cases.

If $x = 5$, then $y$ must be an even number greater than $5$, so $y = 6$. This yields $1$ case.

Adding these up, we get:
$$3 + 2 + 1 = 6$$

The total number of favorable cases is $6 + 0 = 6$. You have successfully navigated the logic, and the final answer is **6**.

Select Answer:

Visualized Solution

Sample Space of Two Dice

Event $A$ : $x < y$

Number of Outcomes in $A$

Event $B$ : Even-Odd Pair

Event $C$ : Odd-Even Pair

Analyzing $(A \cup B) \cap C$

Evaluating $B \cap C$

Evaluating $A \cap C$

Counting Favorable Cases for $A \cap C$

Final Answer

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Surgical Precision of Set Theory

The Clash of Parity

The Final Count: $A \cap C$

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Select Answer:

Visualized Solution

Sample Space of Two Dice

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

Number of Outcomes in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

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

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

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A∪B)∩C

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

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

Counting Favorable Cases for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A∩C