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

Q: Wait, why do we reject the negative root here? Or rather, why is $B \cap C$ empty?

Great question! Let's look at the definitions. Event $B$ requires the first die to be even, and Event $C$ requires the first die to be odd. Since a single die roll results in a single number, that number cannot be both even and odd at the same time. Thus, there are no outcomes that satisfy both conditions, making their intersection the empty set, $\emptyset$.

Q: Why can't we just calculate the probability of $A \cup B$ first and then intersect with $C$?

You absolutely could! However, calculating $A \cup B$ involves finding the union of two sets, which requires careful handling of their intersection to avoid double-counting. Using the Distributive Law, $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$, is a much more elegant and less error-prone path because it breaks the problem into smaller, disjoint pieces.

Q: How do I know when to use the Distributive Law versus just listing all outcomes?

In competitive exams like JEE, time is your most precious resource. While listing outcomes is safe for small sample spaces like $36$ outcomes, it is prone to silly mistakes. The Distributive Law is a powerful algebraic tool that simplifies the logic before you even start counting. Always look for ways to simplify the set expression first; it often reveals that parts of the problem are null or redundant!

Question

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

### The Geometry of Chance: A Journey Through Dice Welcome, future engineers! Today, we are not just solving a probability problem; we are embarking on a journey to visualize the hidden geometry of chance. When we throw two dice, we are essentially navigating a $6 imes 6$ grid of possibilities. Let's break this down, step by step, and see how logic can turn a seemingly complex problem into a simple, elegant solution. ### Mapping the Universe Imagine a grid where the x-axis represents the result of the first die and the y-axis represents the second. Each point $(x, y)$ is a unique outcome. With six faces on each die, our entire universe, the sample space $S$, consists of $6 imes 6 = 36$ points. This is our foundation. Whenever you face a probability problem involving dice, always start by visualizing this grid. It grounds your intuition. ### Defining the Events We have three events defined by specific constraints: 1. **Event $A$**: The first die is strictly less than the second ($x < y$). Geometrically, these are all the points lying strictly above the diagonal line $y = x$. To count these, we can use combinations. We need to choose two distinct numbers from the set $\{1, 2, 3, 4, 5, 6\}$. The number of ways to do this is: $$n(A) = \binom{6}{2} = \frac{6 imes 5}{2} = 15$$ 2. **Event $B$**: The first die is even, and the second is odd. We have 3 choices for the first die $\{2, 4, 6\}$ and 3 choices for the second $\{1, 3, 5\}$. Thus, $n(B) = 3 imes 3 = 9$. 3. **Event $C$**: The first die is odd, and the second is even. Similarly, we have 3 choices for the first die $\{1, 3, 5\}$ and 3 choices for the second $\{2, 4, 6\}$. Thus, $n(C) = 3 imes 3 = 9$. ### The Power of Distributive Logic Now, we face the core of the problem: finding the favorable cases for $(A \cup B) \cap C$. At first glance, this looks intimidating. But here is where the beauty of set theory shines. We apply the Distributive Law: $$(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$$ This transformation is the key. It splits our complex problem into two distinct, manageable parts. Now, let's analyze the second part: $B \cap C$. Event $B$ demands the first die be even, while Event $C$ demands it be odd. Since a number cannot be both even and odd, $B \cap C = \emptyset$. The intersection is empty! ### The Final Count Because $B \cap C$ is the null set, our expression simplifies beautifully to just $A \cap C$. We are now looking for points that satisfy both Event $A$ ($x < y$) and Event $C$ ($x$ is odd, $y$ is even). Let's count them systematically by iterating through the possible values of $x$ (which must be odd): - **If $x = 1$**: $y$ must be even and $y > 1$. The possible values for $y$ are $\{2, 4, 6\}$. That gives us 3 pairs: $(1, 2), (1, 4), (1, 6)$. - **If $x = 3$**: $y$ must be even and $y > 3$. The possible values for $y$ are $\{4, 6\}$. That gives us 2 pairs: $(3, 4), (3, 6)$. - **If $x = 5$**: $y$ must be even and $y > 5$. The only possible value for $y$ is $\{6\}$. That gives us 1 pair: $(5, 6)$. Adding these up, we get $3 + 2 + 1 = 6$ favorable cases. The logic holds, the math is clean, and we have arrived at the answer. Remember, in JEE, the most complex-looking problems often collapse into simple truths if you apply the right logical framework. Keep practicing, keep visualizing, and most importantly, keep falling in love with the process!

Select Answer:

Visualized Solution

Visualizing the Sample Space

Defining Event $A$

Calculating $n (A)$

Defining Event $B$

Defining Event $C$

Applying Distributive Law

Analyzing $B \cap C$

Simplifying the Expression

Finding $A \cap C$

Counting $A \cap C$ (Case $x = 1$ )

Counting $A \cap C$ (Case $x = 3$ )

Counting $A \cap C$ (Case $x = 5$ )

Total Favourable Cases

Checking Other Options

The Sigma Insight: Classical Definition of Probability

The Geometry of Chance

Mapping the Universe

Defining the Events

The Power of Distributive Logic

The Final Count

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Select Answer:

Visualized Solution

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(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