Two fair dice are tossed. Let x be the event that the first die shows an even number and y be the ev

Q: Wait, why aren't they mutually exclusive? They seem like totally different conditions!

Mutual exclusivity isn't about the conditions being 'different'; it's about whether they can happen at the same time. If you roll a 2 and a 1, both events happen. Since they can coexist, they aren't mutually exclusive.

Q: Why does independence mean $P(x \cap y) = P(x) \cdot P(y)$?

Think of independence as 'no influence.' If knowing $x$ happened tells you nothing about the likelihood of $y$, then $P(y|x) = P(y)$. Since $P(x \cap y) = P(x) \cdot P(y|x)$, substituting $P(y)$ for $P(y|x)$ gives you the product rule.

Q: Is 'None of these' a common answer in JEE? It feels like I missed something.

In JEE, 'None of these' is a valid and powerful option. It tests your confidence. If you've proven they are independent and not mutually exclusive, trust your math. Don't force an answer into the other options.

Question

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 :

Options:

(A) Mutually exclusive

(B) Independent and mutually exclusive

(C) Dependent

(D) None of these

Sigma Percentile · Accepted Answer

### The Geometry of Chance: A Journey into Independence

Welcome, future engineers! Today, we are not just solving a probability problem; we are exploring the very architecture of randomness.

When we toss two dice, we are essentially navigating a $6 	imes 6$ grid of possibilities. Imagine this grid as a map where the vertical axis is the first die and the horizontal axis is the second.

Every single point on this map, from $(1,1)$ to $(6,6)$, is a unique reality. There are $36$ such realities in total, and our job is to find where our specific events live on this map.

### Mapping the Events

Let us define our territory. Event $x$ is the condition where the first die shows an even number. The even numbers are $2, 4,$ and $6$.

This means event $x$ occupies three entire rows of our grid. Since each row has $6$ outcomes, we have $3 	imes 6 = 18$ favorable outcomes.

The probability is calculated as:

$$P(x) = \frac{18}{36} = \frac{1}{2}$$

It is a clean, fifty-fifty split. Now, look at event $y$, where the second die shows an odd number. The odd numbers are $1, 3,$ and $5$.

This event occupies three entire columns. Again, there are $18$ outcomes, so:

$$P(y) = \frac{18}{36} = \frac{1}{2}$$

### The Intersection

Here is where the magic happens. We need to know if these events can coexist. This is the intersection, $x \cap y$.

Geometrically, this is where our three rows (from event $x$) overlap with our three columns (from event $y$). If you visualize the grid, you will see a $3 	imes 3$ sub-grid of overlap, which accounts for $9$ outcomes.

Thus, the probability of the intersection is:

$$P(x \cap y) = \frac{9}{36} = \frac{1}{4}$$

Because this probability is not zero, these events are absolutely not mutually exclusive. They can, and do, happen together.

### The Test of Independence

Now, we reach the heart of the problem: Independence. In probability, independence is a beautiful concept. It means that the occurrence of one event provides zero information about the other.

Mathematically, we test this with the product rule:

$$P(x \cap y) = P(x) \cdot P(y)$$

We calculated $P(x) = \frac{1}{2}$ and $P(y) = \frac{1}{2}$. Their product is:

$$\frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$$

Look at our intersection probability: it is also $\frac{1}{4}$. The condition is satisfied perfectly!

### The Conclusion

We have proven that the events are independent, but they are not mutually exclusive. When you look at the options provided, none of them describe this specific combination.

This is a classic JEE moment where the test asks you to trust your derivation over your intuition. The answer is **'None of these.'**

You have navigated the grid, calculated the intersections, and verified the independence. You have mastered the logic. Keep this clarity, and you will conquer any probability problem that comes your way.

Animated Solution for Mathematics - Probability: 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 :

Select Answer:

Visualized Solution

The Sample Space $S$

Defining Event $x$

Visualizing Event $x$

Probability of Event $x$

Defining Event $y$

Visualizing Event $y$

Probability of Event $y$

Testing Mutual Exclusivity

Visualizing the Intersection $x \cap y$

Probability of $x \cap y$

Conclusion on Mutual Exclusivity

Testing Independence

Checking the Independence Condition

Conclusion on Independence

Final Verdict

The Sigma Insight: Addition and Multiplication Theorems

The Geometry of Chance

Mapping the Events

The Intersection

The Test of Independence

The Conclusion

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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

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Visualized Solution

The Sample Space .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

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

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

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

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

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

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