Let Ajay will not appear in JEE exam with probability p = 2/7, while both Ajay and Vijay will appear

Q: Wait, why do we need to find $P(A)$ first? Can't we just use the given $P(A^c)$?

Great question! The formula for the region we want, $P(A \cap V^c)$, relies on the total probability of Ajay appearing, which is $P(A)$. Since we are given the probability of him *not* appearing ($P(A^c)$), we must use the complement rule to 'flip' that information into the probability of him appearing. Think of it as converting your data into the right currency before making the transaction!

Q: Is $P(A \cap V^c)$ the same as $P(A) - P(V)$?

Not quite! $P(A) - P(V)$ would subtract the entire circle V, which might include parts outside of A. The correct logic is to subtract only the *overlap* between A and V. That overlap is $P(A \cap V)$. So, the correct expression is $P(A) - P(A \cap V)$.

Q: Why is the intersection $P(A \cap V)$ subtracted? I thought intersection meant adding?

It is a common trap! Intersection ($A \cap V$) represents the 'both' scenario. When we want the region that is *only* A, we take the whole circle A and 'cut out' the part that it shares with V. Since that shared part is inside A, we subtract it to leave behind only the part of A that is exclusive to Ajay.

Question

Let Ajay will not appear in JEE exam with probability $p = \frac{2}{7}$, while both Ajay and Vijay will appear in the exam with probability $q = \frac{1}{5}$. Then the probability, that Ajay will appear in the exam and Vijay will not appear is :

Options:

(A) $\frac{9}{35}$

(B) $\frac{18}{35}$

(C) $\frac{24}{35}$

(D) $\frac{3}{35}$

Sigma Percentile · Accepted Answer

### Defining the Universe

To solve this, we first define our universe using a Venn diagram. Imagine a large rectangle representing the entire sample space, with two circles inside: circle $A$ for Ajay and circle $V$ for Vijay.

We are told that Ajay will not appear with a probability of $P(A^c) = \frac{2}{7}$. Using the complement rule, $P(A) = 1 - P(A^c)$, we find the total probability for Ajay:

$$P(A) = 1 - \frac{2}{7} = \frac{5}{7}$$

This value, $\frac{5}{7}$, represents the entire weight of circle $A$.

### The Overlap: The 'Both' Scenario

Next, we look at the intersection. The problem states that both Ajay and Vijay will appear with a probability of $P(A \cap V) = \frac{1}{5}$.

In our Venn diagram, this is the almond-shaped region where circle $A$ and circle $V$ overlap. This is a crucial piece of our puzzle, as it represents the portion of Ajay's probability that is shared with Vijay.

### The Target: Isolating the Exclusive

We need the probability that Ajay appears, but Vijay does not. This corresponds to the region strictly inside circle $A$ but outside circle $V$, denoted as $P(A \cap V^c)$.

To find this, we take the entire circle $A$ and 'carve out' the intersection. The logic is expressed as:

$$P(A \cap V^c) = P(A) - P(A \cap V)$$

### The Final Calculation

Substituting our gathered values into the formula, we get:

$$P(A \cap V^c) = \frac{5}{7} - \frac{1}{5}$$

To perform this subtraction, we use a common denominator of $35$:

$$P(A \cap V^c) = \frac{25}{35} - \frac{7}{35} = \frac{18}{35}$$

Thus, the probability that Ajay appears and Vijay does not is **$\frac{18}{35}$**.

### Conclusion

You have successfully navigated the logic of sets and probability. By visualizing the Venn diagram and carefully subtracting the intersection, you have transformed a complex word problem into a clear, elegant result.

Remember, in probability, always start by defining your regions—the math will follow naturally.

Animated Solution for Mathematics - Probability: Let Ajay will not appear in JEE exam with probability $p = \frac{2}{7}$ , while both Ajay and Vijay will appear in the exam with probability $q = \frac{1}{5}$ . Then the probability, that Ajay will appear in the exam and Vijay will not appear is :

Select Answer:

Visualized Solution

Defining the Events

Probability of Ajay Not Appearing

Finding $P (A)$

Both Ajay and Vijay Appear

What do we need to find?

The Logical Formula

Substituting the Values

Final Calculation

The Sigma Insight: Addition and Multiplication Theorems

Defining the Universe

The Overlap

The Target

The Final Calculation

Conclusion

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