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:
- $\frac{9}{35}$
- $\frac{18}{35}$
- $\frac{24}{35}$
- $\frac{3}{35}$
The Correct Option is B
Approach Solution - 1
To solve the problem, let's define the different probabilities given and calculate the required probability step-by-step:
- The probability that Ajay will not appear in the JEE exam is given by \(p = \frac{2}{7}\). Thus, the probability that Ajay will appear in the exam is \(1 - p = 1 - \frac{2}{7} = \frac{5}{7}\).
- The probability that both Ajay and Vijay will appear in the exam is given by \(q = \frac{1}{5}\).
- We need to find the probability that Ajay will appear in the exam and Vijay will not appear. We can denote this as \(P(A \cap V')\).
- Using the total probability for Ajay's appearance:
- \(P(A \cap V) + P(A \cap V') = P(A)\)
- Substituting the known values: \(\frac{1}{5} + P(A \cap V') = \frac{5}{7}\)
- Solve the equation for \(P(A \cap V')\):
- \(P(A \cap V') = \frac{5}{7} - \frac{1}{5}\)
- To subtract these fractions, find a common denominator (35):
- \(\frac{5}{7} = \frac{25}{35} \quad \text{and} \quad \frac{1}{5} = \frac{7}{35}\)
- Therefore, \(P(A \cap V') = \frac{25}{35} - \frac{7}{35} = \frac{18}{35}\)
Thus, the probability that Ajay will appear in the exam and Vijay will not appear is \(\frac{18}{35}\).
The correct answer is \(\frac{18}{35}\).
Approach Solution -2
Given:
\[ P(\text{Ajay does not appear}) = p = \frac{2}{7}, \quad P(\text{Ajay and Vijay both appear}) = q = \frac{1}{5} \]
Let:
\[ P(\text{Ajay appears}) = 1 - p = 1 - \frac{2}{7} = \frac{5}{7} \]
Let \( P(\text{Vijay appears}) = v \). The probability that both Ajay and Vijay appear is given by:
\[ P(\text{Ajay appears}) \times P(\text{Vijay appears}) = q \]
Substituting the given values:
\[ \frac{5}{7} \times v = \frac{1}{5} \]
Solving for \( v \):
\[ v = \frac{1}{5} \times \frac{7}{5} = \frac{7}{25} \]
Thus, the probability that Vijay does not appear is:
\[ P(\text{Vijay does not appear}) = 1 - v = 1 - \frac{7}{25} = \frac{18}{25} \]
Finding the Desired Probability
The probability that Ajay will appear in the exam and Vijay will not appear is given by:
\[ P(\text{Ajay appears}) \times P(\text{Vijay does not appear}) = \frac{5}{7} \times \frac{18}{25} \]
Calculating the product:
\[ P(\text{Ajay appears and Vijay does not appear}) = \frac{5 \times 18}{7 \times 25} = \frac{90}{175} = \frac{18}{35} \]
Conclusion: The probability that Ajay will appear in the exam and Vijay will not appear is \( \frac{18}{35} \).
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Probability Questions
- Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from bag A, if the ball drawn is white, is:
- The probability that Ajay will not go to office is \(\frac{1}{5}\) and probability that Ajay and Vijay will not go to the office is \(\frac{2}{7}\), if their visits to office is independent of each other, then find the probability that Ajay will go to the office, but Vijay will not go, is
- Given set = {1, 2, 3, ..., 50} one number is selected randomly from the set. Find the probability that number is multiple of 4 or 6 or 7.
- A coin is biased so that a head is twice as likely as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is
- An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability that the first draw gives all white balls, and the second draw gives all black balls, is:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,