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 \( \frac{18}{35} \).
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: