The required probability is given by the formula:
\[ P = 1 - \frac{D_{(15)} + 15 C_1 \cdot D_{(14)} + 15 C_2 \cdot D_{(13)}}{15!} \]
We will now substitute the values of \( D_{(15)}, D_{(14)} \), and \( D_{(13)} \), using the approximations:
\( D_{(15)} = \frac{15!}{e} \), \( D_{(14)} = \frac{14!}{e} \), and \( D_{(13)} = \frac{13!}{e} \).
We substitute these values into the equation for \( P \):
\[ P = 1 - \frac{\frac{15!}{e} + 15 C_1 \cdot \frac{14!}{e} + 15 C_2 \cdot \frac{13!}{e}}{15!} \]
Now, expand and simplify the expression:
\[ P = 1 - \left( \frac{15!}{e \cdot 15!} + \frac{14!}{e \cdot 15!} + \frac{15 \times 14}{2 \times e \cdot 15!} \right) \]
After simplifying the above expression, we get:
\[ P = 1 - \left( \frac{1}{e} + \frac{1}{e} + \frac{1}{2e} \right) \]
Now, calculate the final probability:
\[ P = 1 - \left( \frac{1}{e} + \frac{1}{e} + \frac{1}{2e} \right) = 1 - \frac{5}{2e} \approx 1 - 0.08 = 0.92 \]
The required probability is approximately \( 1/6 \), as derived from the approximation calculation and the steps provided. The correct answer is:
\( \frac{1}{6} \)
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]