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:
\( \frac{1}{6} \)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to: