Step 1: Define the random variable \( x \). \( x \) represents the number of defective oranges drawn.
Step 2: Determine probabilities for \( x \). \[ P(x=0) = \frac{\binom{7}{2}}{\binom{10}{2}} = \frac{21}{45} = \frac{7}{15} \] \[ P(x=1) = \frac{\binom{3}{1} \binom{7}{1}}{\binom{10}{2}} = \frac{21}{45} = \frac{7}{15} \] \[ P(x=2) = \frac{\binom{3}{2} \binom{7}{0}}{\binom{10}{2}} = \frac{3}{45} = \frac{1}{15} \]
Step 3: Calculate expected value \( E(x) \). \[ E(x) = 0 \cdot \frac{7}{15} + 1 \cdot \frac{7}{15} + 2 \cdot \frac{1}{15} = \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5} \]
Step 4: Calculate expected value \( E(x^2) \). \[ E(x^2) = 0^2 \cdot \frac{7}{15} + 1^2 \cdot \frac{7}{15} + 2^2 \cdot \frac{1}{15} = 0 + \frac{7}{15} + \frac{4}{15} = \frac{11}{15} \]
Step 5: Calculate the variance \({Var}(x)\). \[ {Var}(x) = E(x^2) - [E(x)]^2 = \frac{11}{15} - \left(\frac{3}{5}\right)^2 = \frac{11}{15} - \frac{9}{25} = \frac{55}{75} - \frac{27}{75} = \frac{28}{75} \]
The probability distribution of the random variable X is given by
X | 0 | 1 | 2 | 3 |
---|---|---|---|---|
P(X) | 0.2 | k | 2k | 2k |
Find the variance of the random variable \(X\).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: