Let \( \{ X_n \}_{n \geq 1} \) be a sequence of independent random variables with \[ \Pr(X_n = -\frac{1}{2^n}) = \Pr(X_n = \frac{1}{2^n}) = \frac{1}{2}, \quad \forall n \in \mathbb{N}. \] Suppose that \( \sum_{i=1}^{n} X_i \) converges to \( U \) as \( n \to \infty \). Then \( 6 \Pr(U \leq \frac{2}{3}) \) equals ___________ (answer in integer).
Step 1: Understanding the distribution of \( X_n \).
Each \( X_n \) is a random variable with two possible outcomes, \( -\frac{1}{2^n} \) and \( \frac{1}{2^n} \), each with probability \( \frac{1}{2} \).
Step 2: Analyze the sum \( S_n = \sum_{i=1}^{n} X_i \).
The random variables \( X_n \) are independent and symmetric, meaning that each step contributes positively or negatively by a decreasing amount, \( \pm \frac{1}{2^n} \). As \( n \to \infty \), the sum \( S_n = \sum_{i=1}^{n} X_i \) converges to the limit \( U \), which is a random variable. The limit \( U \) is the sum of the infinite series: \[ U = \sum_{n=1}^{\infty} X_n. \] This sum converges because the magnitude of each term decreases exponentially.
Step 3: Distribution of \( U \).
The limiting distribution of \( U \) is a uniform distribution on the interval \( [-1, 1] \), since the steps \( X_n \) contribute to the value of \( U \) in a balanced way and the sum converges. Therefore, \( U \) is uniformly distributed on \( [-1, 1] \).
Step 4: Calculate \( \Pr(U \leq \frac{2}{3}) \).
For a uniform random variable \( U \) on \( [-1, 1] \), the probability that \( U \leq \frac{2}{3} \) is simply the proportion of the interval \( [-1, 1] \) that is less than or equal to \( \frac{2}{3} \). The length of the interval from \( -1 \) to \( \frac{2}{3} \) is: \[ \frac{2}{3} - (-1) = \frac{5}{3}. \] Since the total length of the interval \( [-1, 1] \) is 2, the probability is: \[ \Pr(U \leq \frac{2}{3}) = \frac{\frac{5}{3}}{2} = \frac{5}{6}. \]
Step 5: Multiply by 6.
We are asked to find \( 6 \Pr(U \leq \frac{2}{3}) \), so: \[ 6 \times \frac{5}{6} = 5. \]
Thus, the correct answer is \( \boxed{5} \).
Let \( (X_1, X_2, X_3) \) follow the multinomial distribution with the number of trials being 100 and the probability vector \( \left( \frac{3}{10}, \frac{1}{10}, \frac{3}{5} \right) \).
Then \( E(X_2 | X_3 = 40) \) equals:
Let \( X \) follow a 10-dimensional multivariate normal distribution with zero mean vector and identity covariance matrix. Define \( Y = \log_e \sqrt{X^T X} \) and let \( M_Y(t) \) denote the moment generating function of \( Y \) at \( t \), \( t>-10 \). Then \( M_Y(2) \) equals _________ (answer in integer).
Let \( x_1 = 0, x_2 = 1, x_3 = 1, x_4 = 1, x_5 = 0 \) be observed values of a random sample of size 5 from \( {Bin}(1, \theta) \) distribution, where \( \theta \in (0, 0.7] \). Then the maximum likelihood estimate of \( \theta \) based on the above sample is ________ (rounded off to two decimal places).
An electricity utility company charges ₹7 per kWh. If a 40-watt desk light is left on for 10 hours each night for 180 days, what would be the cost of energy consumption? If the desk light is on for 2 more hours each night for the 180 days, what would be the percentage-increase in the cost of energy consumption?
In the context of the given figure, which one of the following options correctly represents the entries in the blocks labelled (i), (ii), (iii), and (iv), respectively?

A bag contains Violet (V), Yellow (Y), Red (R), and Green (G) balls. On counting them, the following results are obtained:
(i) The sum of Yellow balls and twice the number of Violet balls is 50.
(ii) The sum of Violet and Green balls is 50.
(iii) The sum of Yellow and Red balls is 50.
(iv) The sum of Violet and twice the number of Red balls is 50.
Which one of the following Pie charts correctly represents the balls in the bag?