Question

A bag contains 5 white balls and 10 black balls. In a random experiment, \( n \) balls are drawn from the bag one at a time with replacement. Let \( S_n \) denote the total number of black balls drawn in the experiment. The expectation of \( S_{100} \) denoted by \( E[S_{100}] \) is \( \ \ \ \ \ \) (Round off to one decimal place).

Hint:

For binomial distributions, the expectation is given by \( E[X] = np \), and variance is \( \text{Var}(X) = np(1-p) \).

Answer 1

Correct Answer: 66.6

Step 1: Define the probability of drawing a black ball.
The bag contains a total of \( 5 + 10 = 15 \) balls. The probability of drawing a black ball in one draw is: \[ P(B) = \frac{10}{15} = \frac{2}{3}. \] Step 2: Compute the expectation.
Since \( S_{100} \) denotes the number of black balls drawn in 100 trials, it follows a binomial distribution: \[ S_{100} \sim \text{Binomial}(100, \frac{2}{3}). \] The expectation of a binomially distributed random variable \( X \sim \text{Binomial}(n, p) \) is given by: \[ E[X] = n p. \] Substituting the values: \[ E[S_{100}] = 100 \times \frac{2}{3} = 66.7. \]