Key Concepts and Formulas

• Discrete Random Variable and Probability Distribution: A random variable $X$ is discrete if its possible values are countable. Its probability distribution lists each possible value $x_i$ and its corresponding probability $P(X=x_i)$. The sum of all probabilities must be 1: $\sum P(X=x_i) = 1$.
• Expected Value (Mean) $E[X]$: This is the long-run average value of $X$. For a discrete random variable, it's calculated as the sum of each value multiplied by its probability: $E[X] = \sum x_i P(X=x_i)$
• Variance $\sigma^2$: This measures the spread or dispersion of the values of $X$ around its mean. A common and computationally efficient formula for variance is: $\sigma^2 = E[X^2] - (E[X])^2$ where $E[X^2] = \sum x_i^2 P(X=x_i)$.

Step-by-Step Solution

Step 1: Define the Random Variable and Its Possible Values We are given a bag with 4 white (W) balls and 6 black (B) balls, for a total of 10 balls. Three balls are drawn at random. The random variable $X$ represents the number of white balls among the three drawn balls. Since we draw 3 balls and there are 4 white balls available, the possible number of white balls drawn can be 0, 1, 2, or 3.

• $X=0$: 0 white balls, 3 black balls
• $X=1$: 1 white ball, 2 black balls
• $X=2$: 2 white balls, 1 black ball
• $X=3$: 3 white balls, 0 black balls

Step 2: Calculate the Total Number of Outcomes To determine the probabilities, we first need to find the total number of ways to draw 3 balls from the 10 available balls. Since the order of drawing does not matter, we use combinations. The total number of ways to choose 3 balls from 10 is given by $^{10}C_3$. ${^{10}C_3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$ This value, 120, will be the denominator for all our probability calculations.

Step 3: Determine the Probability Distribution of X We calculate $P(X=x)$ for each possible value of $X$. For each case, we select $x$ white balls from the 4 white balls and $(3-x)$ black balls from the 6 black balls.

• For $X=0$ (0 white balls): We choose 0 white balls from 4 ($^4C_0$) AND 3 black balls from 6 ($^6C_3$). Number of ways: $^4C_0 \times ^6C_3 = 1 \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 1 \times 20 = 20$ Probability $P(X=0) = \frac{20}{120} = \frac{1}{6}$

• For $X=1$ (1 white ball): We choose 1 white ball from 4 ($^4C_1$) AND 2 black balls from 6 ($^6C_2$). Number of ways: $^4C_1 \times ^6C_2 = 4 \times \frac{6 \times 5}{2 \times 1} = 4 \times 15 = 60$ Probability $P(X=1) = \frac{60}{120} = \frac{1}{2}$

• For $X=2$ (2 white balls): We choose 2 white balls from 4 ($^4C_2$) AND 1 black ball from 6 ($^6C_1$). Number of ways: $^4C_2 \times ^6C_1 = \frac{4 \times 3}{2 \times 1} \times 6 = 6 \times 6 = 36$ Probability $P(X=2) = \frac{36}{120} = \frac{3}{10}$

• For $X=3$ (3 white balls): We choose 3 white balls from 4 ($^4C_3$) AND 0 black balls from 6 ($^6C_0$). Number of ways: $^4C_3 \times ^6C_0 = 4 \times 1 = 4$ Probability $P(X=3) = \frac{4}{120} = \frac{1}{30}$

Let's verify that the sum of probabilities is 1: $\sum P(X=x_i) = \frac{1}{6} + \frac{1}{2} + \frac{3}{10} + \frac{1}{30} = \frac{5}{30} + \frac{15}{30} + \frac{9}{30} + \frac{1}{30} = \frac{5+15+9+1}{30} = \frac{30}{30} = 1$ The probabilities are correct.

Step 4: Calculate the Expected Value $E[X]$ (Mean) Using the formula $E[X] = \sum x_i P(X=x_i)$: $E[X] = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2)) + (3 \times P(X=3))$ $E[X] = \left(0 \times \frac{1}{6}\right) + \left(1 \times \frac{1}{2}\right) + \left(2 \times \frac{3}{10}\right) + \left(3 \times \frac{1}{30}\right)$ $E[X] = 0 + \frac{1}{2} + \frac{6}{10} + \frac{3}{30}$ $E[X] = \frac{1}{2} + \frac{3}{5} + \frac{1}{10} \quad (\text{simplifying fractions})$ To sum these, we use a common denominator of 10: $E[X] = \frac{5}{10} + \frac{6}{10} + \frac{1}{10} = \frac{5+6+1}{10} = \frac{12}{10} = \frac{6}{5}$

Step 5: Calculate the Expected Value $E[X^2]$ Using the formula $E[X^2] = \sum x_i^2 P(X=x_i)$: $E[X^2] = (0^2 \times P(X=0)) + (1^2 \times P(X=1)) + (2^2 \times P(X=2)) + (3^2 \times P(X=3))$ $E[X^2] = \left(0 \times \frac{1}{6}\right) + \left(1 \times \frac{1}{2}\right) + \left(4 \times \frac{3}{10}\right) + \left(9 \times \frac{1}{30}\right)$ $E[X^2] = 0 + \frac{1}{2} + \frac{12}{10} + \frac{9}{30}$ $E[X^2] = \frac{1}{2} + \frac{6}{5} + \frac{3}{10} \quad (\text{simplifying fractions})$ To sum these, we use a common denominator of 10: $E[X^2] = \frac{5}{10} + \frac{12}{10} + \frac{3}{10} = \frac{5+12+3}{10} = \frac{20}{10} = 2$

Step 6: Calculate the Variance $\sigma^2$ Using the formula $\sigma^2 = E[X^2] - (E[X])^2$: $\sigma^2 = 2 - \left(\frac{6}{5}\right)^2$ $\sigma^2 = 2 - \frac{36}{25}$ To subtract, we use a common denominator of 25: $\sigma^2 = \frac{50}{25} - \frac{36}{25} = \frac{50-36}{25} = \frac{14}{25}$

Step 7: Calculate $100\sigma^2$ The problem asks for the value of $100\sigma^2$. $100\sigma^2 = 100 \times \frac{14}{25}$ $100\sigma^2 = 4 \times 14 = 56$

Common Mistakes & Tips

• Verification of Probabilities: Always sum your calculated probabilities to ensure they add up to 1. This is a crucial check for preventing errors early on.
• Distinguishing $E[X^2]$ and $(E[X])^2$: A common error is to confuse these two terms. Remember that $E[X^2]$ is the expected value of $X$ squared, while $(E[X])^2$ is the square of the expected value of $X$.
• Hypergeometric Distribution: This problem describes a classic hypergeometric distribution scenario. For a population of $N$ items with $K$ successes, if $n$ items are drawn without replacement, the variance is $\sigma^2 = n \frac{K}{N} \left(1 - \frac{K}{N}\right) \frac{N-n}{N-1}$. In this case, $N=10$ (total balls), $K=4$ (white balls), $n=3$ (balls drawn). Plugging these values: $\sigma^2 = 3 \times \frac{4}{10} \times \left(1 - \frac{4}{10}\right) \times \frac{10-3}{10-1} = 3 \times \frac{2}{5} \times \frac{3}{5} \times \frac{7}{9} = \frac{18}{25} \times \frac{7}{9} = \frac{2 \times 7}{25} = \frac{14}{25}$. This confirms our detailed calculation.

Summary

To find the variance of the number of white balls drawn, we first identified the possible values of the random variable $X$ and calculated their respective probabilities using combinations (recognizing this as a hypergeometric distribution). We then used these probabilities to compute the expected value $E[X]$ and the expected value of $X^2$, $E[X^2]$. Finally, we applied the variance formula $\sigma^2 = E[X^2] - (E[X])^2$ to find $\sigma^2 = \frac{14}{25}$, and then multiplied by 100 to get the required value.

The final answer is $\boxed{56}$.