Key Concepts and Formulas

• Law of Total Probability: If $E_1, E_2, \ldots, E_n$ are mutually exclusive and exhaustive events, then the probability of any event $A$ is given by $P(A) = \sum_{i=1}^{n} P(A|E_i)P(E_i)$. This formula is crucial when an experiment has multiple stages or scenarios.
• Combinations ($^nC_r$): The number of ways to choose $r$ items from a set of $n$ distinct items without regard to the order of selection is given by $^nC_r = \frac{n!}{r!(n-r)!}$. This is used for drawing balls "at random" where the order doesn't matter.
• Conditional Probability: The probability of event $A$ occurring given that event $B$ has already occurred is $P(A|B) = \frac{P(A \cap B)}{P(B)}$. In the context of combinations, this often simplifies to the ratio of favorable combinations to total combinations for a specific scenario.

Step-by-Step Solution

This problem involves a two-stage experiment: first, rolling a die to determine the number of balls to draw, and second, drawing balls from a bag. We use the Law of Total Probability to combine the probabilities from these stages.

• Let $A$ be the event that all balls drawn are white.
• Let $E_x$ be the event that the number $x$ is obtained on the die, meaning $x$ balls are drawn. The possible values for $x$ are $\{1, 2, 3, 4, 5, 6\}$. These events $E_x$ are mutually exclusive (only one number can be rolled) and exhaustive (one of these numbers must be rolled).

Step 1: Determine the Probabilities of Each Die Roll Outcome ($P(E_x)$)

• Why this step? The Law of Total Probability requires the probability of each scenario (how many balls are drawn) before we can calculate the probability of drawing white balls within that specific scenario.
• A standard, fair six-sided die has faces numbered 1 to 6. Each outcome is equally likely.
• Therefore, for each $x \in \{1, 2, 3, 4, 5, 6\}$: $P(E_x) = P(\text{die shows } x) = \frac{1}{6}$

Step 2: Calculate the Conditional Probabilities of Drawing All White Balls ($P(A|E_x)$)

• Why this step? This is the core calculation for each specific scenario. For each possible number of balls drawn ($x$), we need to find the probability that all of those $x$ balls are white. This is $P(A|E_x)$, the probability of event $A$ given that event $E_x$ has occurred.
• The bag contains: 6 white balls (W) and 4 black balls (B). Total balls = 10.
• When $x$ balls are drawn from the bag at random, the total number of ways to choose these $x$ balls from the 10 available is $^{10}C_x$.
• The number of ways to choose $x$ white balls from the 6 available white balls is $^6C_x$.
• The conditional probability $P(A|E_x)$ is the ratio of favorable outcomes (drawing $x$ white balls) to the total possible outcomes (drawing any $x$ balls): $P(A|E_x) = \frac{{^6C_x}}{{^{10}C_x}}$

Let's calculate $P(A|E_x)$ for each possible value of $x$:

1. If $x=1$ (1 ball drawn):

   • $P(A|E_1) = \frac{{^6C_1}}{{^{10}C_1}} = \frac{6}{10} = \frac{3}{5}$

2. If $x=2$ (2 balls drawn):

   • $P(A|E_2) = \frac{{^6C_2}}{{^{10}C_2}} = \frac{15}{45} = \frac{1}{3}$

3. If $x=3$ (3 balls drawn):

   • $P(A|E_3) = \frac{{^6C_3}}{{^{10}C_3}} = \frac{20}{120} = \frac{1}{6}$

4. If $x=4$ (4 balls drawn):

   • $P(A|E_4) = \frac{{^6C_4}}{{^{10}C_4}} = \frac{15}{210} = \frac{1}{14}$

5. If $x=5$ (5 balls drawn):

   • $P(A|E_5) = \frac{{^6C_5}}{{^{10}C_5}} = \frac{6}{252} = \frac{1}{42}$

6. If $x=6$ (6 balls drawn):

   • $P(A|E_6) = \frac{{^6C_6}}{{^{10}C_6}} = \frac{1}{210}$

Step 3: Apply the Law of Total Probability to Find $P(A)$

• Why this step? Now that we have the probability of each die roll and the conditional probability of drawing white balls for each die roll, we combine these using the Law of Total Probability to find the overall probability.
• $P(A) = \sum_{x=1}^{6} P(A|E_x)P(E_x)$
• Since $P(E_x) = \frac{1}{6}$ for all $x$: $P(A) = \frac{1}{6} \left( P(A|E_1) + P(A|E_2) + P(A|E_3) + P(A|E_4) + P(A|E_5) + P(A|E_6) \right)$
• Substitute the values calculated in Step 2: $P(A) = \frac{1}{6} \left( \frac{3}{5} + \frac{1}{3} + \frac{1}{6} + \frac{1}{14} + \frac{1}{42} + \frac{1}{210} \right)$
• To sum the fractions, find the Least Common Multiple (LCM) of the denominators $5, 3, 6, 14, 42, 210$. The LCM is $210$.
• Convert each fraction to have a denominator of 210:
  • $\frac{3}{5} = \frac{126}{210}$
  • $\frac{1}{3} = \frac{70}{210}$
  • $\frac{1}{6} = \frac{35}{210}$
  • $\frac{1}{14} = \frac{15}{210}$
  • $\frac{1}{42} = \frac{5}{210}$
  • $\frac{1}{210} = \frac{1}{210}$
• Summing these fractions: $P(A) = \frac{1}{6} \left( \frac{126 + 70 + 35 + 15 + 5 + 1}{210} \right)$ $P(A) = \frac{1}{6} \left( \frac{252}{210} \right)$
• Simplify the fraction $\frac{252}{210}$ by dividing both numerator and denominator by their greatest common divisor, which is 42: $\frac{252}{210} = \frac{6 \times 42}{5 \times 42} = \frac{6}{5}$
• Substitute this back into the equation for $P(A)$: $P(A) = \frac{1}{6} \times \frac{6}{5}$ $P(A) = \frac{1}{5}$

Common Mistakes & Tips

• Incorrectly applying combinations: Ensure you use combinations ($^nC_r$) and not permutations ($^nP_r$) when the order of selection does not matter (e.g., drawing balls from a bag).
• Forgetting the Law of Total Probability: In multi-stage experiments where the outcome of the first stage affects the second, remember to sum the probabilities of each scenario multiplied by their respective conditional probabilities.
• Arithmetic Errors: Calculations involving fractions and binomial coefficients can be tedious. Double-check your arithmetic, especially when finding common denominators and simplifying fractions. A common LCM can save time and reduce errors.
• Boundary Conditions: Remember that $^nC_r = 0$ if $r > n$. While not critical in this specific problem (since the maximum die roll is 6, which is also the number of white balls), it's a vital consideration for similar problems.

Summary

The problem requires calculating the probability of drawing only white balls after a die roll determines the number of balls to be drawn. We applied the Law of Total Probability, summing the products of the probability of each die roll outcome (1/6 for each) and the conditional probability of drawing all white balls for that specific number of draws. Each conditional probability was calculated using combinations, representing the ratio of ways to choose white balls to the total ways to choose any balls. After calculating these values for each possible die roll (1 to 6) and summing them up, we found the total probability to be $\frac{1}{5}$.

Final Answer

The probability that all the balls drawn are white is $\boxed{\frac{1}{5}}$, which corresponds to option (C).