A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the

Q: Wait, why can't we just calculate the probability of drawing white balls without considering the die?

Great question! The die roll is the 'gatekeeper' of this experiment. If you ignore the die, you are assuming the number of balls drawn is fixed, but the problem explicitly states that the number of balls drawn depends on the die. By ignoring the die, you would be solving a completely different problem. We must use the Law of Total Probability to account for all possible scenarios created by the die.

Q: Why do we use combinations $\binom{n}{r}$ instead of permutations?

Because the order in which we draw the balls doesn't matter! Whether you pick a white ball first and then another white ball, or vice versa, the result is the same: you have two white balls. Combinations count the number of ways to select a group, which is exactly what we need here.

Q: Is there a shortcut for the summation $\sum_{i=1}^{6} \frac{\binom{6}{i}}{\binom{10}{i}}$?

While there are advanced combinatorial identities, for a small range like $i=1$ to $6$, direct calculation is often the most reliable path. It prevents errors and helps you build intuition for how the probabilities change as the number of balls drawn increases. Trust the process!

Question

A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is

Options:

(A) $\frac{1}{4}$

(B) $\frac{11}{50}$

(C) $\frac{1}{5}$

(D) $\frac{9}{50}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a table. On this table sits a bag containing six white balls and four black balls. Beside the bag lies a standard six-sided die.

This is a narrative of uncertainty where the outcome of a die roll dictates the nature of our next action. We will use the **Law of Total Probability** to weave these disparate scenarios into a single, elegant solution.

### The Gatekeeper: The Die Roll

Before we touch the bag, we must roll the die. Let $D$ represent the outcome of this roll.

Since it is a fair die, each outcome $i \in \{1, 2, 3, 4, 5, 6\}$ has an equal probability:

$$P(D=i) = \frac{1}{6}$$

We cannot simply guess the outcome; we must respect the randomness of the die as our starting point.

### The Conditional Lens: Combinations

Suppose the die shows a specific number $i$. We are now in a conditional world where we must draw exactly $i$ balls from the bag.

The total number of ways to draw $i$ balls from the ten available is given by the combination formula $\binom{10}{i}$. We are interested in the success of drawing only white balls, which can be done in $\binom{6}{i}$ ways.

Thus, the conditional probability of success, given a specific die roll $i$, is:

$$P(	ext{All White} | D=i) = \frac{\binom{6}{i}}{\binom{10}{i}}$$

### The Grand Summation

To find the total probability, we sum the probabilities of all mutually exclusive scenarios. The Law of Total Probability states:

$$P(	ext{All White}) = \sum_{i=1}^{6} P(D=i) 	imes P(	ext{All White} | D=i)$$

Since $P(D=i)$ is a constant $\frac{1}{6}$, we factor it out:

$$P = \frac{1}{6} \sum_{i=1}^{6} \frac{\binom{6}{i}}{\binom{10}{i}}$$

Calculating each term for $i=1$ to $6$:

For $i=1$: $\frac{\binom{6}{1}}{\binom{10}{1}} = \frac{6}{10} = \frac{3}{5}$

For $i=2$: $\frac{\binom{6}{2}}{\binom{10}{2}} = \frac{15}{45} = \frac{1}{3}$

For $i=3, 4, 5, 6$, the terms are $\frac{1}{6}, \frac{1}{14}, \frac{1}{42}, 	ext{and } \frac{1}{210}$ respectively.

### The Final Reveal

Adding these fractions requires a common denominator of $210$. Summing the numerators:

$$126 + 70 + 35 + 15 + 5 + 1 = 252$$

The sum is $\frac{252}{210}$. Finally, we multiply by our factored $\frac{1}{6}$:

$$P = \frac{1}{6} 	imes \frac{252}{210} = \frac{42}{210} = \frac{1}{5}$$

We have arrived at the answer. The final probability is **$\frac{1}{5}$**.

Animated Solution for Mathematics - Probability: A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is

Select Answer:

Visualized Solution

Understanding the Experiment

Probability of Die Outcomes

Conditional Probability Formula

Total Probability Setup

Calculating for $i = 1$

Calculating for $i = 2$

Calculating for $i = 3$

Calculating for $i = 4$

Calculating for $i = 5$ and $i = 6$

Summing the Fractions

Final Calculation

The Sigma Insight: Total Probability Theorem

Analyzing the Setup

The Gatekeeper

The Conditional Lens

The Grand Summation

The Final Reveal

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