Balls are drawn one-by-one without replacement from a box containing 2 black, 4 white and 3 red ball

Q: Wait, why do we treat the balls as distinct if they are identical in color? Doesn't that change the probability?

This is a classic point of confusion! While it feels counterintuitive, treating identical items as distinct is a powerful mathematical trick that actually preserves the correct probability. If you label all 9 balls (e.g., $B_1, B_2, W_1, W_2, W_3, W_4, R_1, R_2, R_3$), every possible sequence becomes equally likely. When you calculate the ratio of favorable outcomes to total outcomes, the labels cancel out, leaving you with the exact same result as if you had used combinations. It is a safer, more consistent method that prevents double-counting errors.

Q: Can I solve this by multiplying the probabilities of each individual draw?

Absolutely! You can think of it as a sequence of conditional probabilities. For the first draw, the probability of picking a black ball is $\frac{2}{9}$. For the second, it is $\frac{1}{8}$, and so on. If you multiply these: $\frac{2}{9} \times \frac{1}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times \frac{3}{3} \times \frac{2}{2} \times \frac{1}{1}$, you will arrive at the same answer. Both methods are mathematically sound, but the permutation method is often faster for complex ordering problems.

Q: Why is the denominator $9!$?

The denominator represents the total sample space, which is the total number of ways to arrange 9 distinct balls in 9 positions. Since we are drawing all 9 balls one by one without replacement, the number of possible sequences is the number of permutations of 9 distinct objects, which is $9!$. This forms the foundation of our probability ratio.

Question

Balls are drawn one-by-one without replacement from a box containing 2 black, 4 white and 3 red balls till all the balls are drawn. Find the probability that the balls drawn are in the order 2 black, 4 white and 3 red.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a box containing 9 balls: 2 black, 4 white, and 3 red. You are tasked with drawing them one by one until the box is empty.

The question asks for the probability that they emerge in a specific, rigid order: first the 2 black, then the 4 white, and finally the 3 red. This is not just a math problem; it is a study of order within chaos.

### The Power of Distinctness

The first step to mastering this problem is to embrace a powerful perspective shift. Even though the balls of the same color look identical, we must treat them as distinct.

Imagine each ball has a tiny, invisible serial number. By doing this, we ensure that every possible sequence of draws is equally likely. This is the secret to avoiding the traps of probability.

We have 9 balls in total, and we are arranging them in 9 positions. The total number of ways to arrange these 9 distinct balls is $9!$. This is our sample space, the denominator of our probability fraction.

### Building the Favorable Sequence

Now, let us construct our favorable outcome. We need the first 2 positions to be filled by the 2 black balls.

Since we are treating them as distinct, there are $2!$ ways to arrange them in those first two spots. Next, we need the 4 white balls to occupy the next 4 positions.

Just like the black balls, there are $4!$ ways to arrange these 4 distinct white balls. Finally, the 3 red balls must fill the last 3 positions, which can be done in $3!$ ways.

By the fundamental principle of counting, the total number of favorable outcomes is the product of these arrangements: $2! 	imes 4! 	imes 3!$.

### The Final Calculation

We now have our numerator and our denominator. The probability $P(E)$ is the ratio of favorable outcomes to total outcomes:

$$P(E) = \frac{2! 	imes 4! 	imes 3!}{9!}$$

To solve this without getting lost in large numbers, we use the property of factorials. We keep the $4!$ in the numerator and expand the $9!$ in the denominator until we reach $4!$:

$$9! = 9 	imes 8 	imes 7 	imes 6 	imes 5 	imes 4!$$

This allows for a beautiful cancellation:

$$P(E) = \frac{2 	imes 1 	imes 4! 	imes 3 	imes 2 	imes 1}{9 	imes 8 	imes 7 	imes 6 	imes 5 	imes 4!}$$

The $4!$ terms vanish, leaving us with:

$$P(E) = \frac{2 	imes 6}{9 	imes 8 	imes 7 	imes 6 	imes 5}$$

Simplifying this further, we cancel the 6 from the top and bottom, and the 2 with the 8 to leave a 4 in the denominator. We are left with:

$$\frac{1}{9 	imes 4 	imes 7 	imes 5} = \frac{1}{1260}$$

It is a moment of pure mathematical elegance when the complex factorial expression collapses into such a simple, clean fraction. The final probability is **$\frac{1}{1260}$**.

Animated Solution for Mathematics - Probability: Balls are drawn one-by-one without replacement from a box containing 2 black, 4 white and 3 red balls till all the balls are drawn. Find the probability that the balls drawn are in the order 2 black, 4 white and 3 red.

Visualized Solution

Analyze the Inventory

Total Possible Outcomes

Arranging Black Balls

Arranging White Balls

Arranging Red Balls

Total Favorable Ways

Probability Calculation

Expanding the Factorials

Canceling Terms

Final Result

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Power of Distinctness

Building the Favorable Sequence

The Final Calculation

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