A box contains 24 identical balls of which 12 are white and 12 are black. The balls are drawn at ran

Q: Wait, why do we multiply by $1/2$ at the end? Shouldn't the probability of the 7th draw be part of the binomial formula?

That is a fantastic question! The binomial formula $\binom{n}{r} p^r q^{n-r}$ calculates the probability of getting exactly $r$ successes in $n$ trials. If we included the 7th draw in the binomial formula, we would be calculating the probability of getting 4 white balls in 7 trials, which could happen in any order (e.g., the 4th white ball could appear on the 4th, 5th, or 6th draw). By separating the 7th draw, we enforce the specific condition that the 4th white ball *must* be the 7th draw.

Q: Why is the probability of drawing a white ball always $1/2$?

Because the problem states 'with replacement.' This is the magic phrase in probability! It means the composition of the box never changes. Whether it's the 1st draw or the 100th, the ratio of white to black balls remains $12/24$, or $1/2$.

Q: Is there a formula for this specific type of problem, like the Negative Binomial Distribution?

Yes, you have hit on a deeper concept! The Negative Binomial Distribution models the number of trials needed to get a specific number of successes. However, for JEE Advanced, you do not need to memorize complex formulas. Breaking the problem into logical, manageable steps—like we did here—is often safer and helps you avoid 'formula-blindness'.

Question

A box contains $24$ identical balls of which $12$ are white and $12$ are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the $4$th time on the $7$th draw is

Options:

(A) $5/64$

(B) $27/32$

(C) $5/32$

(D) $1/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a box containing 24 identical balls—12 white and 12 black. You are about to embark on a series of draws with replacement.

Because you replace the ball, the box composition never changes. The probability of drawing a white ball, denoted as $p$, remains a constant $1/2$ for every attempt. Similarly, the probability of drawing a black ball, $q$, is also $1/2$.

We are tasked with finding the probability that the 4th white ball appears exactly on the 7th draw.

### The Logical Constraint

To solve this, we must break the timeline into two distinct phases. The 7th draw is our climax; it **must** be a white ball.

For the 4th white ball to land precisely on the 7th draw, two conditions must be met simultaneously:
1. Exactly 3 white balls must have been drawn in the preceding 6 draws.
2. The 7th draw must result in a white ball.

If we have exactly 3 white balls in the first 6 draws, the very next white ball we pull—the 7th—will inevitably be the 4th one.

### The Binomial Engine

To calculate the probability of getting exactly 3 white balls in the first 6 draws, we utilize the Binomial Distribution. We have $n = 6$ trials and we want $r = 3$ successes.

The formula for the Binomial Distribution is:

$$P(X = r) = \binom{n}{r} \cdot p^r \cdot q^{n-r}$$

Plugging in our values, we get:

$$P(X = 3) = \binom{6}{3} \cdot \left(\frac{1}{2}ight)^3 \cdot \left(\frac{1}{2}ight)^3$$

The combination $\binom{6}{3}$ represents the number of ways to arrange 3 white balls in 6 slots:

$$\binom{6}{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$$

The probability term is:

$$\left(\frac{1}{2}ight)^6 = \frac{1}{64}$$

Multiplying these, the probability of having exactly 3 white balls in the first 6 draws is:

$$20 	imes \frac{1}{64} = \frac{20}{64}$$

### The Final Calculation

We now combine the probability of the first 6 draws with the probability of the 7th draw. Since these events are independent, we multiply them:

$$P(	ext{Total}) = P(	ext{3 White in 6}) 	imes P(	ext{White on 7th})$$

$$P(	ext{Total}) = \frac{20}{64} 	imes \frac{1}{2} = \frac{20}{128}$$

Simplifying this fraction by dividing both the numerator and the denominator by 4, we arrive at the final result:

**$$\frac{5}{32}$$**

This problem demonstrates that complex probability questions can be tamed by breaking them down into simple, independent events.

Animated Solution for Mathematics - Probability: A box contains $24$ identical balls of which $12$ are white and $12$ are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the $4$ th time on the $7$ th draw is

Select Answer:

Visualized Solution

Visualizing the 7-Draw Sequence

Single Draw Probabilities

Breaking Down the 7th Draw Condition

Probability of 3 White in 6 Draws

Computing $^{6} C_{3}$ and the Power Terms

Simplifying the First Phase Probability

The Final Draw Requirement

Multiplying the Independent Events

Simplifying to the Final Answer

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Logical Constraint

The Binomial Engine

The Final Calculation

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