An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacem

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

Great question! In this problem, the order in which the balls are drawn *within* each set of four does not matter. We only care about the final composition of the four balls drawn. Combinations are the perfect tool for selecting items where order is irrelevant.

Q: Why can't we just use 15 as the denominator for the second draw?

That is the classic trap! Because the problem states 'without replacement,' the four balls removed in the first draw are gone forever. They no longer exist in the urn for the second draw. You must subtract them from the total, making the new sample space 11, not 15.

Q: Is there a way to solve this using permutations?

You could, but it would be much more tedious. If you used permutations, you would have to account for the order of all 8 balls drawn in sequence. Using combinations is the most elegant and efficient path to the solution.

Question

An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :

Options:

(A) $\frac{5}{256}$

(B) $\frac{5}{715}$

(C) $\frac{3}{715}$

(D) $\frac{3}{256}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before an urn containing fifteen balls in total: six white and nine black. You are tasked with a two-part challenge: first, to draw four white balls, and second, to draw four black balls, all without replacement.

This is a story of changing states and conditional realities. Let us embark on this journey together.

### Phase 1: The First Draw

The first step is to isolate the event $W_1$, where we successfully draw four white balls. We have six white balls available, and we need to choose four.

The number of ways to do this is given by the combination formula $\binom{6}{4}$. Simultaneously, the total number of ways to choose any four balls from the fifteen available is $\binom{15}{4}$.

The probability of our first success is:

$$P(W_1) = \frac{\binom{6}{4}}{\binom{15}{4}}$$

Calculating these, we find $\binom{6}{4} = 15$ and $\binom{15}{4} = 1365$. Simplifying this, we get:

$$P(W_1) = \frac{15}{1365} = \frac{1}{91}$$

### Phase 2: The Crucial Transition

Now, pause. This is where most students stumble. We have removed four white balls from the urn, so the urn is no longer the same as it was.

We started with six white balls; now, only two remain. The nine black balls, however, remain untouched. The total number of balls has dropped from fifteen to eleven.

This update to our sample space is the heartbeat of the problem. If you forget to update the urn, the entire calculation collapses.

### Phase 3: The Second Draw

With our updated urn, we now calculate the conditional probability $P(B_2|W_1)$, the probability of drawing four black balls given that the first draw was all white. We need to select four black balls from the nine available, which is $\binom{9}{4}$, and we are drawing from the new total of eleven balls, which is $\binom{11}{4}$.

The math unfolds as follows:

$$\binom{9}{4} = 126, \quad \binom{11}{4} = 330$$

Thus, the conditional probability is:

$$P(B_2|W_1) = \frac{126}{330} = \frac{21}{55}$$

### Phase 4: The Synthesis

Finally, we bring it all together using the Multiplication Theorem of Probability. The probability of both events occurring in sequence is the product of the individual probabilities:

$$P = P(W_1) 	imes P(B_2|W_1)$$

Substituting our values, we have:

$$P = \frac{1}{91} 	imes \frac{21}{55}$$

Before we multiply, let us look for elegance. We know $91 = 13 	imes 7$ and $21 = 3 	imes 7$. The sevens cancel out, leaving us with:

$$P = \frac{1}{13} 	imes \frac{3}{55} = \frac{3}{715}$$

And there it is—the beauty of the final result, **$\frac{3}{715}$**. Remember, in probability, the system is always evolving; respect the change in the sample space, and the math will always reward you.

Animated Solution for Mathematics - Probability: An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :

Select Answer:

Visualized Solution

Initial State of the Urn

Defining the First Event $W_{1}$

Calculating $P (W_{1})$

Updating the Urn State

Defining the Second Event $B_{2} ∣ W_{1}$

Calculating $P (B_{2} ∣ W_{1})$

Applying the Multiplication Theorem

Final Computation

Conclusion & Key Takeaway

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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