A bag contains six balls of different colours. Two balls are drawn in succession with replacement. T

Q: Wait, why do we use $6^4$ as the total outcomes for the second scenario? Shouldn't it be $6 \times 5 \times 4 \times 3$?

That is a classic trap! The phrase 'with replacement' is the key. Because we put the ball back after each draw, the number of options for each draw remains $6$. If it were 'without replacement', then yes, the options would decrease. But here, each draw is independent, so it is $6 \times 6 \times 6 \times 6 = 6^4$.

Q: Why do we multiply by $5$ for the fourth ball in the calculation for $q$?

Great question. The condition is 'exactly three balls are of the same color'. If we chose the same color for the fourth ball as well, we would have four balls of the same color, which violates the 'exactly three' condition. Therefore, the fourth ball must be any of the other $5$ colors.

Q: Can we use the Binomial Distribution formula $\binom{n}{k} p^k (1-p)^{n-k}$ for $q$?

You absolutely can! If you define 'success' as picking a specific color, the probability of success is $1/6$ and failure is $5/6$. However, you must be careful: the binomial formula gives the probability of getting exactly $k$ successes of a *pre-determined* color. Since the color of the triplet can be any of the $6$ colors, you would need to multiply your result by $6$.

Question

A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is $p$. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colours is $q$. If $p : q = m : n$, where $m$ and $n$ are coprime, then $m + n$ is equal to

Sigma Percentile · Accepted Answer

### The Dance of Probability: A Journey into Combinatorics

Imagine you are standing before a mysterious bag containing six balls, each a unique color: blue, green, red, orange, purple, and gray. This experiment is a study of randomness and order. Let us peel back the layers of this challenge together.

### Phase 1: The Two-Draw Mystery

In our first scenario, we draw two balls in succession with replacement. Because we return the ball after each draw, the bag remains identical for every attempt. The probability of picking any specific color remains a constant $1/6$.

To find the probability $p$ that both balls are the same color, we first determine the total sample space. Since each draw has $6$ possibilities, the total number of outcomes is:

$$6 	imes 6 = 36$$

For the favorable outcomes, we need the first ball to be any color, and the second ball to match it. There are $6$ such pairs: $(C_1, C_1), (C_2, C_2), \dots, (C_6, C_6)$. Thus, the probability is:

$$p = \frac{6}{36} = \frac{1}{6}$$

### Phase 2: The Four-Draw Challenge

Now, we draw four balls with replacement. We seek the probability $q$ that exactly three balls share the same color. We have four slots to fill: $\_ \_ \_ \_$.

First, we choose which three of these four slots will contain the matching balls. This is a combination problem:

$$\binom{4}{3} = 4 	ext{ ways to choose the positions.}$$

Next, we choose the color for this triplet, which offers $6$ choices. Finally, the fourth ball—the 'singleton'—must be a different color to satisfy the 'exactly three' condition. Since we have already used one color for the triplet, we have $5$ colors remaining for this last ball.

The number of favorable outcomes is $4 	imes 6 	imes 5 = 120$. The total number of outcomes for four draws is $6^4 = 1296$. Therefore, the probability $q$ is:

$$q = \frac{120}{1296} = \frac{5}{54}$$

### Phase 3: The Final Synthesis

The problem asks for the ratio $p:q = m:n$. Given $p = 1/6$ and $q = 5/54$, we set up the ratio:

$$\frac{p}{q} = \frac{\frac{1}{6}}{\frac{5}{54}} = \frac{1}{6} 	imes \frac{54}{5}$$

Simplifying the expression, we find that $54$ divided by $6$ is $9$. Thus, the ratio becomes:

$$\frac{p}{q} = \frac{9}{5}$$

Since $9$ and $5$ are coprime, we identify $m = 9$ and $n = 5$. The final result is:

**$m + n = 9 + 5 = 14$**

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Scenario $1$ : Two Draws

Total Outcomes for $2$ Draws

Favorable Outcomes for $p$

Calculating Probability $p$

Scenario $2$ : Four Draws

Total Outcomes for $4$ Draws

Favorable Outcomes for $q$

Choosing Positions

Choosing Colors

Calculating Probability $q$

Setting up the Ratio

Simplifying the Ratio

Final Answer: $m + n$

The Sigma Insight: Classical Definition of Probability

The Dance of Probability

Phase 1

Phase 2

Phase 3

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