Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random vari

Q: Wait, why can't we just use the Hypergeometric distribution formula to find the mean?

You absolutely can! The Hypergeometric distribution is the formal way to model sampling without replacement. However, in a JEE Advanced context, time is your most precious resource. Using the formula $E(X) = n \cdot \frac{K}{N}$ is a shortcut derived from the properties of expectation. It bypasses the need to calculate the probability of every possible outcome ($X=0, 1, 2, 3$), saving you valuable minutes.

Q: Is it always true that $E(X) + E(Y) = n$?

Yes, it is! Because the total number of balls drawn is fixed at $n=3$, we know that $X + Y = 3$ must hold true for every single draw. By the linearity of expectation, $E(X + Y) = E(X) + E(Y) = E(3) = 3$. This is a great way to verify your answers—if your calculated means don't sum to $n$, you know you've made a calculation error.

Q: Why is the probability $5/9$ for every draw? Doesn't the probability change after the first ball is removed?

That is a brilliant observation! You are correct that the probability of the *second* draw depends on the first. However, the *marginal* probability of any specific draw (e.g., the 2nd or 3rd) being a certain color is still $K/N$. Because we don't know the outcome of the previous draws, the symmetry of the problem ensures that the probability of success for any individual draw remains $K/N$. This is why the linearity of expectation works so beautifully here.

Question

Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables $X$ and $Y$ respectively denote the number of blue and yellow balls. If $\bar{X}$ and $\bar{Y}$ are the means of $X$ and $Y$ respectively, then $7\bar{X} + 4\bar{Y}$ is equal to________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a bag containing five blue balls and four yellow balls. You are tasked with drawing three balls at random.

The random variables $X$ and $Y$ represent the number of blue and yellow balls drawn, respectively. Your goal is to find the value of $7\bar{X} + 4\bar{Y}$.

Many students immediately reach for the heavy machinery of combinatorics, calculating the probability of drawing zero, one, two, or three blue balls. While this is mathematically sound, it is a path of high resistance that consumes precious time.

### The Elegant Shortcut: Linearity of Expectation

Instead, let us embrace the elegance of the **Linearity of Expectation**. This is one of the most powerful tools in your probability toolkit.

It states that the expected value of a sum is the sum of the expected values. For sampling without replacement, the expected value of a random variable representing the count of items is simply:

$$E(X) = n \cdot p$$

Here, $n$ is the number of trials and $p$ is the probability of success in a single trial. If you draw three balls, and the probability of any single ball being blue is $\frac{5}{9}$, then on average, you expect to draw $3 \cdot \frac{5}{9}$ blue balls.

### The Calculation: Breaking the Block

Let us apply this to our problem where $n=3$ trials. For blue balls, the total count is $5$ out of $9$.

Thus, the expected number of blue balls is:

$$\bar{X} = E(X) = 3 \cdot \frac{5}{9} = \frac{15}{9} = \frac{5}{3}$$

Similarly, for yellow balls, we have $4$ out of $9$. The expected number of yellow balls is:

$$\bar{Y} = E(Y) = 3 \cdot \frac{4}{9} = \frac{12}{9} = \frac{4}{3}$$

### The Final Act: Bringing it Home

Now, we return to the target expression: $7\bar{X} + 4\bar{Y}$. Substituting our values $\bar{X} = \frac{5}{3}$ and $\bar{Y} = \frac{4}{3}$, we get:

$$7\left(\frac{5}{3}\right) + 4\left(\frac{4}{3}\right)$$

This simplifies to:

$$\frac{35}{3} + \frac{16}{3} = \frac{35 + 16}{3} = \frac{51}{3}$$

Finally, $51$ divided by $3$ gives us the result. The final answer is **17**.

Enter Numerical Value:

Visualized Solution

Visualizing the Bag

Defining Random Variables

Logic of Expectation

Raw Setup for $\overset{ˉ}{X}$

Atomic Compute for $\overset{ˉ}{X}$

Raw Setup for $\overset{ˉ}{Y}$

Atomic Compute for $\overset{ˉ}{Y}$

The Target Expression

Atomic Compute (Multiplication)

Atomic Compute (Addition)

Final Answer

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Elegant Shortcut

The Calculation

The Final Act

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