# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-9/5

Exercises: Try Out - Set 9: Real exam simulation to sharpen your skills and build confidence. - Problem 5

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## Exercise 5

### Question

export const metadata = {
  title: "Problem 5",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Inside a box there are <InlineMath math="1" /> red ball, <InlineMath math="6" /> green balls, and <InlineMath math="2" /> white balls. If <InlineMath math="7" /> balls are to be drawn without replacement, then the probability that the number of green balls drawn is twice the number of white balls drawn is...


### Choices

- [ ] $$\frac{1}{12}$$
- [ ] $$\frac{1}{4}$$
- [x] $$\frac{5}{12}$$
- [ ] $$\frac{1}{2}$$
- [ ] $$\frac{2}{3}$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 5",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Let <InlineMath math="A" /> be the event that the number of green balls drawn is twice the number of white balls drawn.

**Given:**

There are <InlineMath math="9" /> balls in a box (<InlineMath math="1" /> red + <InlineMath math="6" /> green + <InlineMath math="2" /> white = <InlineMath math="9" />).
<InlineMath math="7" /> balls will be drawn at random. The total number of ways to draw the balls is:

<MathContainer>
  <BlockMath math="n(S) = C^9_7 = \frac{9!}{7!(9-7)!} = \frac{9 \times 8}{2 \times 1} = 36" />
</MathContainer>

We want to find the probability that the number of green balls is twice the number of white balls. We divide this into two cases.

**Case 1**

If <InlineMath math="1" /> white ball is drawn, then there must be <InlineMath math="2" /> green balls (since green = 2 times white).
To reach a total of <InlineMath math="7" /> balls drawn, the remaining balls to be drawn must be <InlineMath math="7 - 1 - 2 = 4" /> balls. These must be red balls.
However, there is only <InlineMath math="1" /> red ball available. Therefore, **Case 1 is impossible**.

**Case <InlineMath math="2" />**

If <InlineMath math="2" /> white balls are drawn, then there must be <InlineMath math="4" /> green balls.
To reach a total of <InlineMath math="7" /> balls drawn, the remaining balls to be drawn must be <InlineMath math="7 - 2 - 4 = 1" /> ball. This must be a red ball.
This case is possible because there are <InlineMath math="2" /> white balls, <InlineMath math="6" /> green balls, and <InlineMath math="1" /> red ball available.

The number of ways to draw for this case is:

<MathContainer>
  <BlockMath math="n(A) = C^2_2 \cdot C^6_4 \cdot C^1_1" />
  <BlockMath math="= 1 \cdot 15 \cdot 1" />
  <BlockMath math="= 15" />
</MathContainer>

**Conclusion**

Thus, the probability of event <InlineMath math="A" /> is:

<MathContainer>
  <BlockMath math="P(A) = \frac{n(A)}{n(S)} = \frac{15}{36} = \frac{5}{12}" />
</MathContainer>

Therefore, the probability that the number of green balls drawn is twice the number of white balls drawn is <InlineMath math="\frac{5}{12}" />.


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