# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-6/3

Exercises: Try Out - Set 6: Real exam simulation to sharpen your skills and build confidence. - Problem 3

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## Exercise 3

### Question

export const metadata = {
  title: "Problem 3",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/23/2025",
};

A family has <InlineMath math="3" /> children. It is known that at least one of the children is a girl. The probability that the family has <InlineMath math="2" /> girls and <InlineMath math="1" /> boy is


### Choices

- [ ] $$\frac{1}{2}$$
- [ ] $$\frac{1}{7}$$
- [ ] $$\frac{1}{4}$$
- [x] $$\frac{3}{7}$$
- [ ] $$\frac{8}{9}$$

### Answer & Explanation

export const metadata = {
  title: "Problem 3 Solution",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/23/2025",
};

Let <InlineMath math="B" /> be a boy and <InlineMath math="G" /> be a girl.

The sample space for a family with <InlineMath math="3" /> children (without conditions) has <InlineMath math="2^3 = 8" /> possibilities:

<MathContainer>
  <BlockMath math="S_{total} = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG\}" />
</MathContainer>

It is given that "at least one child is a girl". This means the event "all children are boys" (<InlineMath math="BBB" />) is impossible.

The new sample space (<InlineMath math="S" />) consists of all possibilities except <InlineMath math="BBB" />:

<MathContainer>
  <BlockMath math="S = \{BBG, BGB, BGG, GBB, GBG, GGB, GGG\}" />
</MathContainer>

The number of elements in the sample space is <InlineMath math="n(S) = 7" />.

The expected event (<InlineMath math="A" />) is having <InlineMath math="2" /> girls and <InlineMath math="1" /> boy.

From the sample space <InlineMath math="S" />, the outcomes that satisfy this condition are:

<MathContainer>
  <BlockMath math="A = \{BGG, GBG, GGB\}" />
</MathContainer>

The number of expected outcomes is <InlineMath math="n(A) = 3" />.

The probability of the event is:

<MathContainer>
  <BlockMath math="P(A) = \frac{n(A)}{n(S)} = \frac{3}{7}" />
</MathContainer>

Therefore, the probability that the family has <InlineMath math="2" /> girls and <InlineMath math="1" /> boy is <InlineMath math="\frac{3}{7}" />.


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