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In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is

Options

A2146\displaystyle \frac{21}{46}
B2517\displaystyle \frac{25}{17}
C150\displaystyle \frac{1}{50}
D325\displaystyle \frac{3}{25}
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Correct Answer

✅ Option a — 2146\displaystyle \frac{21}{46}

All Options:

  • A2146\displaystyle \frac{21}{46}
  • B2517\displaystyle \frac{25}{17}
  • C150\displaystyle \frac{1}{50}
  • D325\displaystyle \frac{3}{25}

Detailed Solution & Explanation

Let the class composition be:
- Number of boys = 15\displaystyle 15
- Number of girls = 10\displaystyle 10
- Total number of students = 15+10=25\displaystyle 15 + 10 = 25

We select 3 students at random from 25 students. The total number of ways to choose 3 students is:
N(S)=(253)=25×24×233×2×1=25×4×23=2300N(S) = \binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 25 \times 4 \times 23 = 2300
We want to find the probability of selecting exactly 1 girl and 2 boys. The number of favorable outcomes is:
n(E)=(101)×(152)n(E) = \binom{10}{1} \times \binom{15}{2} Calculate the combinations:
(101)=10\binom{10}{1} = 10 (152)=15×142×1=105\binom{15}{2} = \frac{15 \times 14}{2 \times 1} = 105 n(E)=10×105=1050n(E) = 10 \times 105 = 1050
The probability of the event is:
P(E)=n(E)N(S)=10502300=105230=2146P(E) = \frac{n(E)}{N(S)} = \frac{1050}{2300} = \frac{105}{230} = \frac{21}{46}
Hence, **Option A** is the correct answer.

About This Chapter: Probability

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Probability Operations, Expected Value

A logic-heavy chapter dealing with random experiments, events (mutually exclusive, exhaustive), set theory probability, conditional probability, and Bayes' Theorem. It forms the basis for Theoretical Distributions.

View Official ICAI Syllabus

Exam Strategy Tip

Always draw a quick Venn Diagram or tree when faced with 'At least one' or 'Only A but not B' wording. It saves you from double-counting.

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