ProbabilityPYQ Jan 26Question 4239 of 187
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Ms. Radhika appeared in interview at three different companies. In the first company there are 5 candidates, in second company there are 12 candidates and in third company there are 15 candidates. What is probability that Ms. Radhika would be selected?

Options

A231375\displaystyle \frac{231}{375}
B321375\displaystyle \frac{321}{375}
C154225\displaystyle \frac{154}{225}
D71225\displaystyle \frac{71}{225}
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Correct Answer

✅ Option d — 71225\displaystyle \frac{71}{225}

All Options:

  • A231375\displaystyle \frac{231}{375}
  • B321375\displaystyle \frac{321}{375}
  • C154225\displaystyle \frac{154}{225}
  • D71225\displaystyle \frac{71}{225}

Detailed Solution & Explanation

Let us assume that the selection decisions of the three companies are independent of each other.

For each company, Ms. Radhika is one of the candidates. The probability of her selection in each company is:
- Probability of selection in the 1st\displaystyle 1^{\text{st}} company (P(C1)\displaystyle P(C_1)) = 15\displaystyle \frac{1}{5}
So, probability of rejection in the 1st\displaystyle 1^{\text{st}} company (P(C1′)\displaystyle P(C_1')) = 1−15=45\displaystyle 1 - \frac{1}{5} = \frac{4}{5}
- Probability of selection in the 2nd\displaystyle 2^{\text{nd}} company (P(C2)\displaystyle P(C_2)) = 112\displaystyle \frac{1}{12}
So, probability of rejection in the 2nd\displaystyle 2^{\text{nd}} company (P(C2′)\displaystyle P(C_2')) = 1−112=1112\displaystyle 1 - \frac{1}{12} = \frac{11}{12}
- Probability of selection in the 3rd\displaystyle 3^{\text{rd}} company (P(C3)\displaystyle P(C_3)) = 115\displaystyle \frac{1}{15}
So, probability of rejection in the 3rd\displaystyle 3^{\text{rd}} company (P(C3′)\displaystyle P(C_3')) = 1−115=1415\displaystyle 1 - \frac{1}{15} = \frac{14}{15}

The probability that Ms. Radhika is selected by at least one company is given by:
P(Selected)=1−P(Selected by none)P(\text{Selected}) = 1 - P(\text{Selected by none})
P(Selected by none)=P(C1′)×P(C2′)×P(C3′)P(\text{Selected by none}) = P(C_1') \times P(C_2') \times P(C_3')
P(Selected by none)=45×1112×1415P(\text{Selected by none}) = \frac{4}{5} \times \frac{11}{12} \times \frac{14}{15}

Let us simplify this product:
P(Selected by none)=4×11×145×12×15=616900=154225P(\text{Selected by none}) = \frac{4 \times 11 \times 14}{5 \times 12 \times 15} = \frac{616}{900} = \frac{154}{225}

Now, calculate the probability of being selected by at least one company:
P(Selected)=1−154225=225−154225=71225P(\text{Selected}) = 1 - \frac{154}{225} = \frac{225 - 154}{225} = \frac{71}{225}

Hence, **Option D** 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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