Theoretical DistributionsMCQMTP Dec 2023 Series IIQuestion 3531 of 230
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The probability that a man aged 45\displaystyle 45 years will die within a year is 0.012\displaystyle 0.012. What is the probability that of 10\displaystyle 10 men, at least 9\displaystyle 9 will reach their 46\displaystyle 46th birthday? [Given e−0.12=0.88692\displaystyle e^{-0.12} = 0.88692 ]

Options

A0.0933\displaystyle 0.0933
B0.9335\displaystyle 0.9335
C0.9333\displaystyle 0.9333
D0.9555\displaystyle 0.9555
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Correct Answer

✅ Option b — 0.9335\displaystyle 0.9335

All Options:

  • A0.0933\displaystyle 0.0933
  • B0.9335\displaystyle 0.9335
  • C0.9333\displaystyle 0.9333
  • D0.9555\displaystyle 0.9555

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Detailed Solution & Explanation

**Probability of At Least 9 Men Surviving Out of 10** **Given:** - Probability of a man dying within a year (p\displaystyle p) = 0.012\displaystyle 0.012 - Number of men (n\displaystyle n) = 10\displaystyle 10 - Given value: e−0.12=0.88692\displaystyle e^{-0.12} = 0.88692 **Step 1: Define the Poisson parameter (m\displaystyle m)** Since n\displaystyle n is small and p\displaystyle p is small, we can model the number of deaths X\displaystyle X using a Poisson distribution with parameter: m=np=10times0.012=0.12m = np = 10 \\times 0.012 = 0.12 **Step 2: Understand the condition** We want the probability that **at least 9 men** reach their 46th birthday (survive). This means either 9\displaystyle 9 or 10\displaystyle 10 men survive, which is equivalent to saying that **at most 1 man dies** (Xle1\displaystyle X \\le 1): P(Xle1)=P(X=0)+P(X=1)P(X \\le 1) = P(X = 0) + P(X = 1) **Step 3: Calculate the probabilities** Using the Poisson probability mass function: P(X=x)=frace−mmxx!P(X = x) = \\frac{e^{-m} m^x}{x!} - For x=0\displaystyle x = 0 (no deaths): P(X=0)=frace−0.12(0.12)00!=e−0.12P(X = 0) = \\frac{e^{-0.12} (0.12)^0}{0!} = e^{-0.12} - For x=1\displaystyle x = 1 (exactly 1 death): P(X=1)=frace−0.12(0.12)11!=0.12e−0.12P(X = 1) = \\frac{e^{-0.12} (0.12)^1}{1!} = 0.12 e^{-0.12} Summing these probabilities: P(Xle1)=e−0.12+0.12e−0.12=e−0.12(1+0.12)=1.12e−0.12P(X \\le 1) = e^{-0.12} + 0.12 e^{-0.12} = e^{-0.12} (1 + 0.12) = 1.12 e^{-0.12} **Step 4: Substitute the given value** Substitute e−0.12=0.88692\displaystyle e^{-0.12} = 0.88692: P(Xle1)=1.12times0.88692=0.9933504approx0.99335P(X \\le 1) = 1.12 \\times 0.88692 = 0.9933504 \\approx 0.99335 **Discrepancy Note:** The mathematically derived probability is 0.99335\displaystyle 0.99335. However, the textbook contains a typographical error where 0.99335\displaystyle 0.99335 is written as 0.9335\displaystyle 0.9335 (Option B). To align with the textbook's intended choice, we note this decimal transposition error. Hence, **Option B** is the correct answer.

About This Chapter: Theoretical Distributions

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Binomial, Poisson, Normal Distribution

This chapter covers Binomial, Poisson, Normal Distribution and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

Key Concepts to Understand

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

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