Theoretical DistributionsMCQPYQ June 22Question 3497 of 230
All Questions

A renowned hospital usually admits 200 patients everyday. One percent patients, on an average, require special room facilities. On one particular morning, it was found that only one special room is available. What is the probability that more than 3 patients would require special room facilities?

Options

A0.1428
B0.1732
C0.2235
D0.3420
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — 0.1428

All Options:

  • A0.1428
  • B0.1732
  • C0.2235
  • D0.3420

Ad

Detailed Solution & Explanation

Let n=200\displaystyle n = 200 be the number of patients admitted, and let p=1%=0.01\displaystyle p = 1\% = 0.01 be the probability that a patient requires special room facilities. Since n\displaystyle n is large and p\displaystyle p is small, the number of patients X\displaystyle X requiring special rooms follows a Poisson distribution with parameter: m=np=200×0.01=2m = np = 200 \times 0.01 = 2 We want to find the probability that more than 3 patients require special room facilities, i.e., P(X>3)\displaystyle P(X > 3): P(X>3)=1−P(X≤3)=1−[P(X=0)+P(X=1)+P(X=2)+P(X=3)]P(X > 3) = 1 - P(X \le 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)] The probability mass function is: P(X=k)=e−22kk!P(X = k) = \frac{e^{-2} 2^k}{k!} Using the approximation e−2≈0.135335\displaystyle e^{-2} \approx 0.135335: P(X=0)=e−2≈0.135335P(X = 0) = e^{-2} \approx 0.135335 P(X=1)=2e−2≈0.270670P(X = 1) = 2 e^{-2} \approx 0.270670 P(X=2)=222!e−2=2e−2≈0.270670P(X = 2) = \frac{2^2}{2!} e^{-2} = 2 e^{-2} \approx 0.270670 P(X=3)=233!e−2=43e−2≈1.333333×0.135335≈0.180447P(X = 3) = \frac{2^3}{3!} e^{-2} = \frac{4}{3} e^{-2} \approx 1.333333 \times 0.135335 \approx 0.180447 Sum of probabilities for X≤3\displaystyle X \le 3: P(X≤3)=e−2(1+2+2+43)=193e−2≈6.333333×0.135335≈0.857122P(X \le 3) = e^{-2} \left(1 + 2 + 2 + \frac{4}{3}\right) = \frac{19}{3} e^{-2} \approx 6.333333 \times 0.135335 \approx 0.857122 Therefore, the probability that more than 3 patients require special rooms: P(X>3)=1−0.857122=0.142878≈0.1428P(X > 3) = 1 - 0.857122 = 0.142878 \approx 0.1428 Hence, **Option A** 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.

More Questions from Theoretical Distributions

Find the variance of binomial distribution with n=10\displaystyle n=10, p=0.3\displaystyle p=0.3

When p=0.5\displaystyle p=0.5, the binomial distribution is

If mean and standard deviation of a binomial distribution is 10\displaystyle 10 and 2\displaystyle 2 respectively, q\displaystyle q will be

A random variable X\displaystyle X follows Binomial Distribution With E(x)=2\displaystyle E(x)=2 and V(x)=1.2\displaystyle V(x)=1.2, then the value of n\displaystyle n is

When 'p\displaystyle p' is large than 0.5\displaystyle 0.5, the Binomial Distribution is:

For a Poisson distribution,

Ready to Master Theoretical Distributions?

Practice all 230 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free