Question

Suppose the probability that an individual suffers an adverse reaction from a particular drug is known to be \(0.001\). The probability that out of \(2000\) individuals, exactly three will suffer an adverse reaction is ____.

• A. \(\dfrac{3e^{-3}}{2}\)
• B. \(e^{-3}\)
• C. \(\dfrac{3e^{-2}}{2}\)
• D. \(\dfrac{4e^{-2}}{3}\)

Hint:

Use Poisson approximation to Binomial when \[ \boxed{ n\ge20,\quad p\le0.05,\quad \lambda=np. } \] Then, \[ \boxed{ P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!}. } \]

Answer 1

The Correct Option is D

Concept: Since \[ n=2000,\qquad p=0.001, \] where \(n\) is very large and \(p\) is very small, the Binomial distribution can be approximated by the Poisson distribution. The Poisson parameter is \[ \boxed{\lambda=np=2000\times0.001=2.} \] The Poisson probability is \[ \boxed{ P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!}. } \]

Step 1: Find the mean of the Poisson distribution. \[ \lambda=np=2. \]

Step 2: Calculate the probability of exactly three adverse reactions. For \(r=3\), \[ P(X=3) = \frac{e^{-2}(2)^3}{3!} = \frac{8e^{-2}}{6} = \frac{4e^{-2}}{3}. \] Hence, \[ \boxed{ P(X=3)=\frac{4e^{-2}}{3}. } \] Therefore, the correct option is \[ \boxed{(D)\;\dfrac{4e^{-2}}{3}.} \]