Question

If the probability that a person suffers a bad reaction from an injection is $0.001$, then the probability that out of $2000$ individuals, exactly $3$ will suffer a bad reaction is:

• A. $\frac{4}{3} e^{-2}$
• B. $4 e^{-2}$
• C. $\frac{1}{3} e^{-2}$
• D. $2 e^{-2}$

Hint:

For very large $n$ and small $p$, the Poisson approximation $\lambda = np$ is the fastest and most efficient way to compute probabilities.

Answer 1

The Correct Option is A

Step 1: Choose the right model.
Here the number of people $n = 2000$ is very large and the chance $p = 0.001$ is very small. In such cases the Poisson distribution is the easy approximation to use.

Step 2: Write the Poisson formula.
The chance of exactly $r$ events is given by a standard formula using the mean $\lambda$.
\[ P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!} \]

Step 3: Find the mean $\lambda$.
The mean equals $n$ times $p$.
\[ \lambda = np = 2000\times 0.001 = 2 \]

Step 4: Set $r = 3$.
We want exactly $3$ bad reactions, so put $r = 3$ and $\lambda = 2$ into the formula.
\[ P(X = 3) = \frac{e^{-2}\cdot 2^3}{3!} \]

Step 5: Work out the numbers.
We have $2^3 = 8$ and $3! = 6$.
\[ P(X = 3) = \frac{8\,e^{-2}}{6} \]

Step 6: Simplify the fraction.
Reduce $\frac{8}{6}$ to $\frac{4}{3}$.
\[ P(X = 3) = \frac{4}{3}e^{-2} \]
\[ \boxed{\tfrac{4}{3}e^{-2}} \]