Question-70
A small pharmaceutical company manufactures a rare vaccine for a specific disease. Based on historical data, the average failure rate of the vaccine is 8 per 10,000 doses due to improper storage or handling. In one year, the company distributes 5,000 doses. What is the probability that more than 2 doses fail in a given year\(?\)
Use the concept of “Poisson distribution is a limiting case of the binomial distribution under the following conditions:
(i). \(n\), the number of trials is indefinitely large, i.e., \(n \to \infty\).
(ii). \(p\), the constant probability of success for each trial is indefinitely small, i.e., \(p \to 0\).
(iii). \(np = \lambda,\) (say) is finite.
The problem states that the failure rate of the vaccine is 8 per 10,000 doses. For a distribution of 5,000 doses, the expected number of failures \((\lambda)\) is calculated as :
\(\lambda = n p = 5,000 \times \dfrac{8}{10,000} = 4\), since \(p\) is small and \(n\) is large.
Thus, the mean number of failures is \(\lambda = 4\).
Suppose \(X\) be the random variable as number of vaccine failures. Now, we need to calculate \(P(X > 2) = 1 - P(X \leq 2) ~~ \ldots (1)\).
Now, \(P(X \leq 2)= P(X=0) + P(X = 1) + P(X=2)\)
\(P(X \leq 2) = \dfrac{e^{-4}4^0}{0!} + \dfrac{e^{-4}4^1}{1!} + \dfrac{e^{-4}4^2}{2!} = 0.238\)
Thus, \(P(X > 2) = 1 - 0.238 = 0.762\)