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If the probability that an individual suffers a bad reaction from an injection of a given serum is 0.001, then using Poisson's distribution, the proba
Question

If the probability that an individual suffers a bad reaction from an injection of a given serum is 0.001, then using Poisson's distribution, the probability that out of 2000 individuals, exactly 3 will suffer a bad reaction is:

A.

​43e−2\frac{4}{3} e^{-2} \\​​

B.

​92e−2\frac{9}{2} e^{-2} \\​​

C.

​23e−3 \frac{2}{3} e^{-3} \\​​

D.

​43e−3\frac{4}{3} e^{-3} \\​​

Correct option is A

Given:
​p=0.001,n=2000λ=np=2000×0.001=2k=3p = 0.001, \quad n = 2000 \\\lambda = np = 2000 \times 0.001 = 2 \\k = 3 \\​​
Formula used:
​P(X=k)=λk⋅e−λk!P(X = k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!} \\​​
Solution:
​P(X=3)=23⋅e−23!=8⋅e−26=43e−2P(X = 3) = \frac{2^3 \cdot e^{-2}}{3!} = \frac{8 \cdot e^{-2}}{6} = \frac{4}{3} e^{-2} \\​​
Correct answer is (a)

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