Question

Suppose that the probability of suffering a side effect from acertain flu vaccine is 0.005. If 1000 persons are inculated, findapproximately the probability that:
a) At most 1 person suffers
b) 4,5,or 6 persons suffer

Answer 1

Answer:

The answers to the question are

a) The probability that at most 1 person suffers side effect from the flu vaccine is 4.043×10⁻²

b) The probability that 4,5,or 6 persons suffer side effect from the flu vaccine is 0.49716.

Step-by-step explanation:

We solve the question using the formula for Poisson probability as follows

Let the probability of suffering a side effect from a certain flu vaccine be p

we have n = 1000 and p = 0.005

then by binomial theorem μ = 1000*0.005 = 5

using  Poisson probability formula, we have$P(X=k) =\frac{\lambda ^k e^{-\lambda}}{k!}$

where λ = μ

for P(X=0) we have $\frac{5^0e^{-5}}{0!} =$ 6.73×10⁻³

P(X=1) we have $\frac{5^1e^{-5}}{1!} =$ 3.37×10⁻²

Using the addition rule, we have

P (X≤1) = P(X=0) + P(X=1) =  6.73×10⁻³ + 3.37×10⁻² = 4.043×10⁻²

b) Similarly, we have for k = 4, 5, 6 we have

P(X=4) we have $\frac{5^4e^{-5}}{4!} =$0.1755

P(X=5) we have $\frac{5^5e^{-5}}{5!} =$ 0.1755

P(X=6) we have $\frac{5^6e^{-5}}{6!} =$0.14622

Therefore P(X=4) or P(X=5), P(X=6)  is given by

0.1755 + 0.17546 + 0.14622 = 0.49716