Question

food allergies affect an estimated 17% of children under age 5 in the us. what is the probability that in a kindergarten class of 11 kids under age 5 at least one of them has food allergies?

Explanation:

Step1: Find the probability of a child not having food - allergies

The probability that a child under age 5 has food - allergies is $p = 0.17$. So the probability that a child under age 5 does not have food - allergies is $q=1 - p=1 - 0.17 = 0.83$.

Step2: Find the probability that none of the 11 children have food - allergies

Assuming that the food - allergy status of each child is an independent event, the probability that none of the $n = 11$ children have food - allergies is given by the binomial probability formula $P(X = 0)=C(n,0)\times p^{0}\times q^{n}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$. Since $C(11,0)=1$ and $p^{0}=1$, we have $P(X = 0)=0.83^{11}$.
\[0.83^{11}\approx0.149\]

Step3: Find the probability that at least one child has food - allergies

The probability that at least one child has food - allergies is the complement of the event that none of the children have food - allergies. Let $P(X\geq1)$ be the probability that at least one child has food - allergies. Then $P(X\geq1)=1 - P(X = 0)$.
\[P(X\geq1)=1 - 0.83^{11}\approx1 - 0.149 = 0.851\]

Answer:

$0.851$