Question

2. According to the Center for Disease Control, about 1 in 80 children in the U.S. have been diagnosed with autism. Suppose you consider a group of 10 children. Find the probability that: a) None have autism; b) 4 have autism; c) at least 2 has autism.

Answer 1

Answer: (a) 88.18%

(b) 0.00000226%

(c) 10.7%

Explanation:

`\text{Have Autism}: (1)/(80)=0.0125\qquad \text{Don't have Autism}: (79)/(80)=0.9875`

a) 0 "Have Autism" and 10 "Don't have Autism"

(0.0125)⁰ (0.09875)¹⁰ = 0.8818

= 81.18%

b) 4 "Have Autism" and 6 "Don't have Autism"

(0.0125)⁴ (0.9875)⁶ = 0.0000000226

= 0.00000226%

c) 2 (or more) Have Autism and 8 (or less) Don't Have Autism

It is best to find the opposite and subtract from 100%

0 "Have Autism" and 10 "Don't have Autism" = 0.8818 (see part a)

or

1 "Have Autism" and 9 "Don't have Autism" = (0.0125)¹ (0.9875)⁹

= 0.0112

1 - (0.8818 + 0.0112) = 0.1070

= 10.7%