Answer to Question #133871 in Statistics and Probability for Koketso Maripane

Question

Autism south africa has found that 50 %0f people with autism spectrum disorder (asd) struggle with social interaction. Assume we randomly select 5 people living with asd. What is the probability that at most 1 person with asd will struggle with social interaction?

Answer 1

we can solve this problem using binomial distribution

here given,

50 % of people with autism spectrum disorder (asd) struggle with social interaction

that is $p = 0.5$

50 % of people with autism spectrum disorder (asd) not struggle with social interaction

that is $q = 0.5$

randomly select 5 people living with asd

that is $n = 5$

the probability that at most 1 person with asd will struggle with social interaction =

(probability that 0 person with asd will struggle with social interaction )+

(probability that 1 person with asd will struggle with social interaction)

= $[5C0 * (0.5)^0 * (0.5)^5 ] + [5C1 * (0.5)^1 * (0.5)^4]$

=[0.03125] + [0.15625]

= 0.1875

= 18.75% ANSWER