Question

According to a study done by Otago University, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267. Suppose you sit on a bench in a mall and observe people's habits as they sneeze. What is the probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth

Answer 1

Answer:

47.52% probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth

Step-by-step explanation:

For each individual, there are only two possible outcomes. Either they cover their mouth when sneezing, or they do not. The probability of an individual covering their mouth when sneezing is independent of other individuals. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

According to a study done by Otago University, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267.

This means that $p = 0.267$

What is the probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth

10 individuals, so n = 10.

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.267)^{0}.(0.733)^{10} = 0.0448$

$P(X = 1) = C_{10,1}.(0.267)^{1}.(0.733)^{9} = 0.1631$

$P(X = 2) = C_{10,2}.(0.267)^{2}.(0.733)^{8} = 0.2673$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0448 + 0.1631 + 0.2673 = 0.4752$

47.52% probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth