Question

A survey reports that the probability a person has blue eyes is 0.10. Assume that 4 people are randomly selected at Miramar College and asked if they have blue eyes, find the probability that at least 1 of them have blue eyes. Round to 3 decimal places. 0.291

Answer 1

Answer:

0.344 = 34.4% probability that at least 1 of them have blue eyes.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have blue eyes, or they have not. The probability of a person having blue eyes is independent of any other person. 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.

A survey reports that the probability a person has blue eyes is 0.10.

This means that $p = 0.1$

4 people are randomly selected at Miramar College

This means that $n = 4$

Find the probability that at least 1 of them have blue eyes.

Either none of them have blue eyes, or at least one do. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$. So

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.656$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.656 = 0.344$

0.344 = 34.4% probability that at least 1 of them have blue eyes.