Question

approximately 55% of high school students participate in athletic programs. if we choose 5 high school students at random, what is the probability that at least one of them

Answer 1

Answer:

98.15% probability that at least one of them participate in athletic programs.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they participate in athletic programs, or they do not. The probability of a student participaing in an athletic program is independent of other students, 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.

55% of high school students participate in athletic programs.

This means that $p = 0.55$

5 high school students

This means that $n = 5$

What is the probability that at least one of them participate in a athletic program?

Either none participate, or at least one of them does. The sum of the probabilities of these events is 1. So

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

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

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

In which

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

$P(X = 0) = C_{5,0}.(0.55)^{0}.(0.45)^{5} = 0.0185$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0185 = 0.9815$

98.15% probability that at least one of them participate in athletic programs.