Question

A study conducted at a certain college shows that 54% of the school's graduates move to a different state after graduating. Find the probability that among 7 randomly selected graduates, at least one moves to a different state after graduating.

Answer 1

Answer:

99.56% probability that among 7 randomly selected graduates, at least one moves to a different state after graduating.

Step-by-step explanation:

For each graduate, there are only two possible outcomes. Either they move to a different state, or they do not. The probability of a graduate moving to a different state is independent of other graduates. 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.

54% of the school's graduates move to a different state after graduating.

This means that $p = 0.54$

7 randomly selected graduates

This means that $n = 7$

Find the probability that among 7 randomly selected graduates, at least one moves to a different state after graduating.

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

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

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

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

In which

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

$P(X = 0) = C_{7,0}.(0.54)^{0}.(0.46)^{7} = 0.0044$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0044 = 0.9956$

99.56% probability that among 7 randomly selected graduates, at least one moves to a different state after graduating.