Question

"A study conducted at a certain college shows that 56% of the school's graduates find a job in their chosen field within a year after graduation. Find the probability that among 6 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating."

Answer 1

Answer:

99.27% probability that among 6 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they find a job in their chosen field within one year of graduating, or they do not. The probability of a student finding a job in their chosen field within one year of graduating 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.

56% of the school's graduates find a job in their chosen field within a year after graduation.

This means that $p = 0.56$

Find the probability that among 6 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating.

This is $P(X \geq 1)$ when n = 6.

Either none find a job, or at least one does. The sum of the probabilities of these events is decimal 1. So

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

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

In which

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

$P(X = 0) = C_{6,0}.(0.56)^{0}.(0.44)^{6} = 0.0073$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0073 = 0.9927$

99.27% probability that among 6 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating.