A study conducted at a certain college shows that "53%" 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. 0.989 0.978 0.927 0.167 0.530

Respuesta :

Answer:

0.989

Step-by-step explanation:

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

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

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

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A study conducted at a certain college shows that "53%" of the school's graduates find a job in their chosen field within a year after graduation.

This means that [tex]p = 0.53[/tex]

6 randomly selected graduates

This means that [tex]n = 6[/tex]

Probability that at least one finds a job in his or her chosen field within a year of graduating:

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

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

We want [tex]P(X \geq 1)[/tex]

So

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{6,0}.(0.53)^{0}.(0.47)^{6} = 0.011[/tex]

So

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.011 = 0.989[/tex]