Question

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

Answer 1

0.989

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.

`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) = (n!)/(x!(n-x)!)`

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
`p = 0.53`

6 randomly selected graduates

This means that
`n = 6`

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

`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_(6,0).(0.53)^(0).(0.47)^(6) = 0.011`

So

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.011 = 0.989`