It is known that 60% of the students at a large university have a job and 40% do not have a job. If three of these students are randomly selected, what is the probability at least one does not have a job? (Hint: the compliment of this event is all three have jobs.)

78.4% probability at least one does not have a job

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they have a job, or they do not have a job. The probability of a student having a job 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.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

60% of the students at a large university have a job

This means that

Three of these students are randomly selected

This means that

What is the probability at least one does not have a job?

Either all of them have a job, or at least one does not. The sum of the probabilities of these events is decimal 1. So

We want P(X < 3). Then

In which

78.4% probability at least one does not have a job

your mom x 3

step-by-step explanation:

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.