Question

ct Question 1 0/3 pts A study conducted at a certain college shows that 57% of the school's graduates find a job in their chosen field within a year after graduation. Find the probability that among 9 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating. Round your answer to nearest ten-thousandth. 0.9994 rrect Question 2 0/3 pts A sample of 4 different calculators is randomly selected from a group containing 47 that are defective and 29 that have no defects. What is the probability that all four of the calculators selected are defective? Round your answer to nearest ten- thousandth. 0.1359

please help with work and answer!

Answer 1

(1)

The probability of getting job within one year is 0.57(i.e, 0.57)

Here we select randomly 9 students.

Let the a random variable X that represent number of students getting job within one year.

$\therefore X\sim Binomial (9,0.57)$

The probability mass function of X is given by,

n P(X = z) = p*(1 – p)***; x = 0, 1, 2,..

We need to find the probability of atleast one student will getting job within one year.

$i.e, P(X\geq 1)$

= 1- P(X <1)

$=1-P(X=0)$

$=1-\binom{9}{0}0.57^0(1-0.57)^{9-0}$

$=1-0.43^9$

$=1-0.0005025926119$

$=0.999497407$

$\simeq 0.9995$[ round to tenth thousand place]

The probability is 0.9995

(2) A group containing 47 defective calculators and 29 no defective calculators.

We randomly selecte 4 calculators.

We need to find the probability that all the calculator are defective.

If we select 4 calculators from 47 defective calculators then we will get all defective calculators.

$\therefore P(all\ calculators\ are \ defective)=\frac{\binom{47}{4}}{\binom{76}{4}}$

$\therefore P(all\ calculators\ are \ defective)=\frac{178365}{1282975}$

$\therefore P(all\ calculators\ are \ defective)=0.139024532$

$\therefore P(all\ calculators\ are \ defective)\simeq 0.1390$[ round to tenth thousand place]

The probability is 0.1390