Question

work for a publishing company. The company wants to send two employees to a statistics conference. To be​ fair, the company decides that the two individuals who get to attend will have their names randomly drawn from a hat. ​(a) Determine the sample space of the experiment. That​ is, list all possible simple random samples of size n equals 2. ​(b) What is the probability that Roberto and John attend the​ conference? ​(c) What is the probability that Clarice attends the​ conference? ​(d) What is the probability that John stays​ home?

Answer 1

Answer:

(a) S = {MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC}

(b) The probability that Roberto and John attend the​ conference is 0.10.

(c) The probability that Clarice attends the​ conference is 0.40.

(d) The probability that John stays​ home is 0.60.

Step-by-step explanation:

It is provided that :

Marco (M), Roberto (R), John (J), Dominique (D) and Clarice (C) works for the company.

The company selects two employees randomly to attend a statistics conference.

(a)

There are 5 employees from which the company has to select two employees to send to the conference.

So the total number of ways to select two employees is:

${5\choose 2}=\frac{5!}{2!(5-2)!}=\frac{5\times 4\times 3!}{2\times 3!}=10$

The 10 possible samples are:

MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC

(b)

The probability of the event E is:

$P(E)=\frac{n(E)}{N}$

Here,

n (E) = favorable outcomes

N = Total number of outcomes.

The variable representing the selection of  Roberto and John is, RJ.

The favorable number of outcomes to select Roberto and John is, 1.

The total number of outcomes to select 2 employees is 10.

Compute the probability that Roberto and John attend the​ conference as follows:

$P(RJ)=\frac{n(RJ)}{N}=\frac{1}{10}=0.10$

Thus, the probability that Roberto and John attend the​ conference is 0.10.

(c)

The favorable outcomes of the event where Clarice attends the conference are:

n (C) = {MC, RC, JC and DC} = 4

Compute the probability that Clarice attends the​ conference as follows:

$P(C)=\frac{n(C)}{N}=\frac{4}{10}=0.40$

Thus, the probability that Clarice attends the​ conference is 0.40.

(d)

The favorable outcomes of the event where John does not attends the conference are:

n (J') = MR, MD, MC, RD, RC, DC

Compute the probability that John stays​ home as follows:

$P(J')=\frac{n(J')}{N}=\frac{6}{10}=0.60$

Thus, the probability that John stays​ home is 0.60.