Suppose that three executives bump into each other in an elevator and drop their identical cellular phones as the doors are closing, leaving them with no alternative but to pick up a phone at random.

a. Describe in detail how you could conduct a simulation of this situation to produce empirical estimates of the probabilities involved.
b. List all the possible outcomes in the sample space for this situation.
c. Use the sample space to determine the probability that

• Nobody gets the correct phone.
• Exactly one person gets the correct phone.
• Exactly two people get the correct phone.
• All three people get the correct phone.
• At least one person gets the correct phone.

d. If this situation were to befall a group of five people, you could calculate the probability that nobody gets the correct phone as 11/30. Explain in your own words what this probability means about the likelihood of nobody getting the correct phone. Include the long-run interpretation of probability in your answer and relate your response to the situation’s context.

a. See Explanation Below

b. The complete list of possibilities is:

E1,E2,E3

E1,E3,E2

E2,E1,E3

E2,E3,E1

E3,E1,E2

E3,E2,E1

c. P(X = 0) = ⅓

P(X = 1) = ½

P(X = 2) = 0

P(X = 3) = 1/6

P(X ≥ 1) = ⅔

d. See Explanation Below

Step-by-step explanation:

a.

First, we list out the sample space.

To do this, we need to list out all possible events or occurrence.

Then the number of required or expected outcomes is divided by the total sample space.

With the above illustration, the estimates of each probabilities involved is solved.

b. Given

Let E1 represent first executive picking a phone.

Let E2 represent second executive picking a phone

Let E3 represent third executive picking a phone

So, E1E2E3 means E1 picked up his own cell phone, E2 picked up his own cell phone and E3 picked up his own cell phone.

So the complete list of possibilities is:

E1,E2,E3

E1,E3,E2

E2,E1,E3

E2,E3,E1

E3,E1,E2

E3,E2,E1

c.

i. Let X = 0 be the event that nobody gets the correct phone.

From the list of the possibilities;

Only E3,E1,E2 and E2,E3,E1 satisfy this condition

So, number of required outcome ,= 2

Number of possible outcome ,= 6

P(X,=0) = 2/6

P(X = 0) = ⅓

ii. Let X = 1 be the event that exactly one person gets the correct phone.

From the list of the possibilities;

Only E1,E3,E2 ; E2,E1,E3 and E3,E2,E1 satisfy this condition

So, number of required outcome ,= 3

Number of possible outcome ,= 6

P(X = 1) = 3/6

P(X = 1) = ½

iii. Let X = 2 be the event that exactly two people get the correct phone.

The probability that exactly two get the correct phone doesn't exist because if two people gets their correct phone then the third person will also get the correct phone.

So, P(X = 2) = 0

iv. Let X = 3 be the event that all three people get the correct phone.

From the list of the possibilities;

Only E1,E2,E3 satisfy this condition

So, number of required outcome ,= 1

Number of possible outcome ,= 6

P(X = 3) = 1/6

v. Let X ≥ 1 be the event that at least one person gets the correct phone.

This can be calculated by; 1 - Probability that Nobody gets the correct phone.

P(X ≥ 1) = 1 - P(X = 0)

P(X ≥ 1) = 1 - ⅓

P(X ≥ 1) = ⅔

d. For a situation that it involves 5 people

It means the probability about the likelihood of nobody getting the correct phone is indefinitely.

And in the long-run it's expected that none of the five people getting the correct phone in approximately 36.67% of the trials.

fourth month

step-by-step explanation:

it is the one with the most keychains.

so, converting firstly the mixed fractions to improper fractions and then "adding".