pdf Ron, Sue, and Ted arrive at the beginning of a professor’s office hours. The amount of time they will stay is exponentially distributed with means 1, 1/2, and 1/3 hours. (a) What is the expected time until only one student remains? (b) For each student, find the probability they are the last student left. (c) What is the expected time until all three students are gone?

Answer:

a) 0.45 hours

b)

P( Ron ) = 7/12

P( sue ) = 4/15

P(Ted ) = 3/20

c)  1.22 hours

Step-by-step explanation:

Ron , sue and Ted

mean values = 1, 1/2 , 1/3  hours

a) determine the expected time until only one student remains

E(x) = 9/20 = 0.45 hours

b) Probability that each student is the last student left

P( Ron ) = 7/12

P( sue ) = 4/15

P(Ted ) = 3/20

C) Time until all students are gone

= 146 / 120 =  1.22 hours

attached below is the detailed solution

This question is based on the concept of probability.Therefore, the answers are as follows:

(a) 0.45 hours

[tex]\bold{(b) \, P(Ron) = \dfrac{7}{12} ,P (Sue) = \dfrac{4}{15} ,P(Ted) = \dfrac{3}{20}\beta}[/tex]  

(c)  1.22 hours

Given:

The amount of time they will stay is exponentially distributed with means 1, 1/2, and 1/3 hours.

According to the question,

The amount of time they will stay is exponentially distributed with means 1, 1/2, and 1/3 hours.

(a) The expected time until only one student remains,

[tex]E(X) = \dfrac{9}{20} = 0.45 \,hours[/tex]

(b)  Now, find the probability they are the last student left,

[tex]P(Ron) = \dfrac{7}{12} \\P (Sue) = \dfrac{4}{15} \\P(Ted) = \dfrac{3}{20}[/tex]

(c) The expected time until all three students are gone is,

[tex]= \dfrac{146}{120} \\\\= 1.22 hours[/tex]

Therefore, the answers are as follows:

(a) 0.45 hours

[tex](b) P(Ron) = \dfrac{7}{12} ,P (Sue) = \dfrac{4}{15} ,P(Ted) = \dfrac{3}{20}\beta[/tex]  

(c)  1.22 hours

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