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Sagot :
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
For more details, prefer this link:
https://brainly.com/question/23044118
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