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Suppose you have a machine that randomly prints out tickets with numbers on them. 17% of the time the ticket has number 3. 24% of the time the ticket has number 8. 37% of the time the ticket has number 15. 22% of the time the ticket has number 22. There are no other ticket numbers. Let the random variable X be the number on the ticket.

Suppose you have a machine that randomly prints out tickets with numbers on them. 17% of the time the ticket has number 3. 24% of the time the ticket has number 8. 37% of the time the ticket has number 15. 22% of the time the ticket has number 22. There are no other ticket numbers. Let the random variable X be the number on the ticket.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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if someone could help me solve this one problem that would be great :) 

Ticket Machine Setup Suppose you have a machine that randomly prints out tickets with numbers on them. 17% of the time the ticket has number 3. 24% of the time the ticket has number 8. 37% of the time the ticket has number 15. 22% of the time the ticket has number 22. There are no other ticket numbers. Let the random variable X be the number on the ticket.
Transcribed Image Text:Ticket Machine Setup Suppose you have a machine that randomly prints out tickets with numbers on them. 17% of the time the ticket has number 3. 24% of the time the ticket has number 8. 37% of the time the ticket has number 15. 22% of the time the ticket has number 22. There are no other ticket numbers. Let the random variable X be the number on the ticket.
20. Roll ten unfair six-sided dice where, for each die, P(one) = 0.08, P(two) = 0.12, P(three) = 0.10, P(four) = 0.22, P(five) = 0.28, P(siz) = 0.20. Find P(1 – 2 – 3 – 1 – 2 – 1). %3D A. 0.00010086 B. 0.00020086 C. 0.00040086 D. 0.00060086
Transcribed Image Text:20. Roll ten unfair six-sided dice where, for each die, P(one) = 0.08, P(two) = 0.12, P(three) = 0.10, P(four) = 0.22, P(five) = 0.28, P(siz) = 0.20. Find P(1 – 2 – 3 – 1 – 2 – 1). %3D A. 0.00010086 B. 0.00020086 C. 0.00040086 D. 0.00060086
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