Question

In a class of 125 students, 28 are computer science majors, 52 are mechanical engineering majors, 10 are civil engineers and the rest are general engineering majors.Assume students only have one major If a student is chosen at random what is the probability they are: Round your answers to 3 decimal places a clvil engineering major? o8 a civil engineering major or mechanical engineering major?496 a general engineering major? 280 not a computer science major? 776 Suppose six students from the class are chosen at random what is the probability none are mechanical engineering majors? OShow My Work (optional 3. 0.33/1 points 1/3 Submissions Used My Notes Ask Your Teach Suppose that 45% of all adults regularly consume coffee, 60% regulary consume carbonated soda and 40% regularly consumes both coffee and soda. (a) what is the chance a randomly selected adult regularly drinks coffee but doesn't drink soda? (b) What is the probability that a randomly selected adult consumes coffee, soda or both? 65 (c) What is the probability that a randomly selected adult doesn't regularly consume at least one ofthese two products? 60 Show My Work (optional)

Answer 1

Question 1:

• Civil engineering major = 10/125 = 0.08
• Civil engineering or mechanical engineering major = (52+10)/125 = 0.496
• General engineering major = (125 - (28+52+10))/125 = 0.28
• Not a computer science major = (125 - 28)/125 = 0.776
• Probability of a student not being from mechanical engineering major = (125 - 52)/125 = 0.584. So if six students are picked up randomly then the probability that none of are from the mechanical engineering major = $0.584^6$ = 0.03967

Question 2 (3 as per the question no given)

Probability of having coffee, P(C) = 0.45,

Probability of having soda, P(S) = 0.6

​​​​​​​Probability of having both, $\inline P (C \cap S) = 0.4$

• Desired Probability = $\inline P(C) - P (C \cap S) = 0.45 - 0.4 = 0.05$ ​​​​​​​
• Desired Probability = $\inline P(C \cup S) = P(C) + P(S) - P (C \cap S) = 0.45 + 0.6 - 0.4 = 0.65$
• Desired Probability = $\inline 1-P(C \cup S) = 1 - 0.65 = 0.35$

** If any of these answers does not match please comment.