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MATH 110 Module 3 Problem Set 2025-
M3: Problem Set
- Due No due date
- Points 5
- Questions 11
- Time Limit None
Instructions
Be sure to familiarize yourself with the following tools. They will be available during your exams.
Statistics_Equation_Sheet (https://previous.nursingabc.com/Help%20Files/ Math%20110%20Introduction%20to%20Statistics%20Formula%20Sheet.pdf)
Scientific Calculator (https://www.desmos.com/scientific)
Attempt History
Time Score for this quiz: 5 out of 5 Submitted Apr 17 at 9:13pm This attempt took 381 minutes.
Question 1 0 / 0 pts According to the National Oceanic and Atmospheric Administration (NOAA), between 1851 and 2013 there were 290 hurricanes that hit the U.S. Coast. Of these, 117 were Category 1 hurricanes, 76 were Category 2 hurricanes, 76 were Category 3 hurricanes, 18 were Category 4 hurricanes, and 3 were Category 5 hurricanes.
Make a probability distribution for this data. If a hurricane hits the U.S. coast, what is the probability that the hurricane will be a Category 1 hurricane. Your Answer:
Category Calculation Probability
Attempt Score LATEST Attempt 1 381 minutes 5 out of 5
If a hurricane hits the U.S. coast, there is a .403 probability that
it is a category 1 hurricane Solution:
We can see from the probability distribution, that if a hurricane hits the U.S. coast, there is a. probability that it is a category 1 hurricane.
Question 2 0 / 0 pts
Find the answer to each of the following by first reducing the fractions as much as possible:
a) P(17,8)=
b) C(19,15)=
Your Answer:
a)
a) P(17,8)= P(n,r)=n!/(n- r)! P(17,8)=17! =17/9!(16)(15)(14)(13)(12)(11)(10)= 980,179,200 b) C(n,r)= n!/r!(n-r)!
Question 4 0 / 0 pts Suppose that 10 countries submit bids for the summer Olympics. The Olympic committee will select 4 finalists out of these 10 countries. In how many ways can the Olympic committee pick these 10 finalists? Your Answer: C(n,r)=n!/r!(n-r)! C(10,4)= 10!/4!(10-4)! =10!/4!(6!) = 10(9)(8)(7)/4(3)(2)(1) = 210
Solution.
Here order does not matter. So, we will use combinations. There are ten countries to choose from, so n=10. Four countries will be chosen, so r=4. We will use
Question 5 0 / 0 pts
Suppose A and B are two events with probabilities:
Find the following:
Your Answer:
a)
P(A U B) = P(A) + (B) - P(A n B)
= .40 + .45 -.
=.
Therefore, P(A U B) =.
b) P (Ac)
P (A ) = 1- P(A)c
1-.40 = .60 Therefore,
P (Ac) =.
c) P(Bc)
P(Bc)= 1 -P(B)
1-.45= .55P
Therefore, P(Bc) = 55
Solution.
We are given:
Question 6 0 / 0 pts
Question 7 0 / 0 pts
Suppose A and B are two events with probabilities:
Your Answer:
a) (A|B)= P(AnB)/P(B) = .25/. =. b) (B|A)= P(AnB)/P(A) .25/. =.
Solution.
We are given:
Question 8 0 / 0 pts
Suppose A and B are two events with probabilities:
Your Answer:
To find (A n B), we will first find P(A) using P(A) = 1 - P(Ac)
P (A) = 1- .30 =.
Therefore, P(A n B) = P(B|A) x P(A)
(A n B) = .40 x .70 = 0.
(A n B) = 0.
Solution.
We are given:
Question 10 0 / 0 pts In a manufacturing plant, three machines A,B, and C produce 40 %, 35 %, and 25 %, respectively, of the total production. The company's quality control department determined that 1 % of the items produced by machine A, 1.5 % of the items produced by machine B, and 2 % of the items produced by machine C are defective. If an item is selected at random and found to be defective, what is the probability that it was produced by machine B? Your Answer:
We first define the events as follows:
A: item is produced by machine A B: item is produced by machine BWe are asked to find the C: item is produced by machine C D: item is defectiveprobability that an item was produced by machine B given that it is defective, or P(B|D)
According to Bayes' theorem, we can express this as: P(B|D) = P(D|B) × P(B) / P(D)
the following probabilities from the problem: P(A) = 0.40 (40% of items are produced by machine A) P(B) = 0.35 (35% of items are produced by machine B) P(C) = 0.25 (25% of items are produced by machine C) P(D|A) = 0.01 (1% of items produced by machine A are defective) P(D|B) = 0.015 (1.5% of items produced by machine B are defective) P(D|C) = 0.02 (2% of items produced by machine C are defective)
Total probability =
P(D|A) × P(A) + P(D|B) × P(B) + P(D|C) × P(C) 0.01 x 0.40 + 0.015 x 0.35 + 0.02 x 0.
0.004 + 0.00525 + 0.005= 0.
Substitute these values into Bayes' theorem, we get P(B|D) = P(D|B) × P(B) / P(D) = 0.015 × 0.35 / 0. = 0.
Therefore, If an item is selected at random and found to be defective, the probability that it was produced by machine B is 0.0368 or 3.68%
Solution
In a manufacturing plant, three machines A, B, and C produce 40 %, 35 %, and 25 %, respectively, of the total parts production. The company's quality control department determined that 1 % of the parts produced by machine A, 1.5 % of the parts produced by machine B, and 2 % of the parts produced by machine C are defective. If a part is selected at random and found to be defective, what is the probability that it was produced by machine B?
If we use Def to designate “defective”.
We are told that given that a part was produced by machine A, the probability that it has a defect is: P ( Def | A ) = .01.
We are told that given that a part was produced by machine B, the probability that it has a defect is: P ( Def | B ) =.015.
We are told that given that a part was produced by machine C, the probability that it has a defect is: P ( Def | C ) =.02.
Furthermore, we are told that the probability that a part was produced by machine A, B, and C, are respectively:
P ( A ) = .40, P ( B ) = .35, P ( C ) = .25.
We want to find P ( B | Def ), so use:
Question 11 5 / 5 pts As a reminder, the questions in this review quiz are a requirement of the course and the best way to prepare for the module exam. Did you complete all questions in their entirety and show your work? Your Answer:
Yes I did
