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Don't have my book or audio? You can download a free sample of my book and audio lectures containing all of Lesson 1:
Did you read my tips on how to study and learn this course? If not, here is a link to those important suggestions: The department posted SOLUTIONS for Assignment 1 (these are not my solutions). Click here.The department posted SOLUTIONS for Assignment
2 (these are not my solutions). Click here.The department posted SOLUTIONS for Assignment 3 (these are not my solutions). Click here.
Tips for Distance Assignment 4
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Here is a link to the actual assignment, for those of you who don't have it: Study Lesson 5: Introduction to Probability and Lesson 6: The Binomial Distribution in my book, if you have it, to prepare for this assignment.
Of
course, always seek out assistance from my book, your course notes, etc. if you ever hit a question you don't understand, but try not to be learning things as you do an assignment. Learn first, then put your learning to the test.
Although this question appears to be a 3-circle Venn diagram problem (similar to my Lesson 5, #18), they have left key information out that renders it not worth the effort to try to make the diagram right away.
To make a
three-circle Venn diagram, for event A, B and C, among the givens you need the following: - The probability an outcome belongs to all three of A, B and C.
- The three "and" pairs, which is to say, the P(A and B), the P(A and C), and the P(B and C).
- Then you can assume that everything will go smoothly filling in the 8 sections of the Venn diagram.
If the above is not given, you should use the
probability formulas to solve what is missing above first.
In this problem, you have been given the probability an outcome belongs to all three of H, F and C; you have been given P(H and C); and, you have been given P(H and F). The problem is that you have not been given P(F and C). Before I make a Venn, I would use a probability formula to solve that
missing value.
But that is actually one of their questions below, so, you are better off just taking each question as it comes and answering it directly.
This is tailor-made for the General Addition Rule P(A or B) = P(A) + P(B) - P(A and B). You have been given everything you need.
Again, use the General Addition Rule P(A or B) = P(A) + P(B) - P(A and B) and a little algebra. Note that you have been given P(H or C), P(H and C), and P(H). So, use those to figure out
P(C).
Now, they want P(F and C).
Don't you dare use the formula P(F and C) = P(F)*P(C)!!!!
THAT FORMULA IS ONLY TRUE WHEN TWO EVENTS ARE
INDEPENDENT!!! You have no idea whether F and C are independent or not!!!!
One of the most common mistakes students make, is thinking that, anytime they want to know P(A and B), they merely have to compute P(A)*P(B). WRONG!!!! You can only do that when you are told that A and B are INDEPENDENT.
Here is the proper thing to do: The General Addition Rule, P(A or
B) = P(A) + P(B) - P(A and B), is ALWAYS TRUE. Use that formula to solve either "or" or "and" unknowns whenever possible. That formula has four terms: P(A), P(B), P(A and B), and P(A or B). If you are given any three of those four terms, you can use this formula and a little algebra to solve the missing fourth term. That is what you should have done to solve #2 above, and that is what can also be done here to solve #3. You can solve P(F and C) if you
know P(F), P(C) and P(F or C).
Now, you finally know everything you need to make the three-circle Venn diagram. So, I would make the Venn and use it to solve this question.
Now, you should know P(C) from #2 and P(F and C) from #3. Now you can prove whether any two events in this problem are independent.
Two events are independent if and only if they satisfy the Independence Multiplication
Rule,
P(A and B) = P(A)*P(B).
For example, you were told P(H)=0.49, P(F)=0.62, and P(H and F)=0.27.
If H and F are independent, then P(H and F) = P(H)*P(F).
0.27 should equal 0.49*0.62.
But 0.49*0.62= 0.3038, not 0.27, proving that H and F are dependent, not independent.
You can do a similar test for the other pairs.
Method 1
I would use a two-way table to set up and solve this problem. Similar to what I do in Lesson 5. Note that they want the probability distribution of X, the number of heads, similar to my
questions 3(f) and 4(f) where I use the earlier two-way table and distribution I had come up with to then do a probability distribution for X, the number of boys. BE CAREFUL! This is not the usual 50-50 H or T fair coin. They tell you there is a 0.6 chance of H.
Method 2
This certainly can be solved using binomial methods, as taught in Lesson 6, #1, for example. Keep in mind that
X is the number of heads. What is n and p? They then want you to solve the probabilities for X=0, X=1, etc. using the binomial formula. Obviously, those probabilities should match your answers in Method 1.
Lesson 6. If you are ever asked to decide if a particular situation is binomial or not, remember, to be binomial, four conditions must be satisfied: - There must be a fixed number of trials, n.
- Each trial can have only two
possible outcomes, success or failure, and the probability of success on each trial must have a constant value, p.
- Each trial must be independent.
- X, the number of successes, is a discrete random variable where X = 0, 1, 2, ... n.
Hints: - If you are reading off numbers from a randomly selected row in the random number table, note that every row has 40 digits. That is like 40 trials looking for whatever
digit you may be looking for. What is the probability that, at any moment on the table, the next digit is a 0, or a 1, or a 2, etc..
- If you are selecting objects, are you sampling with replacement (independent trials) or without replacement (dependent trials)?
- Ask yourself, "Is each trial independent?" Does the result for the next individual or trial depend on what happened earlier?
- If you are given a Normal
population, but are selecting a sample of size n, and want to see how many of them are greater than 62 (for example), THAT IS A BINOMIAL DISTRIBUTION! You can use Table A to find what proportion are greater than 62. That is your p. Each trial, the person/thing either is greater than 62, or they are not. And the chance they are greater than 62 is p, the proportion you shaded on the bell curve.
- If you are
ever conducting trials until you get a desired result, that will never be binomial because you do not have a fixed number of trials, n. For example, if I am rolling a die until I get a six and X= the number of rolls until I get a six, that is not binomial, because there are no fixed number of trials, n.
- Even if they haven't given you a percentage p, there may still be a fixed value for p.
For example, if I buy a Tim Horton's coffee during their Roll Up The Rim To Win contest, I know there is a chance that I will win a prize. There is a fixed percentage of cups that are winners, but I don't know what that percentage is. However, if I buy 10 cups of coffee, then X= the number of winners, is binomial, n=10, p=some percentage that exists but just not known to me.
You can solve this problem using two-way tables.
Hint:
Don't be distracted by the three colours of the lights. They are counting stops, which is to say, counting red lights.
At each intersection, the light either is red or it is not.
You have been given P(HH). But that is P(H)*P(H) = h*h=h^2.
Lesson 6, binomial, like my #1-6.
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