 # Exercise #1 According to a recent article from the New England Journal of Medical Stuff, 63% of cowboys suffer from saddle sores, 52% of cowboys suffer.

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Part 3 Module 4 Probability of disjunctions, complements, and conditional probability

Exercise #1 According to a recent article from the New England Journal of Medical Stuff, 63% of cowboys suffer from saddle sores, 52% of cowboys suffer from bowed legs, and 40% suffer from both saddle sores and bowed legs. What is the probability that a randomly selected cowboy has saddle sores or bowed legs? Note that these two events are not mutually exclusive. A B. .35 C D. None of these

A special formula An alternative way to answer the previous question is by applying this special formula for the probability of the disjunction of events: For any events E, F, P(E or F) = P(E) + P(F) – P(E and F)

P(E or F) = P(E) + P(F) – P(E and F)
There are a couple of things to be said about this special formula. Recall that in Part 3 Module 3 we saw that if E, F are mutually exclusive, then P(E or F) = P(E) + P(F) The new formula contains this formula from Part 3 Module 3 as a special case, because, if the events are mutually exclusive, then the third term, P(E and F), is zero, and so disappears. The advantage to using this formula is that it allows us to avoid having to draw the Venn diagram, if the data is in percent form, and the event is a disjunction. Generally, drawing the Venn diagram is a good activity, because there are other questions for which the diagram is helpful.

Exercise #2 www.math.fsu.edu/~wooland/prob1/prob0.html
Among a certain group of Vikings, 11 of them like to pillage, 12 of them like to plunder, while 6 of them like to pillage and like to plunder and 18 of them don't like to pillage and don't like to plunder. If one of these Vikings is randomly selected, find the probability that he/she likes to pillage or likes to plunder. Round to four decimal places. A B C D. None of these.

Follow-up question Referring to the answer to the previous question:
If the probably that a randomly selected Viking like to pillage or plunder is , what is the probability that a randomly selected Viking doesn’t like to pillage and doesn’t like to plunder? Solution The Venn diagram shows that the answer is 18/35, or However, there is another way to answer this question, without looking at the diagram. Recall from logic and set mathematics (DeMorgan’s Laws) that “doesn’t plunder and doesn’t pillage” is the opposite (complement) of “plunders or pillages.” This means that we can also get the answer by applying the complements rule to the answer to the previous question: P(doesn’t plunder and doesn’t pillage) = 1 – P(plunders or pillages) = 1 – = .5143

Exercise #3 At the Forest Folks Gathering there are 39 jolly hobbits, 41 grumpy hobbits, 19 jolly leprechauns and 25 grumpy leprechauns. If one person is randomly selected, find the probability that he/she is a leprechaun or grumpy. A. .532 B. .202 C. .669 D. .685

Exercise #4 The table below summarizes preferences of diners at the Hearty Guys Breakfast. If one of these hearty guys is randomly selected, find the probability that he is a construction worker or prefers pancakes. pancakes waffles French totals toast lumberjack construction worker totals A B C D. 0.56

Part 3 Module 4 advice Most of the probability questions from Part 3 Module 4 do not require fancy calculations. You just have to think of probability in its most basic sense (a fraction representing “this out of that”), and understand the presentation of data to find the number that goes in the denominator (the total population) and the number that goes in the numerator (those who satisfy the event that you are asked about). Drawing Venn diagrams may be helpful for some problems. You may have to work with percents.

Conditional probability
Refer back to the simple example of rolling one die. Let F be the event that the result is a “2.” We know that P(F) = 1/6, because there are 6 equally likely outcomes, one of which is a “2.” Here is a variation on that theme: Suppose that the die roll has already been performed, and we are told that the result was an even number (event E). Would we still say that the P(F) = 1/6? No. Given that we know that an even number was rolled, then there are only three possible outcomes, not six. {2, 4, 6} Of these three possible outcomes, one of them is a “2” so the probability of having rolled a “2,” given that we rolled an even number, is 1/3.

Conditional probability
This is an example of conditional probability. Conditional probability arises when we know something about the outcome of an experiment. We must take into account the effect of the given outcome on the probabilities of other events. In the previous case, we say: P(F, given E) = 1/3

Calculating conditional probability
Conditional probability questions are questions of the form, “What is the probability of this event, given that this other event has occurred?” There are special formulas that can be used to calculate P(A, given B). P(A, given B) = n(A and B)/n(B) P(A, given B) = P(A and B)/P(B) The first of the two formulas is useful when we are presented with data in the form of raw numbers. The second formula is useful when we are presented with data in percent form.

Example In a box we have a bunch of puppies: 4 brown bulldogs
2 gray bulldogs 5 brown poodles 3 gray poodles If one puppy is selected, what is the probability that the puppy is... 1. ...brown? 2. ...a poodle? 3. ...gray or a bulldog? 4. ...brown and a bulldog? 5. ...a bulldog, given that it is gray? 6. ...brown, given that it is a poodle?

Exercise #5 At the Annoying Persons Conference there are 22 charming telemarketers, 30 irritating telemarketers, 16 charming life insurance sellers and 21 irritating life insurance sellers. If one person is randomly selected, find the probability that he/she is a life insurance seller given that he/she is charming. A. .42 B. .18 C. .58 D. 1.38 E. .25

Exercise #6 A survey of Gators indicates that 7% are charming, 4% are modest, and 3% are both charming and modest. Find the probability that a Gator is modest, given that he/she is charming. A. .75 B. .03 C. .43 D. .25

Exercise #7 A survey of Gators indicates that 7% are charming, 4% are modest, and 3% are both charming and modest. Find the probability that a Gator is charming, given that he/she is not modest. A .0417 B C D

Exercise #8 The table below summarizes preferences of diners at the Hearty Guys Breakfast. (This is the same table that we used in in earlier question.) If one of these hearty guys is randomly selected, find the probability that he prefers waffles, given that he is a lumberjack. Pa W F totals L C totals A B C D. 0.14

Exercise #9 www.math.fsu.edu/~wooland/prob1/prob18.html
The table below shows the distribution according to annual household income of households in the town of Boonies. income percent \$0 - \$9, % \$10,000 - \$19, % \$20,000 - \$29, % \$30,000 - \$39, % \$40,000 - \$49, % \$50,000 - \$59, % \$60,000 or more 16% If one household is randomly selected, what is the probability that it has an income less than \$40,000, given that its income is greater than \$19,999? A B C D

Exercise #10 At the Annoying Persons Conference there are 22 charming telemarketers, 30 irritating telemarketers, 16 charming life insurance sellers and 21 irritating life insurance sellers. If one person is randomly selected, find the probability that he/she is a life insurance seller given that he/she is charming. A. .42 B. .18 C. .58 D. 1.38 E. .25

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