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UNIT 3 MODULE 5
UNIT 3 MODULE 5 PROBABILITIES INVOLVING NEGATIONS, DISJUNCTIONS, and CONDITIONAL PROBABILITY The following facts follow from our discussions of counting in UNIT 3 MODULE 3 and probability in UNIT 3 MODULE 4. P(E or F) = P(E) + P(F) – P(E and F) P(not E) = 1 – P(E) (Note: these problems can frequently be analyzed with Venn diagrams as well.)
EXAMPLE 3.5. According to a recent article from the 63% of cowboys suffer from saddle sores, New England Journal of Medical Stuff , 52% of cowboys suffer fro 40% suffer from both saddle sores and bowed legs.m bowed legs,
What is the probability that a randomly selected cowboy...
- ...has saddle sores or bowed legs? 2. ...doesn't have saddle sores?
- ...has saddle sores but doesn't have bowed legs? 4. ...has saddle sores and bowed legs?
- ...has neither of these afflictions?
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EXAMPLE 3.5.1 SOLUTIONS
As with counting problems, when a probability problem refers to two overlapping categories, we can organize the information with a Venn diagram. Since the data was given in terms of percentages, we will pretend that the total population is 100. Then, each of the percentages is just a raw number.
saddle sores bowed legs 23 40 12 25
- The number of cowboys who have saddle sores or bowed legs = 23 + 40 + 12 = 75. So, P(saddle sores or bowed legs) = 75/100 =.
- From the diagram, the number of cowboys who don't have saddle sores is 12 + 25 = 37, so P(doesn't have saddle sores) = 37/100 =. We could also get this answer from the complements rule. Since 63% of the cowboys have saddle sores, P(has saddle sores) = .63. Then, P(don't have saddle sores) = 1 - .63 = .37.
- The diagram shows us that there are 23 cowboys out of 100 who have saddle sores but don't have bowed legs, so P(has saddle sores but not bowed legs) = 23/100 =.
- The diagram shows that 40 cowboys out of 100 have both conditions (this information was also stated directly at the beginning of the problem), so P(has saddle sores and bowed legs) = 40/100 =.
- The diagram shows that there are 25 cowboys out of 100 who have neither affliction, so P(has neither affliction) = 25/100 = .25 EXAMPLE 3.5. A survey of 50 Yugo drivers revealed the following: 30 enjoy waiting for tow trucks 35 enjoy hitchhiking 25 enjoy waiting for tow trucks and hitchhiking
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EXAMPLE 3.5.
The table below shows the distribution of guests on the Jerry Slinger show. S: screams obscenities P: punches somebody
P 14^ S^ S ′^ Totals P ′ 52 %^ 8%^ 22% Totals 66 %^26 %^ 78% %^34 %^ 100%
- What is the probability that a guest screams obscenities or punches somebody?
- What is the probability that a guest doesn't scream obscenities and doesn't punch anybody?
MUTUALLY EXCLUSIVE EVENTS Events E and F are mutually exclusive if it is not possible for both E and F to occur simultaneously. This means that P(E and F) = 0. If events E, F are mutually exclusive, then P(E or F) = P(E) + P(F)
UNIT 3 MODULE 5
EXAMPLE 3.5.
In a certain class, 45% of the students are freshmen (F), 30% are sophomores (So) 20% are juniors (J), 5% are seniors (Se) What is the probability that a randomly chosen student is a junior or senior?
EXAMPLE 3.5. A university awards scholarships on the basis of student performance on a certain placement test. The table below indicates the distribution of scores on that test.
0 Score- 200 ScholarshipNone Percentage13% 201300 - None 23% 301400 - None 26% 401500 - Partial 12% 501600 - Partial 11% 601700 - Partial 9% 701800 - Full 6%
If one student is randomly selected, find the probability 1. ...received a partial scholarship. that he/she...
- ...didn't have a score in the 201 3. ...had a score less than 501 - 300 range.
- ...received some kind of scholarship If 800 students are selected, how many would we expect... 5. ...received no scholarship?
- ...had scores higher than 600?
UNIT 3 MODULE 5
CONDITIONAL PROBABILITY Suppose we roll one die. Let A be the event that the result is the number "2." Then we know that P(A) = 1/ However, suppose that before I reveal the result of the die roll, I tell you t has occurred (event E). hat an even number Would you still say that P(A) = 1/6?
If we know that and even number has been rolled, then there are only three possible outcomes ([2, 4, 6}), not six, so given this special
information it would be reasonable to rolled a “2” is 1/3. say that the probability that we
We say that^ This is an example of CONDITIONAL probability. "The probability that the die roll is '2,' given that the die roll is 'even,' is 1/3. “ Notation: P(A, given E) =1/ or P(A|E) = 1/3 The vertical bar separating the names of the events reads “given that.” General fact: For any events E, F
P(E, given F) = P(E and F) P(F) which is the same as
P(E, given F) = n(E and F) n(F) If we are referring to population statistics,
P(E, given F) = portion of population satisfying both conditions E and F portion of population satisfying condition F As a practical matter, conditional probability problems tend to be simpler than these formulas imply. Usually we can solve them simply by thinking in terms of basic probability facts and taking into account the significance of the “given” condition.
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EXAMPLE 3.5.7 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...
- ...brown? 2. ...a poodle?
- ...gray or a bulldog? 4. ...brown and a bulldog?
- ...a bulldog, 6. ...brown, given given that it is a poodle? that it is gray?
SOLUTION TO EXAMPLE 3.5.7 #5 and #6 5. We want to find the probability that a puppy is a bulldog, given that it is gray. This means that we have Among the five gray puppies, two of them are bulldogs, so already selected the puppy, and we know that it is one of the five gray puppies. P(bulldog|gray) = 2/
- We want to find the probability that a puppy is brown, given that it is a poodle. This m that we have already selected the puppy and we know that it is one of the eight poodles. eans Among the eight poodles, five of them are brown, so P(brown|poodle) = 5/
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- What is the probability that a randomly chosen student received a scholarship, given that he/she had a score less than 501?
EXAMPLE 3.5.10 The table below shows the distribution according to cumulative GPA of juniors at Normal University.
0.00^ GPA –^ % of Juniors 2.00^ 1.99 –^ 16% 2.50^ 2.49 –^ 24% 3.00^ 2.99 –^ 28% 3.50^ 3.49 –^ 22% 4.00^ 10%
- Find the probability that a randomly selected junior's GPA is greater than 1.9 is less than 3.50. 9, given that it A. .22 B. .88 C. .82 D..
- Referring to the data above, find the probability that a randomly selected junior's GPA is in the 2.50 - 3.49 range, given that it is greater than 1.99.
UNIT 3 MODULE 5
EXAMPLE 3.5.11 Recently, Gomer took his Yugo to Honest Al's Yugo Repair Shop for a brake job. Later, while driving home, a wheel fell off of the car. When Gomer returned to Honest Al's to complain that the wheel must have fallen off because of the brake job was done incorrectly, Honest Al produced a ream of statistics from NHTSA that showed that for this type of brake job, the probabilty that the wheel will fall off, even if the work is done incorrectly, is only about 0.008. Based on that data, Honest Al graciously offered to cover 1% of the cost of repairing the damage to Gomer's car. What question should Gomer have asked? C: work done correctly F: wheel fell off
EXAMPLE 3.5.11 SOLUTION The statistic that Honest Al cited would be useful if we were tried to predict whether Gomer’s wheel would fall off. Since the wheel has definitely fallen off, that statistic is meaningless. This illustrates a common and fundamental error in the use of statistics: treating events as random and uncertain even though those events ha asked a question that takes into account the fact that the wheel has already fallen off, such asve already occurred. Gomer should have “Given that the wheel has fallen off, what’s the probability that the work was done incorrectly?” If he had looked at Honest Al’s data, here’s what he would have seen:
C: work done correctly F: wheel fell off incorrectly, given that the wheel has fa^ The data shows that the probability that the work was donellen of, is 5/6 or roughly .833. Rather than paying 1% if the cost of replacing Gomer’s vehicle, it would be more reasonable for Honest Al to pay 83% of the cost.
C F 1 F 860 ′^ Totals 861 C Totals′ (^56 6501510 )
UNIT 3 MODULE 5
EXAMPLE 3.5.14 The conventional test for tuberculosis (TB) is only about 50% accurate. Does this mean that if you test that the table below summarizes the results of the TB screening for a sample of 500 people. In positive for TB, then the probability that you actually have TB is about .5? Suppose this table, TB means "A person has tuberculosis," and P means "A person tes TB." ts positive for
Use this information to find the probability that a person who tests positive for TB actually has the disease.
P TB 9 TB 250 ′ Totals 259 P Totals′ (^110 240490 )
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PRACTICE EXERCISES Table A below shows the distribution of undergraduate students at Normal University according to the number of credit hours for which they are registered this semester. Table B below shows the distribution of students at Normal University according to cumulative G.P.A.
of credit hours^ TABLE A % of students cumulative G.P.A.TABLE B % of students
11 or fewer 12 12%31% 0.000.81 - - 0.801.60 14%16% 1314 6%8% 1.612.41 - - 2.40 3 .20 38%17% 1516 21%9% 3.21 - 4.00 15% 18 or more^17 11%2% (^1) student: - 4: Refer to the appropriate table to determine the probability that a randomly selected
1. has a G.P.A. greater than 0.80. A. .16 B. .86 C. .81 D.. 2. is re A. .516gistered for 12 or 13 credit hours. B. .186 C. .37 D.. 3. is registered for more than 16 credit hours. A. .13 B. .22 C. .31 D.. 4. has a G.P.A. that is not in the 0.81 A. .14 B. .15 C. ..71 - 3.20 range. D.. (^5) 1% of the time, win a small teddy bear 4% of the time, win a feather attached to an alligator - 6: Statistics for a certain carnival game reveal that the contestants win a large teddy bear clip 35% of the time, and lose the rest of the time. What is the prob selected player… ability that a randomly 5. A. .4 …wins a teddy bear. B. .05 C. .5 D.. 6. A. .65 …doesn’t lose. B. .35 C. .4 D.. 7. A survey of 50 informed voters revealed the following: 32 believe that Earth has been visited by space aliens 28 believe that Elvis is still alive 20 believe that Earth has been visited by space aliens and Elvis is still alive. According to this data, what is the probability that a randomly selected informed voter believes that Earth has been visited by space aliens or Elvis is still alive?
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11 : corporation. The table below shows the distribution according to salary of the employees of a large annual salary $0 - 9,999 % of employees4% 10,000 30,000 - - 29,99959,999 38%32% 100,000 or more^ 60,000^ -^ 99,999^ 17%9%
11. range or in the $60,000 Find the probability that a randomly chosen employee’s salary is in the $0,000 - $99,999 range. - $9, A. .2032 B. .0068 C. .57 D.. 12. randomly selected, find the probability that it doesn't have fleas or is a bulldog. The table below summarizes the distribution of a number a dogs. If one of these dogs is
fleas beagle 21^ poodle 17^ bulldog 9^ totals 47 no fleas totals 309 1330 145 2774
A. 0.49 B. 0.36 C. 0.41 D. 0.
13. Among a group of 900 bulldogs how many would we expect to agree with the statement “cats A survey of bulldogs reveals that 28% of them agree with the statement “cats are yummy.” are yummy?” A. 572 B. 25 C. 648 D. 252 14. leprechauns and 27 grumpy leprechauns. At the Wee Folks Gathering there are 45 jolly hobbits, 27 grumpy hobbits, 5 jolly If one person is randomly selected, find the probability that he/she is a leprechaun or jolly. A. .77 B. .74 C. .048 D.. 15. A survey of 50 informed voters rev 32 believe that Earth has been visited by space aliensealed the following: 28 believe that Elvis is still alive 20 believe that Earth has been visited by space aliens and Elvis is still alive. According to this data, what is the probability that a ra that Elvis is still alive, given that he/she believes that Earth has been visited by space aliens?ndomly chosen voter doesn’t believe A. .375 B. .6 C. .24 D..
UNIT 3 MODULE 5
16. have?” The responses are summarized in the following table: A group of Harley-Davidson enthusiasts were recently asked “How many tattoos do you # of tattoos 0 % of respondents2% 12 4%3% 4 or more^3 86%5% What is the probability that a randomly chosen Harley tattoo, given that he/she has fewer than 4 tattoos? -Davidson enthusiast has more than one A. .08 B. .04 C. .57 D.. 17. corporation. The table below shows the distribution according to salary of the employees of a large annual salary $0 - 9,999 % of employees4% 10,000 30,000 - - 29,99959,999 38%32% 100,000 or more^ 60,^000 -^ 99,999^ 17%9%
Find the probability that a randomly chosen employee’s salary is more than $9,999, given that it is less than $60,000. A. 1.297 B. .946 C. .543 D..
18. apples. If one fruit is randomly chosen, find the probability that it is a peach, given that it is In a basket, there are 10 ripe peaches, 8 unripe peaches, 12 ripe apples, and 4 unripe unripe. A. .50 B. 44 C. .67 D..
