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Probability II. “Baseball is 90% mental. The other half is physical.” Yogi Berra. Denoted by P(Event). Probability. This method for calculating probabilities is only appropriate when the outcomes of the sample space are equally likely.
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Probability II “Baseball is 90% mental. The other half is physical.” Yogi Berra
Denoted by P(Event) Probability This method for calculating probabilities is only appropriate when the outcomes of the sample space are equally likely.
As the number of repetitions of a chance experiment increase, the difference between the relative frequency of occurrence for an event and the true probability approaches zero. Law of Large Numbers
Rule 1.Legitimate Values For any event E, 0 < P(E) < 1 A probability is a number between 0 and 1. The probability of rain must be 110%! Basic Rules of Probability
Rule 2. Sample space If S is the sample space, P(S) = 1 “Something Has to Happen Rule” The probability of the set of all possible outcomes must be 1. I’m 100% sure you are going to have a boy… or a girl.
Rule 3. Complement • For any event E, • P(E) + P(not E) = 1 • P(E) = 1 – P(not E) If the probability that you get to class on time is .8, then the probability that you do not get to class on time is .2.
Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occurs A randomly selected student is female - What is the probability she plays soccer for SHS? A randomly selected student is female - What is the probability she plays football for SHS? Independent Independent Dependent
Rule 4.Multiplication If two events A & B are independent, General rule: If the probability of rolling a 5 on a fair dice is 1/6, what is the probability of rolling a 5 three times in a row? P( three 5’s in a row) = (1/6) x (1/6) x (1/6) = 1/216 or .004629
What does this mean? Independent? Yes
Given a deck of cards and a die, one card is drawn and the dice is rolled. What is the probability that an ace is drawn and an even is rolled? P(ace and even) = P(ace) * P(even) =
Independent? Yes No
Given a deck of cards, two cards are drawn without replacement. What is the probability that they are both hearts? P(heart and heart) =
Ex 6) Suppose I will pick two cards from a standard deck without replacement. What is the probability that I select two spades? Are the cards independent? NO P(A & B) = P(A) · P(B|A) Read “probability of B given that A occurs” P(Spade & Spade) = 1/4 · 12/51 = 1/17 The probability of getting a spade given that a spade has already been drawn.
Rule 5.Addition • If two events E & F are disjoint, • P(E or F) = P(E) + P(F) If the probability that a randomly selected student is a junior (A) is .2 and the probability that the student is a senior (B) is .5, what is the probability that the student is either a junior or a senior? P(A υ B) = P(A) + P(B), if A and B are disjoint. P(A υ B) = .2 + .5 = .7
Two events that have no common outcomes are said to be disjoint or mutually exclusive. A and B are disjoint events
Rule 5.Addition • If two events E & F are disjoint, P(E or F) = P(E) + P(F) • (General) If two events E & F are not disjoint, • P(E or F) = P(E) + P(F) – P(E & F) Probability of owning a MP3 player: .50 Probability of owning a computer: .90 So the probability of owning a MP3 player or a computer is 1.40? Not disjoint events!
What does this mean? Mutually exclusive? Yes
Given a deck of cards, one card is drawn. What is the probability that it is a 3 or a 4? P(3 or 4) = P(3) + P(4) =
Mutually exclusive? Yes No
Given a deck of cards, one card is drawn. What is the probability that it is an ace or a red card? P(ace or red) = P(ace) + P(red) – P(ace and red) = - = +
Mutually exclusive? Yes No Independent? Yes
Ex 5) If P(A) = 0.45, P(B) = 0.35, and A & B are independent, find P(A or B). Is A & B disjoint? NO, independent events cannot be disjoint If A & B are disjoint, are they independent? Disjoint events do not happen at the same time. So, if A occurs, can B occur? Disjoint events are dependent! P(A or B) = P(A) + P(B) – P(A & B) If independent, multiply How can you find the probability of A & B? P(A or B) = .45 + .35 - .45(.35) = 0.6425
In a class, there are 12 boys made up of 8 Seniors and 4 Juniors . There are also 8 girls, made up of 3 Seniors and 5 Juniors. Find the probability of choosing a boy or a Senior. Note that choosing a boy and choosing a Senior are not disjoint (they can occur simultaneously). P(boy or a senior) = P(Boy) + P(Senior) – P(Senior boy) =
Rule 6.At least one The probability that at least one outcome happens is 1 minus the probability that no outcomes happen. P(at least 1) = 1 – P(none)
Dr. Pepper For a sales promotion the manufacturer places winning symbols under the caps of 10% of all Dr. Pepper bottles. You buy a six-pack. What is the probability that you win something? P(at least one winning symbol) = 1 – P(no winning symbols) 1 - .96 = .4686
Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that it is not US made? P(not US made) = 1 – P(US made) = 1 - .4 = .6
Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that it is made in Japan or Germany? P(Japanese or German) = P(Japanese) + P(German) = .3 + .1 = .4
Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that you see two in a row from Japan? P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = .3 x .3 = .09
Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that none of three cars came from Germany? P(no Germany in three) = P(not G) x P(not G) x P(not G) = .9 x .9 x .9 = .729
Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that at least one of three cars is US made? P(at least one US in three) = 1 – P(no US in three) = 1 – (.6)(.6)(.6) = .784
Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that the first Japanese car is the fourth one you choose? P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (.7)3 (.3) = .1029
Watch out for: • probabilities that don’t add up to 1 • don’t add probabilities of events if they are not disjoint • don’t multiply probabilities of events if they are not independent • don’t confuse disjoint and independent
A probability that takes into account a given condition Rule 7: Conditional Probability
In a class, there are 12 boys made up of 8 Seniors and 4 Juniors . There are also 8 girls, made up of 3 Seniors and 5 Juniors. Find the probability of choosing a boy given that he is a Senior. P(Boy|Senior) = 8/11 P(Senior|Boy) = 8/12 = 2/3
Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 12) What is the probability that the driver is a student?
Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 13) What is the probability that the driver drives a European car?
Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 14) What is the probability that the driver drives an American or Asian car? Disjoint?
Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 15) What is the probability that the driver is staff or drives an Asian car? Disjoint?
Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 16) What is the probability that the driver is staff and drives an Asian car?
Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 17) If the driver is a student, what is the probability that they drive an American car? Condition
Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 18) What is the probability that the driver is a student if the driver drives a European car? Condition
Definition of Independent Events • Two events E and F are independent if and only if • P(F | E) = P(F) or P(E | F) = P(E)
EXAMPLE Illustrating Independent Events The probability a randomly selected murder victim is male is 0.7515. The probability a randomly selected murder victim is male given that they are less than 18 years old is 0.6751. Since P(male) = 0.7515 and P(male | < 18 years old) = 0.6751, the events “male” and “less than 18 years old” are not independent. In fact, knowing the victim is less than 18 years old decreases the probability that the victim is male.
I draw one card and look at it. I tell you it is red. What is the probability it is a heart? P( heart | red) =
Are “red card” and “spade” mutually exclusive? Are they independent? A red card can’t be a spade so they ARE mutually exclusive
Are “red card” and “ace” mutually exclusive? Are they independent? 2 aces are red cards so they are NOT mutually exclusive
Are “face card” and “king” mutually exclusive? Are they independent? Kings are Face cards so they are NOT mutually exclusive
“Slump? I ain’t in no slump. I just ain’t hittin.” Yogi Berra