Probability Chapter 14 - AP Statistics
Law of Large Nubers If we repeat an event (for instance, flipping a coin) a LARGE nuber of ties, the OBSERVED probability approaches the TRUE ( theoretical ) probability. For exaple: The true (theoretical) probability of flipping a coin and landing on HEADS is 0.5 (50%). If we flip a coin FOUR ties, are we absolutely going to land on HEADS exactly twice? However - if we flip a coin 100 ties, we expect to land on HEADS very close to 50 ties.
Law of Averages (I put law in quotes because it s not a real law! This is a isinterpretation of the law of large #s.) People use the law of averages when they use long-ter probabilities to ake short-ter predictions. (This is what gets people into trouble at casinos.) For exaple: If we flip a coin 9 ties, and all 9 ties we land on HEADS H H H H H H H H H you ight istakenly think that we are due for a TAILS. WE HAVE THE SAME CHANCE OF LANDING ON HEADS OR TAILS EVERY SINGLE TIME WE FLIP THE COIN! BUT THIS IS NOT HOW PROBABILITY WORKS!
Another exaple of is-use of the law of large nubers: A couple has 5 children, and ALL of the are girls. (Let s assue that the probability of having a boy/girl is 50/50.) The couple of obviously due for a boy, right? NO! The probability of having a boy or girl is independent each tie.
Quick Probability Rules: 1) If we select 3 M&Ms at rando WITHOUT REPLACEMENT, what is the probability that all 3 are red? Each tie we pick a red M&M, there are less red M&Ms, and less total M&Ms, to choose fro! 5 12 (1st M&M) (2nd M&M) (3rd M&M) 2) If we select 3 M&Ms at rando WITH REPLACEMENT, what is the probability that all 3 are red? Since we ARE replacing the M&Ms, the probability stays the sae each tie we pick! 5 12 (1st M&M) (2nd M&M) (3rd M&M) 12 total x 4 3 11 x 10 = 0.0455 x 5 5 12 x 12 = 0.0723
Exaple 1: Rolling Pigs!
Feet (aka Trotter ) Back (aka Razerback ) Side-dot (aka Naptie ) Side-NO-dot (aka Bacon ) Snout (aka Snouter ) Snout-ear (aka Leaning Jowler )
P(each side) = 1/6 P(each side) =??
Rolling Pigs! Now let s find our experiental probability of each position: In class, we rolled pigs a large nuber of ties to estiate the probabilities. For these notes, we ll use these: Position feet snout back side-dot side-no-dot snout-ear Probability 0.10 0.06 0.28 0.25 0.30 0.01 Now, assuing these probabilities represent the true ( theoretical ) probability of each position...
Ex. 1: Rolling Pigs! Questions 1. If we roll a pig die once, what is the probability that it does NOT land on its feet? P(NOT feet) = 1 - P(feet) = 1-0.10 = 0.90 2. If we roll a pig die once, what is the probability of rolling a snout OR side-dot? (ADDITION RULE) P(snout OR side-dot) = P(snout) + P(side-dot) = 0.06 + 0.25 = 0.31 3. If we roll a pige die twice (or roll two different pig dice), what is the probability of rolling a snout AND then a side-dot? (MULTIPLICATION RULE) P(snout AND side-dot) = P(snout) x P(side-dot) = 0.06 x 0.25 = 0.015
Ex. 1: Rolling Pigs! Questions 4. If we roll a pig die twice, what is the probability that it lands on back both ties, OR side-dot both ties? P(back both ties OR side-dot both ties) = P(back both ties) + P(side-dot both ties) = (0.28)(0.28) + (0.25)(0.25) = 0.1684 5. What is the probability of rolling back four ties in a row? P(back 4 ties) = (0.28)(0.28)(0.28)(0.28) = (0.28) 4 = 0.0061
Ex. 1: Rolling Pigs! Questions 6. If we roll a pig die 4 ties, what is the probability that it NEVER lands on back? P(never back) = P(not back 4 ties) = (1-0.28) 4 = 0.2687 7. If we roll a pig die 4 ties, what is the probability that it lands on its back AT LEAST ONCE? P(at least one) = 1 - P(never) = 1-0.2687 = 0.7313
Exaple 2: The Colorblind Proble
Red-green colorblindness is a condition that affects about 6% of the ale population. For this proble, assue that the probability that a randoly selected ale is red-green colorblind is exactly 0.06. Find each probability: 1. that a randoly selected ale is NOT colorblind P(not colorblind) = 1 - P(colorblind) = 1-0.06 = 0.94 2. two people (both ale) are chosen at rando; BOTH are colorblind P(both colorblind) = (0.06)(0.06) = 0.0036
Red-green colorblindness is a condition that affects about 6% of the ale population. For this proble, assue that the probability that a randoly selected ale is red-green colorblind is exactly 0.06. Find each probability: 3. two ales are chosen at rando; the first IS colorblind, and the 2nd is NOT P(colorblind and not colorblind) = (0.06)(1-0.06) = 0.0564 4. in an AP Statistics class, there are 12 ales; NONE of the ales are colorblind P(none colorblind) = P(not colorblind 12 ties) = (1-0.06) 12 = 0.4759 5. in the sae class; AT LEAST ONE of the ale students is colorblind P(at least one colorblind) = 1 - P(none colorblind) = 1-0.4759 = 0.5241
Exaple 3: the 8th-grader proble An Algebra II class has 5 freshen, 17 sophoores, and 3 eighth-graders. 1. If you select one student at rando, what is the probability that you will select an 8th-grader? P(8th grader) = 3 = 0.12 12 2. If you select a group of 3 students fro the class, what is the probability that you will pick ONLY freshen? P(3 9th graders) = 5 x 4 x 3 = 0.0043 25 24 23 3. If you select 2 students at rando fro the class, what is the probability that both are freshen OR both are sophoores? P(both 9th OR both 10th) = P(both 9th) + P(both 10th) 5 x 4 + 17 x 16 = 0.4867 25 24 25 24