STATS: unit 5
if independent
- P(A) x P(B) = P(A n B) - if P(A I B) = P(A) then they are independent - n = and - I = given
intuitive multiplication rule
- "and" means multiply - account for dependent events
conditional probability intuition
- "given" means divide by the given
simple addition rule
- "or" means add - the union P(AuB) brings together A, B, and AnB. so you can add the probabilities together - think: unions -> bring together
the probability that a randomly chosen person is a left-handed baseball player is 0.014 and the probability that a randomly chosen baseball player is left-handed is 0.260 What is the probability that a randomly chosen person from this country plays baseball?
- .054 𝑃(baseball player)= 𝑃(baseball player and left-handed) DIVIDED BY 𝑃(left-handed∣baseball player)
probability
- A number that describes how likely it is that an event will occur - long run relative frequency - always between 0 and 1 - short term: unpredictable - long term: predictable
disjoint
- Mutually Exclusive
how to write dependence
- No, mothers surveyed in 1976 are more likely to have 4+ children than in the mothers not surveyed in 1976 - P (C I S) doesn't equal P(C I S^c) - no they are not independent because they are not equal
if mutually exclusive
- P (A n B) = 0 - the probability of A and B is zero
general addition rule
- P (A or B) = P(A) + P(B) - P(A and B) - AuB (union) - AuB (intersection) - if events A and B are mutually exclusive, they can't occur together - P (A and B) = 0, so P(A or B) = P(A) + P(B)
always true
- P (A u B) = P(A) + P(B) - P(A n B) - P (A I B) = [P (A n B)]/ P(B) - u is or - n is and - I = given
general multiplication rule
- P(A and B) = P(A) x P(B|A) - conditional probability - if A and B are independent - P(AIB) = P(B) - P(A and B) = P(A) + P(B) - P(at least 1) = 1 - P (none)
formal addition rule
- P(A or B) = P(A) + P(B) - P(A and B) - or means add - - P(A and B) subtracts double counting
P(AuB)
- P(A) + P(B) - P(AnB) - probability of A or B occurring - A, B, or both - u = or
conditional probability formula
- P(A|B) = P(A and B) / P(B) - event/given - so the denominator is a new number of possibilities vs the original total number of possibilities
probability of making the first two and missing the last one
- P(makes 2 then misses 1) - P(makes 1st) and P(makes 2nd) and P(makes 3rd) - 0.906 x 0.906 x 0.094 = 0.077 = 7.7%
P(AnB)
- Probability of A and B occurring - upside-down u - BOTH - intersection of A and B
Joint Probability
- The probability that a randomly chosen member of the population is a left-handed baseball player is the probability that the person is both a baseball player and left-handed. This is known as a joint probability. 𝑃(baseball player and left-handed)=0.014P(baseball player and left-handed)=0.014 - Joint probabilitities are computed using the multiplication rule, which is also known as the general multiplication rule. When 𝐴A and 𝐵B represent two events, the probability that 𝐴A and 𝐵B both occur is 𝑃(𝐴 and 𝐵)=𝑃(𝐴)⋅𝑃(𝐵∣𝐴)P(A and B)=P(A)⋅P(B∣A) - The last term, 𝑃(𝐵∣𝐴)P(B∣A), represents the conditional probability of event 𝐵B occurring given that event 𝐴A has occurred. Therefore, 𝑃(baseball player and left-handed)=𝑃(baseball player)⋅𝑃(left-handed∣baseball player)P(baseball player and left-handed)=P(baseball player)⋅P(left-handed∣baseball player) - You are calculating the probability that a randomly chosen member of the population is a baseball player, so dividing both sides of the equation by the conditional probability gives the following - 𝑃(baseball player)=𝑃(baseball player and left-handed)𝑃(left-handed∣baseball player)P(baseball player)=P(baseball player and left-handed)P(left-handed∣baseball player) - The probability that a randomly chosen member of the population is left-handed and a baseball player is 0.014.0.014. 𝑃(left-handed and baseball player)=0.014P(left-handed and baseball player)=0.014 - The conditional probability that a randomly chosen member of the population is left-handed given that he or she is a baseball player is 0.260. 𝑃(left-handed∣baseball player)=0.260P(left-handed∣baseball player)=0.260 - Therefore, 𝑃(baseball player)=0.0140.260=5.4P(baseball player)=0.0140.260=5.4 - It does not matter which event is labeled 𝐴A and which is labeled 𝐵B. - 𝑃(𝐴 and 𝐵)=𝑃(𝐴)⋅𝑃(𝐵∣𝐴)=𝑃(𝐵)⋅𝑃(𝐴∣𝐵)P(A and B)=P(A)⋅P(B∣A)=P(B)⋅P(A∣B) - When 𝐴A and 𝐵B are independent, 𝑃(𝐵∣𝐴)=𝑃(𝐵)P(B∣A)=P(B) and 𝑃(𝐴∣𝐵)=𝑃(𝐴)P(A∣B)=P(A); the formula simplifies to the multiplication rule for independent events - 𝑃(𝐴 and 𝐵)=𝑃(𝐴)⋅𝑃(𝐵)P(A and B)=P(A)⋅P(B) - Selecting a baseball player and selecting someone who is left-handed are not independent, because the conditional probability of selecting a baseball player given that the person is left-handed is not equal to the probability of selecting a baseball player
when doing FRQs
- WRITE JUSTIFICATION WITH THE EQUATIONS AND SUCH LIKE WHY IT IS OR ISN'T MUTUAL EXCLUSIVE OR IS OR ISN'T INDEPENDENT
tree diagram
- a diagram used to show the total number of possible outcomes - at the end it would be the last 2 multiplied so if 1st being ace is 2/5 chance and 2nd being ace is 1/4 then do 2/5 times 1/4 = .1 and so on - the probability of winning would be adding both the ways to win - it goes A to B given A to P (A and B) and so on - when you label the parts, make sure to show what the labels mean, like if you do YT instead of has a youtube account make a key - label the ages section as "age of the users" - show the calculations you do for at least 3 of the branches
interpret probability
- after many many times ( in context of question ), we will get ( context of question ) about ____% of the time
n
- and
complement
- c - none - A^c = 1-P (A) - P (X greater than or equal to 1) = 1-P (x<1) - probability of an event NOT happening - P (A^c) = 1-P(A) - mutually exclusive: events that cannot occur together - general addition rule - P (A or B) = P(A) + P(B) - P(1 child) = .22
two way table and venn diagram
- can be used to display the sample space for a chance process - put the overlap in the middle and the other stuff on left and right - 4 on the outside
find the probability of neither A nor B
- do 1 minus the probability of A or B (AUB)
probability of two of the same
- do WITHOUT replacement unless stated otherwise - so (16/24) x (15/23)
how to figure out the values based on the totals
- do something like x/200 = 480/600 - then figure it out from there by looking at the ratios
writing probability
- ex. The probability of selecting a student at Fisher University from a sample of 200 students who had taken math is .429 - make sure to show the P( ), like P(C I M)
intuitive addition rule
- find the sum of # of ways A can occur & B can occur but don't count when they occur together; divide by total # of outcomes in sample space - "or" means add - avoid double counting
formal addition rule
- for all events: P(AnB) = P(A) x P(B I A) - for all independent events: P(AnB) = P(A) x P(B)
interpret conditional probability
- given the mother was surveyed in 1976, there is a 40% probability that she has 4+ children
not independent
- has association - P(BW|M)≠ P(BW)
testing for independence
- if P(A I B) = P(A) then they are independent - I = given
simulation
- imitation of chance behavior based on a model that accurately reflects the situation - ex. dice, flip coin, RNG
independent
- knowing outcome of one event doesn't help you predict outcome of second event - P(A|B) = P(A) - P(BW|M)=P(BW) - no association
probability model
- list showing all possible outcomes and their probabilities - must add to 1 - each probability is between 0 and 1
do not count twice
- make venn diagram and for OR count the left right and middle, making sure the left and right subtracted the middle
P(O ^c)
- means do NOT do what O means - c means eliminate
tree diagrams
- models multiple dependent or successive events (events that depend on each other that happen right after each other) - each branch represents a different level ex. 1. you have 0.65 probability of getting higher than average GPA and ACT scores 2. if you have a higher than average GPA/ACT, you have a 0.83 chance of being admitted 3. if you have a below average GPA/ACT score, you have a 0.39 chance of being admitted - start with higher than avg GPA/ACT and below avg GPA/ACT then branch off to the admitted and not admitted for each section - event 2 (admission) depends on event 1 (GPA/ACT) - let H = event of getting higher than avg GPA/ACT - let A = event of being admitted to your dream school - P(HnA) = .65 x .83 = .54 (and means multiply) - P(A) = .68 (path 1 + path 2) (p1 = HnA, p2 = H^cnA (.35 x .39 = .14) (.54 +.14) - P(H I A) = .79 (path 1/given = .54/.58) (given = divide by given)
probability of having heads and tails
- must find getting heads first then tails and getting tails first then heads - or P(Ace and King) = adding
Independent vs. Mutually Exclusive
- mutually exclusive: when events occur that have no intersection (ie. they cannot both occur) - P(A) = .80 - P(A I B) = 0 (A given that B happened) - mutually exclusive events are not independent. Knowing that one event occurs greatly affects the probability of the other event (lowers it to 0)
Sampling without Replacement
- once an element has been included in the sample, it is removed from the population and cannot be selected a second time - ex. P (blue, red, blue) = (12/20) x (8/19) x (11/18) = 0.154 - "and" means multiply
u
- or
P(A)
- probability of A occurring
P(B)
- probability of B occurring
P(A^c)
- probability of event A not occurring
P((AuB)^c)
- probability of neither A nor B occurring
3 way venn diagram
- put the middle first and go out from there, subtracting from the total of each category
Successive Independent Events
- successive: events are successive if one happens after the other (ie. a repeated process) - independent: events are independent if knowing the outcome of one event doesn't change the probability of another event - ex. P (5 heads in a row) = (.5)^5 = 0.03 - since the probabilities don't change, you don't need a tree diagram
P(T I H^c)
- tall given that they are not a higher earner - 22/43 + 51.2% - vs. P(T) = 123/192 = 64.1%
Law of Large Numbers
- the larger the number of individuals that are randomly drawn from a population, the more representative the resulting group will be of the entire population - if we ever do something many many times the proportion of desired outcomes will approach its probability
odds of not happening
- the odds of B are .14 so the odds of NOT be is .86 out of 100
Conditional Probability
- the probability that one event happens given that another event is already known to have happened - P (A I B) = P(A and B) / P(B) - "A given B"
conditional probability
- the probability that one event happens given that another event is already known to have happened - condition: a "given" in a problem - P (A I B) = probability of A occurring given that B has already occurred
sample space
- the set of all possible outcomes
If they are independent
- then P(A) = P(A I B) = P(A I B^c) - when knowing one event has occurred or has not occurred does not affect the probability of the 2nd event - if P(A I B) = P(A I B^c) then A and B are independent
independence
- two events (A&B) are independent if knowing the outcome of one event does not affect the probability that the other event will occur - P (A I B) = P(A) - A given B - are selecting a tall man and selecting a man who is a low-earner independent? -- ex. response: When we knew the person was low-earner (when H^c was "given"), the probability of selecting a tall person shrunk. So, the events of selecting a low earner and someone who is tall are not independent. In other words, their earnings gave us information about their height. So, earnings and height are not independent.
describe how to carry out a simulation
- use a random number generator to get 50 integers between 1 and 9 - let 1 be bankrupt and let 2-9 be other outcomes - count how many 1s and record - do this many times to estimate the probability of getting 10+ bankrupts
Theoretical Probability
- what we expect to happen - ex. if the dice are rolled many many times, player 1 will be expected to win about 44.4% of the time.
The "hot hand" theory
- when a player starts to make many basketball shots in a row, they have a "hot hand" - their probability of making basketball shots is higher than normal - your basketball shots are not independent - your current basketball shot probability depends on your previous basketball shots - you can assume it does not exist and find the chances of making 8 in a row by doing their probability of making it to the power of 8 and multiplying this by how many basketball shots they have made in their career - 7 to get the expected number of 8-streaks by chance alone - (# of possible 8-streaks) = (total basketball shots) -7 - you can then compare this to the actual number of 8-streaks to see if it is possible by chance
"At least one" scenarios
- when asked to find the probability of "at least one" occurrence of a successive independent event, take the complement of the event that none occur - P(at least one) = 1 - P(none) - ex. P(at least one) = 1 - P(makes none) - 1 - (.094) ^4 = 0.99992
mutually exclusive
- when events that have no intersection (ie. they cannot both occur) - P (AuB) = P(A) + P(B) - don't have to worry about double counting - both events can't occur at same time - a card can't be both an ace and a king - a cis man cannot be pregnant - a king and hearts aren't mutually exclusive because there can be a king of hearts
5 Intuitive Probability Rules for a 5
1. Probabilities are between 0-1 (inclusive) 2. Complement rule: P(A^c) = 1- P(A) 3. "Or" means add, beware of double-counts 4. "Given" means divide by the given 5. "And" means multiply, adjust for dependence
check that it is a valid probability model
1.) all probabilities are between 0 and 1 (inclusive) 2.) all probabilities add to 1
simulation process
1.) describe how you will simulate one trial (one repetition) 2.) perform many trials (repetitions) 3.) use the results to answer the question