Single Pick Probability AND vs. OR Sequential Probability With Replacement Conditional Disjoint vs. Non Disjoint Unit 4 – Probability – Part 1

Single Pick

Unit 4 – Probability – Part 1 Single Pick AND Ex: What is the probability of selecting a face card that’s also a heart from a deck of 52 cards?

Unit 4 – Probability – Part 1 Single Pick AND Ex: What is the probability of selecting a face card that’s also a heart from a deck of 52 cards? P(A) = Probability(Face Card) P(B) = Probability(Heart) P(A∩ B) = Probability(Face Card and Heart) P(A∩ B) = 3/52 Concept Check: What is the probability of rolling a prime number that’s also less than 10 on a 20 sided dice?

Unit 4 – Probability – Part 1 Single Pick AND OR Ex: What is the probability of selecting a face card that’s also a heart from a deck of 52 cards? P(A∩ B) = 3/52 Ex. What is the probability of selecting a 7 or a heart from a deck of 52 cards? P(A ∪ B) = 16/52 Concept Check: What is the probability of rolling either an even number or a prime number on a 14 sided dice?

Unit 4 – Probability – Part 1 Single Pick Sequential AND With Replacement OR Ex: What is the probability of selecting a face card that’s also a heart from a deck of 52 cards? P(A∩ B) = 3/52 Ex. What is the probability of selecting a 7 or a heart from a deck of 52 cards? P(A ∪ B) = 16/52 Ex. What is the probability of rolling a 6 on a single dice, twice in a row? (1/6)(1/6) = 1/36 Concept Check: What is the probability of beating your friend twice at chess if you have a known 80% win rate?

Unit 4 – Probability – Part 1 Single Pick Sequential AND Without Replacement With Replacement OR Ex: What is the probability of selecting a face card that’s also a heart from a deck of 52 cards? P(A∩ B) = 3/52 Ex. What is the probability of selecting a 7 or a heart from a deck of 52 cards? P(A ∪ B) = 16/52 Ex. What is the probability of rolling a 6 on a single dice, twice in a row? (1/6)(1/6) = 1/36 Ex. What is the probability of getting a pair of Aces when drawing the top 2 cards of a regular deck of cards? (4/52)(3/51) = 12/2652 Concept Check: What is the probability of randomly selecting 4 spades from a standard deck of cards?

Unit 4 – Probability – Part 1 Disjoint vs. Non Disjoint Disjoint Example: What is the probability of selecting an Ace and a King in a single pick from a normal deck of cards? Note: Both disjoint and non disjoint automatically imply you are in a single pick environment.

Unit 4 – Probability – Part 1 Disjoint vs. Non Disjoint Disjoint Example: What is the probability of selecting an Ace and a King in a single pick from a normal deck of cards? The events “Ace” and “King” are disjoint, meaning they do not overlap. In this case, we had an “and” scenario which means it is impossible. Because the two events are non overlapping, both cannot occur at the same time. P(A∩ B) = 0/52 = Note: Both disjoint and non disjoint automatically imply you are in a single pick environment.

Unit 4 – Probability – Part 1 Disjoint vs. Non Disjoint Disjoint Example: What is the probability of selecting an Ace or a King in a single pick from a normal deck of cards? Note: Both disjoint and non disjoint automatically imply you are in a single pick environment.

Unit 4 – Probability – Part 1 Disjoint vs. Non Disjoint Disjoint Example: What is the probability of selecting an Ace or a King in a single pick from a normal deck of cards? The events “Ace” and “King” are disjoint, meaning they do not overlap. In this case, we have an “or” scenario which means either is acceptable. Because the two events are non overlapping, we simply add. P(A ∪ B) = P(A) + P(B) = 8/52 Note: Both disjoint and non disjoint automatically imply you are in a single pick environment.

Unit 4 – Probability – Part 1 Disjoint vs. Non Disjoint Non Disjoint Example: What is the probability of selecting a Face card and a Heart in a single pick from a normal deck? Note: Both disjoint and non disjoint automatically imply you are in a single pick environment.

Unit 4 – Probability – Part 1 Disjoint vs. Non Disjoint Non Disjoint Example: What is the probability of selecting a Face card and a Heart in a single pick from a normal deck? Note: Both disjoint and non disjoint automatically imply you are in a single pick environment. The events “Face Card” and “Heart” are non disjoint, meaning they overlap. In this case, we had an “and” scenario which means both conditions must be satisfied. P(A∩ B) = 3/52

Unit 4 – Probability – Part 1 Disjoint vs. Non Disjoint Non Disjoint Example: What is the probability of selecting a Face card or a Heart in a single pick from a normal deck? Note: Both disjoint and non disjoint automatically imply you are in a single pick environment.

Unit 4 – Probability – Part 1 Disjoint vs. Non Disjoint Non Disjoint Example: What is the probability of selecting a Face card or a Heart in a single pick from a normal deck? Note: Both disjoint and non disjoint automatically imply you are in a single pick environment. The events “Face Card” and “Heart” are non disjoint, meaning they overlap. In this case, we had an “or” scenario which means either condition can be true. P(A ∪ B) = 22/52

Unit 4 – Probability – Part 1 Disjoint vs. Non Disjoint Non Disjoint Example: What is the probability of selecting a Face card or a Heart in a single pick from a normal deck? Note: Both disjoint and non disjoint automatically imply you are in a single pick environment. P(A ∪ B) = 22/52 P(A ∪ B) = P(A) + P(B) – P(A∩ B) P(A ∪ B) = 12/ /52 – 3/52 P(A ∪ B) = 22/52

Unit 4 – Probability – Part 1 Disjoint vs. Non Disjoint Concept Check part 1: Say if the events are disjoint or non disjoint - Aces and Hearts - Even numbers and primes - Spades and 7s - Juniors and Seniors - Teenagers and people with drivers licenses

Discrete Probability Continuous Probability Binomial Distributions Geometric Distributions Unit 4 – Probability – Part 2

Discrete Unit 4 – Probability – Part 2 Discrete probability distributions involve integer units which cannot be divided – examples in real life include people and objects. For example, the following distribution shows the probability that an individual that is randomly selected has 0, 1, 2, 3, 4, or 5 children. Children are discrete individual units which are indivisible so we have a blocky shape to our curve.

Discrete Unit 4 – Probability – Part 2 Here are additional examples of discrete distributions. Notice that they can have many shapes:

Discrete Unit 4 – Probability – Part 2 Calculating probabilities on discrete probability curves is generally pretty straightforward. Let X be defined as the roll, ex. 3 Find P(X < 3) = 1 and 2 are successes = 2/6 Find P(X≤3) Find P(X=2) Find P(X>1)

Discrete Unit 4 – Probability – Part 2 Calculating probabilities on discrete probability curves is generally pretty straightforward. Let X be defined as the roll, ex. 3 Find P(X < 3) = 1 and 2 are successes = 2/6 Find P(X≤3) Find P(X=2) Find P(X>1)

Discrete Unit 4 – Probability – Part 2 The distribution below shows the number of heads recorded out of 4 consecutive flips and the probability of each. Let X be defined as the number of heads. Find P(X < 1) = 0 is successes = 1/16 Find P(X≤3) Find P(X=2) Find P(X>1)

Discrete Unit 4 – Probability – Part 2 The distribution below shows the number of heads recorded out of 4 consecutive flips and the probability of each. Let X be defined as the number of heads. Find P(X < 1) = 0 is successes = 1/16 Find P(X≤3) Find P(X=2) Find P(X>1)

Continuous Unit 4 – Probability – Part 2 Continuous probability distributions involve non integer units – examples in real life include distance, weight, and time. For example, the following distribution shows the distribution of mile times for a sample of 100 middle school students.

Continuous Unit 4 – Probability – Part 2 Without knowing the function (or calculus) to find the area of the interval, we would have no way to find the P(A). In Stats we get around this by hoping we’re working with a Normal Distribution. Non Normal = Normal =

Binomial Unit 4 – Probability – Part 2 Binomial setting questions are really obvious. They almost always look something like this: “Find the probability that Stephen Curry makes exactly 7 of his free throw attempt out of 9, given he has a 91% free throw make percentage.” Notice: Events result in either a success or failure n = number of shots attempted (given) P = Probability of each shot is the same (given) x = number of makes (given)

Binomial Unit 4 – Probability – Part 2 “Find the probability that Stephen Curry makes exactly 7 of his free throw attempt out of 9, given he has a 91% free throw make percentage.” Solution: binompdf(n,P,x) = binompdf(9,.91,7) = %

Binomial Unit 4 – Probability – Part 2 “Find the probability that a basketball player makes exactly 3 of his free throw attempt out of 7, given he has a 73% free throw make percentage.” “Find the probability that a basketball player makes 8 or fewer shots out of 10, given they have a 43% field goal percentage.” “Find the probability that a basketball player makes more than 10 attempts of 16, given they have a 33% success rate.”

Geometric Unit 4 – Probability – Part 2 Geometric setting questions are really obvious. They almost always look something like this: “Find the probability that Stephen Curry’s FIRST miss occurs on his 5 th attempt, given he has a 91% free throw make percentage.” Notice: Events result in either a success or failure P = Probability of each shot is the same (given) x = shot on which the first failure will occur (given) Although these are typically not tested on the AP exam, they are easy.

Geometric Unit 4 – Probability – Part 2 “Find the probability that Stephen Curry misses his free throw attempt on his 5 th attempt, given he has a 91% free throw make percentage.”