At the Iron Bank, 36% of customers invest in Stocks, 22% invest in Futures, and 15% invest in both Stocks and Futures. You could have used a contingency table to answer each part of this question. A contingency table is perfectly sufficient to show work (and thus to earn partial credit on any incorrect responses). P(Stocks) = 36% P(Futures) = 22% P(Stocks,Futures) = 15% 1- what proportion of Iron Bank customers invest in neither Stocks nor Futures? Use the negation rule in combination with the addition rule. P (neither stocks nor futures) = 1 −P (stocks or futures) = 1 −[P (stocks + P (futures) −P (stocks, futures)] = 1 −[0.36 + 0.22 −0.15] = 1 −0.43 = 0.57 OR 2- if the Iron Bank customers who invest in Futures, what percentage invest in Stocks? Recognize that this problem is asking for a conditional probability and use the multiplication rule: P (stocks |futures) = P (stocks, futures)/P (futures) = 0.15/0.22 ≈0.68 3- What is the probability that a randomly selected Iron Bank customer invests in Stocks and does not invest in Futures? P (stocks) = P (stocks, no futures) + P (stocks, futures) 0.36 = P (stocks, no futures) + 0.15 0.36 −0.15 = P (stocks, no futures) = 0.21 OR Numerical V. Categorical - Numerical variables are represented by a number with units. Numerical variables have consistent intervals (dollars, years, percentage points). They answer the questions, “how much?” or “how many?” - Categorical variables are represented by a word, category or number without units. you can’t do arithmetic with them (15 * 9 rank + 2 rank= 137 ranks?) Categorical variables have no or inconsistent intervals (genre has no intervals, rank has inconsistent intervals). Lesson 3 Counting PROBABILITY NOTATION- The probability of some event is a number between zero and one: π ≤ π(π¬π¨π¦π ππ―ππ§π) ≤ π An event with probability 1 is certain to occur. An event with probability 0 is certain not to occur. For example: - π (ππππ πππππ βππππ ) = 0.5 - π (ππππβπ‘ ππππππ‘π ππ π‘πππ) = 0.812 - π (ππππ πππ¦ ππ π»πππ) = 0.000000000000001 WHERE DO PROBABILITIES COME FROM? - Careful counting Ex- Draw a card at random from a standard 52-card deck What is π (πππππππ)? What is π (ππππ ππ ππ’πππ)? π (πππππππ) = 13/52 = 0.25 π (ππππ ππ ππ’πππ) = 8/52 ≈ 0.154 THE COUNTING PRINCIPLE: π = # ππ’π‘πππππ ππ πππ‘ππππ π‘/ π‘ππ‘ππ # ππ ππ’π‘πππππ = π/π Key assumption: All outcomes are equally likely Careful Counting Example 2: Rolling two fair dice Let π be the sum of the two numbers. Clearly, π is random: it varies from one roll to the next. What is π(π ≥ 9)? π (π ≥ 9) = 10/36 ≈ 0.278 WHERE DO PROBABILITIES COME FROM? - Data Ex- π(πππ€ππππ ππππ¦ ππ π ππππ) = 100/206 ≈ 0.485 Ex- P (A college student will be in a car accident this year) ≈ 0.009 WHERE DO PROBABILITIES COME FROM? - Subjective judgment Ex- P (ππππππ π€πππ ππ ππππ) = ? Ex- π (π΄ππππ π π‘πππ π’π πππ₯π‘ π¦πππ) = 0.7 ? (buy) π (π΄ππππ π π‘πππ π’π πππ₯π‘ π¦πππ)= 0.4 ? (Sell) WHERE DO PROBABILITIES COME FROM? - Other probabilities - THREE RULES OF PROBABILITY - Negation Rule In general, for any event π΄: π π§π¨π π = π − π(π) - Also known as the “complement rule” Let’s pretend that the probability of rain tomorrow is 0.3 - Then clearly the probability that it will NOT rain tomorrow is 0.7 Say the probability that Serena Williams wins the U.S. Open tournament is 0.15. - Thus, the probability that Serena does NOT win is 0.85. - ADDITION RULE - Consider two events A and B. -Let π(π΄, π΅) be the joint probability that both A and B happen. Then π·(π¨πππ©) = π·(π¨) +π· (π©) −π·(π¨,π©) Ex- ADDITION RULE EXAMPLE - Draw a card at random. What is (π ππππ ππ π΄ππ ππ πππππ)? Key insight: “Ace” and “Spade” are not mutually exclusive outcomes. - Therefore, we cannot simply add π π΄ππ) + π(πππππ like we did with π (King or Queen) because we will double count the Ace of Spades. Here’s what we know... Of the 52 cards: 4 are aces: π (π΄ππ) = 4/52 13 are spades: π (πππππ) = 13/52 1 is both an ace and a spade: π (π΄ππ,πππππ)= 1/52 Total= π π΄ππ ππ πππππ = 4/52 + 13/52 − 1/52 = ππ/ππ ≈ 0.308 CONDITIONAL PROBABILITY π· (π¨/π©) is the “probability of A given B” -examples: π( ππππ π‘βππ πππ‘ππππππ /ππππ’ππ¦ π‘βππ πππππππ) π (ππ ππππ‘π ππ / ππ πβπππ ππ¦ π π‘ππ’πβπππ€π ππ‘ βππππ‘πππ) π (ππππππ‘ππ π‘π πππππππ π πβπππ/ πππππππ πΊππ΄ > 3.6) Instagram:P (follow @KatyPerry /follow @TaylorSwift) Amazon:P (buy organic dog food / bought GPS dog collar) Netflix:P(watch πππ’ππ πΊππππ / watch ππ‘ππππππ πβππππ ) Perhaps the single most important fact to remember about conditional probabilities: π(π΄|π΅)≠π(π΅|π΄) WHERE DO (conditional probabilities) PROBABILITIES COME FROM? EX- Mammograms - P cancer = 15/200 - P die, cancer = 3/200 - P die|cancer = 3/15 In general, we can estimate π(π΄|π΅) as: π· (π¨/π©) = (πππππππππ ππ π¨ πππ π© ππππ πππππππππ / πππππππππ ππ π© πππππππππ) MULTIPLICATION RULE The probability of A given B is equal to the frequency of A and B both happening divided by the frequency of B happen. - The multiplication rule expresses this idea in general terms: π· (π¨/π©) = π·(π¨,π©)/ π·(π©) - We can also use the alternate version below if we want to go in reverse, from a conditional probability to a joint probability. - This says the same thing with the terms re-arranged: π·(π¨,π©) =π·(π¨/π©∗π·(π©) -