Multiplication Principle for Counting Outfits (Top and Bottom) Total Number of Outfits = 3 x 2 = 6 Say I have 3 shirts and 2 pairs of bottoms. How many possible outfits do I have? Each line going from a shirt to a pair of bottoms represents one outfit.

Multiplication Principle for Counting Outfits (Top, Bottom, Shoes) Total Number of Outfits = 3 x 2 x 4 = 24 Now, let’s add in some shoes as well. Say, 4 pairs. Now, how many outfits do we have? Now, an outfit is represented by a pair of lines that link the pieces of clothing together. Think of it as a path. How many paths are there?

Multiplication Principle for Counting License Plates 0-9 for each digit 10 possible numbers for each of 6 spots 10 x 10 x 10 x 10 x 10 x 10 = 106 = 1,000,000 A-Z for first spot (26 choices) 0-9 for second to seventh spots (10 choices each) 26 x 10 x 10 x 10 x 10 x 10 x 10 = 26 x 106 = 26,000,000 Now, let’s consider the possible license plate numbers in a state (assuming no custom plates). A-Z for first, second, and third spots (26 choices each) 0-9 for fourth to seventh spots (10 choices each) 26 x 26 x 26 x 10 x 10 x 10 x 10 = 263 x 104 = 175,760,000

Independent Events Jar of cookies: 7 chocolate chips and 5 oatmeal raisins (12 cookies total) You randomly grab 4 cookies, one at a time. But each time, you just record the type of cookie and put it back without eating it. What is the probability that you grabbed 4 chocolate chips in a row? Total number of ways you can get 4 chocolate chips in a row 7×7×7×7=2401 Total number of ways you can get any sequence of 4 cookies 12×12×12×12=20736 Probability of 4 chocolate chips in a row 2401 20736 ≈0.12 7 12 × 7 12 × 7 12 × 7 12 = 7 12 4 ≈0.12

Independent Events If events A and B are independent, then 𝑃 𝐴∩𝐵 =𝑃 𝐴 𝑃(𝐵) The probability of one event does not effect the probability of the second event. In other words, knowing something about one event tells you nothing about the second event. Not the same as mutually exclusive events! 𝑃 𝐴∩𝐵 =0 If A happens, B cannot happen (and vice versa) In other words, mutually exclusive events are dependent For example, on a single die roll, you cannot have both 1 and 2 happen at the same time. If 1 happens on a given roll, 2 (or any other value) cannot happen on that same roll. Also independent: 𝐴 and 𝐵 , 𝐴 and 𝐵, 𝐴 and 𝐵 𝑃 𝐴∩ 𝐵 =𝑃 𝐴 𝑃( 𝐵 ) 𝑃 𝐴 ∩𝐵 =𝑃 𝐴 𝑃(𝐵) 𝑃 𝐴 ∩ 𝐵 =𝑃 𝐴 𝑃( 𝐵 )

Independent Events If events A and B are independent, then 𝑃 𝐴∩𝐵 =𝑃 𝐴 𝑃(𝐵) 𝑃 𝐴∩ 𝐵 =𝑃 𝐴 𝑃( 𝐵 ) 𝑃 𝐴 ∩𝐵 =𝑃 𝐴 𝑃(𝐵) 𝑃 𝐴 ∩ 𝐵 =𝑃 𝐴 𝑃( 𝐵 ) Independence means that the occurrence of one event does not affect the probability of subsequent events. Or that knowing something about one event tells you nothing about another event.

Independent Events Suppose there are 20 freshmen and 30 sophomores in a classroom. Out of the group, 5 freshmen and 10 sophomores have cars, respectively: 𝑃 𝐹𝐶 = 5 20 = 1 4 𝑃 𝑆𝐶 = 10 30 = 1 3 If one freshmen and one sophomore were randomly chosen, what is the probability that… 1) the freshmen and sophomore both have cars? 𝑃 𝐹𝐶 ∩𝑆𝐶 =𝑃 𝐹𝐶 𝑃 𝑆𝐶 = 1 4 1 3 = 1 12 2) the freshmen has a car but the sophomore doesn’t have a car? 𝑃 𝐹𝐶 ∩ 𝑆𝐶 =𝑃 𝐹𝐶 𝑃 𝑆𝐶 = 1 4 1− 1 3 = 1 4 2 3 = 1 6 3) neither the freshmen nor the sophomore have a car? 𝑃 𝐹𝐶 ∩ 𝑆𝐶 =𝑃 𝐹𝐶 𝑃 𝑆𝐶 = 1− 1 4 1− 1 3 = 3 4 2 3 = 1 2

Independent Events Extremely important concept, both statistically and practically Common misconception: Assuming events are independent when they are actually not Sally Clark Case (1998) First child died at 11 weeks of age Second child died at 8 weeks of age Sally claimed they both died of Sudden Infant Death Syndrome (SIDS) Expert witness Roy Meadow (pediatrician) claimed the probability of both infants dying of SIDS is astronomically small, 1 out of 73 million Incidence of SIDS: 1 out of 8550 Probability of two SIDS: 1 8550 2 ≈ 1 73000000 Critical misconception: Infant deaths by SIDS are independent Reality: Infant deaths by SIDS most certainly not independent for same mother Due to genetic predisposition, probability of multiple SIDS among infants of same mother much higher Sally sent to prison in 1999 solely based on Dr. Meadow’s incorrect calculation Sally released 4 years later in 2003 after the Royal Statistical Society pointed out the mistake Sally developed serious psychiatric problems and died another 4 years later in 2007 from alcohol poisoning

Independent Events 𝑃 𝐴∩𝐵∩𝐶∩⋯ =𝑃 𝐴 𝑃 𝐵 𝑃(𝐶)⋯ Not just limited to 2 events If events A, B, C … are all independent of each other, 𝑃 𝐴∩𝐵∩𝐶∩⋯ =𝑃 𝐴 𝑃 𝐵 𝑃(𝐶)⋯ This is called mutual independence.

Independent Events Common misconception: Assuming events are dependent when they are actually independent Gambler’s Fallacy: Belief that you are due for a win after a long losing streak Simple Example: Win if a coin flips heads Probability of this particular sequence is 1 2 5 = 1 32 However, the probability of heads at each coin toss remains the same at 1 2 Coin does not have a memory, nor does the universe keep a record of your previous results 1 2 𝑃(𝐻)=

Independent Events Experiment: Roll a fair die 5 times What is the probability of rolling 5 ones in a row? A one is rolled each time 1 6 , which happens 5 times independently: 1 6 × 1 6 × 1 6 × 1 6 × 1 6 = 1 6 5 = 1 7776 What is the probability of rolling no 5 and no 6 on any roll? Either a 1, 2, 3, or 4 is rolled each time 4 6 , which happens 5 times independently: 4 6 × 4 6 × 4 6 × 4 6 × 4 6 = 4 6 5 = 1024 7776 ≈ 0.13 What is the probability of rolling a 3 on at least one of the rolls? = P[(3 on just 1 roll) ∪ (3 on exactly 2 rolls) ∪ (3 on exactly 3 rolls) ∪ (3 on exactly 4 rolls) ∪ (3 on all 5 rolls)] = 1 – P(3 on no rolls) = 1- P(1, 2, 4, 5, or 6 on all five rolls)= 1− 5 6 × 5 6 × 5 6 × 5 6 × 5 6 =1− 5 6 5 =1− 3125 7776 ≈ 0.60

Dependent Events Jar of cookies: 7 chocolate chips and 5 oatmeal raisins (12 cookies total) You randomly grab 4 cookies, one at a time. What is the probability that you grabbed 2 oatmeal raisins then 2 chocolate chips? (order matters) Number of ways you can grab 2 oatmeal raisins and then 2 chocolate chips, in that order 5×4 × 7×6 =20×42=840 Number of ways you can grab any 4 sequence of cookies 12×11×10×9=11880 Probability of 2 oatmeal raisins then 2 chocolate chips 840 11880 = 7 99 ≈0.07 5 12 × 4 11 × 7 10 × 6 9 = 5×4×7×6 12×11×10×9 = 840 11880 = 7 99 ≈0.07