1. 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.

2. 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?

3. 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

4. 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
≈0.12
7 12 × × × = ≈0.12

5. 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 )

6. 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.

7. 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 = 𝑃 𝑆𝐶 = = 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 12
2) the freshmen has a car but the sophomore doesn’t have a car?
𝑃 𝐹𝐶 ∩ 𝑆𝐶 =𝑃 𝐹𝐶 𝑃 𝑆𝐶 = − 1 3 = = 1 6
3) neither the freshmen nor the sophomore have a car?
𝑃 𝐹𝐶 ∩ 𝑆𝐶 =𝑃 𝐹𝐶 𝑃 𝑆𝐶 = 1− − 1 3 = = 1 2

8. 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: ≈
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

9. 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.

10. 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 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)=

11. 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 , which happens 5 times independently:
1 6 × 1 6 × 1 6 × 1 6 × 1 6 = =
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 , which happens 5 times independently:
4 6 × 4 6 × 4 6 × 4 6 × 4 6 = = ≈ 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 × =1− =1− ≈ 0.60

12. 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
= 7 99 ≈0.07
5 12 × 4 11 × 7 10 × 6 9 = 5×4×7×6 12×11×10×9 = = 7 99 ≈0.07