3.3 - Compound Events

Compound Events

Events as Disctinct Cases

It is ofter helpful for understanding the problem and computation to think of events as mutually exclusive and exhaustive subsets - distinct cases

Conducting Inferential Statistics

Population - ${A, B, C, D, E, F}$
Interest - proportion of vowels in population
Sampling method - SRS with $n = 3$

Identify Cases

Sample contains vowels
Case 1 - contains exactly one vowel
$| A | = \overset{v o w e l s}{2} \cdot \overset{c o n s o n a n t s}{_{4} C_{2}} = 2 \cdot 6 = 12$
Case 2 - contains exactly two vowels
$| B | = 1 \cdot 4 = 4$
Case 3 - contains 3 vowels $\leftarrow$ not needed, there are only 2 vowels in set

Probability of the sample

P (has vowels) = \frac{| A | + | B |}{_{6} C_{3}} = \frac{12 + 4}{20} = 80 %

Sample contains at least 2 consonants
Case 1 - contains 2 consonants
$| M | =_{4} C_{2} \cdot 2 = 6 \cdot 2 = 12$
Case 2 - contains 3 consonants
$| N | =_{4} C_{3} = 4$

Probability

P (\geq 2 consonants) = \frac{12 + 4}{20} = 80 %

Examples

A small population of 100 members. 3 strata - size of 20, 30, and 50. A cluster random sample of size 10 preserves the population proportions.
From strata $n = 20$
$| A | = (\binom{20}{2}) = 190$
From strata $n = 30$
$| B | = (\binom{30}{3}) = 4060$
From strata $n = 50$
$| C | = (\binom{50}{5}) = 2118760$
Total - $| A | \cdot | B | \cdot | C | = 1.63 \cdot 10^{12}$

A coed basketball league requires 2 of 5 players to be women. Team has 6 men and 4 women. How many different sets of plays could they have on the court.
Case 1 - 2 women players
$| A | = (\binom{4}{2}) \cdot (\binom{6}{3}) = 6 \cdot 20 = 120$
Case 2 - 3 women players
$| B | = (\binom{4}{3}) \cdot (\binom{6}{2}) = 4 \cdot 15 = 60$
Case 3 - 4 women players
$| C | = (\binom{4}{4}) \cdot (\binom{6}{1}) = 1 \cdot 6 = 6$
Case 4 - 5 women players $\leftarrow$ not possible
Total - $| A | + | B | + | C | = 120 + 60 + 6 = 186$

A small university of 250 students has 150 STEM majors. A SRS of 10 students. How many samples will over-represent STEM majors
Base - $p = \frac{150}{250} = 0.6$
To overrepresent:
Case 1 - 7 STEM
$| A | = (\binom{150}{7}) \cdot (\binom{100}{3}) = 4.76 \cdot 10^{16}$
Case 2 - 8 STEM
$| B | = (\binom{150}{8}) \cdot (\binom{100}{2}) = 2.60 \cdot 10^{16}$
Case 3 - 9 STEM
$| C | = (\binom{150}{7}) \cdot (\binom{100}{3}) = 8.29 \cdot 10^{15}$
Case 4 - 10 STEM
$| D | = (\binom{150}{7}) \cdot (\binom{100}{3}) = 1.17 \cdot 10^{15}$
Total - $| A | + | B | + | C | + | D | = 8.30 \cdot 10^{16}$

Definitions

An event is understood as compound when described using multiple events

Outcomes in A, B, or in both, is $A \cup B$ (A or B)
Outcomes in A and B simultanously, is $A \cap B$ (A and B)
Outcomes of A that are in B, is $A | B$ (A given B)
Events A and B are disjoint if the $A \cup B = \emptyset$
Cases are example of disjoints

Counting Compound Events

Consider a single card from a standard deck of cards Determine the number of outcomes
Drawing a red card
$| R | = 13 + 13 = 26$
Drawing a face card
$| F | = 4 + 4 + 4 = 12$
Drawing a queen or a heard
$| A | = 4 + 13 - \overset{we counted queen of hearts twice}{1} = 16$
Drawing an even or a multiple of 5
$| B | = 20 + 4 = 24$

Counting with OR

The number of outcomes in the compound event $A \cup B$ is equal to number of outcomes which are in $A$ or $B$ summed, subtracted by $A \cap B$

| A \cup B | = | A | + | B | - | A \cap B |

Consider drawing two cards from a deck of carts
A red card first and a black card second
$26 \cdot 26 = 26^{2}$
Two face cards
$12 \cdot 11 = 132$
Cards of opposite color
Case 1 - black first
$| B | = 26 \cdot 26 = 26^{2}$
Case 2 - red first
$| R | = 26 \cdot 26 = 26^{2}$
Total - $| B | + | R | = 2 \cdot 26^{2}$
Same suit
$52 \cdot 12 = 624$

Counting with And Then

The number of outcomes of event $A$ and then $B$ , is equal to number of outcomes in $A$ times $B | A$

| A \cap B | = | A | \cdot | B | A |

Adding vs Multiplying Events

When we're dealing with events in sequence (Fundamental Counting Principle), we multiply values

When we're dealing with cases, or A or B, we add values

Probability of Compound Events

Unusual Event

When the probability of an event happening is $< 5 %$ , we call it an unusual event

Probability Rules and Definition

Let A and B are events

P (A and B) = P (A) \cdot P (B | A)

P (A or B) = P (A) + P (B) - P (A and B)

Definition

If $P (A | B) = P (A)$ then $A$ and $B$ are independent

Exercises

A bag of coins has 17 quarters, 34 dimes, 13 nickels and 46 pennies. Draw two coins randomly.
Total coins - $110$
$| S | = 110 \cdot 109 = 11990$
What is the probability of:
Two quarters - $\frac{17 \cdot 16}{11990} = \frac{272}{11990} = 0.023 = 2.3 %$
Unusual event, as the probability $< 5 %$
It can be shown as $\frac{17 \cdot 16}{110 \cdot 109} = \overset{P (Q_{first})}{\frac{17}{110}} \cdot \overset{P (Q_{second})}{\frac{16}{109}}$
Quarter and then a nickle
Case 1 (QN) - $\frac{17 \cdot 13}{110 \cdot 109}$
Case 2 (NQ) - $\frac{13 \cdot 17}{110 \cdot 109}$
Total - $\frac{17 \cdot 13 + 13 \cdot 17}{110 \cdot 109} = 0.037 = 3.7 %$
Still an unusual event
Coins valuing at least 3 cents in total
Case 1 (QQ) - $0.023$ (calculated previously)
Case 2 (QD) - $\frac{17 \cdot 34}{110 \cdot 109} = 0.048$
Case 3 (DQ) - $\frac{17 \cdot 34}{110 \cdot 109} = 0.048$
Case 4 (QN) - $0.018$
Case 5 (NQ) - $0.018$
Total = $0.023 + 2 \cdot 0.048 + 2 \cdot 0.018 = 0.156 = 15.6 %$

A BBQ joint offers 7 types of meat, 11 sides, 3 types of toast, and 3 sauces. If a 2 meat combo comes with 2 different meats, 2 different sides, 1 piece of toast, and 2 sauces (can be the same)
How many different 2 meat combos can be made
$\underset{m e a t}{(\binom{7}{2})} \cdot \underset{s i d e s}{(\binom{11}{2})} \cdot \underset{t o a s t}{(\binom{3}{1})} \cdot \underset{s a u c e}{3^{2}} = 21 \cdot 55 \cdot 3 \cdot 6 = 20790$
To choose sauces, we have $3$ cases of the same sauces, and $(\binom{3}{2}) = 3$ of the different sauces, so $6$ in total
What is the probability that a randomly selected 2 meat combo will have two of the same BBQ sauces
We only really care about the sauces, so it's $\frac{3}{6} = 50 %$

Six sided die is biased towards odd numbers, so that each odd number has a probability of $0.25$ . Even numbers have equal probability.
determine the probability that when rolling the die three times in a row, the rolls land on three consecutive numbers in increasing order
Possible outcomes - $123, 234, 345, 456$
Case 1 (OEO) - $2 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{12} = \frac{2}{192} = \frac{1}{96}$
Case 2 (EOE) - $2 \cdot \frac{1}{4} \cdot \frac{1}{12} \cdot \frac{1}{12} = \frac{2}{576} = \frac{1}{288}$
Total - $\frac{1}{96} + \frac{1}{288} = \frac{4}{288} = \frac{1}{72} = 0.0139 = 1.39 %$

Assuming male and female children are equally likely, determine the probability of family with 4 children having at least 1 boy and and least 1 girl
Compliment 1 (only girls) - ${0.5}^{4} = 0.0625$
Compliment 2 (only boys) - ${0.5}^{4} = 0.0625$
Total - $1 - 2 \cdot 0.0625 = 0.875 = 87.5 %$

Two-Way Tables

A common way of presenting data

	Rep.	Dem.	Ind.
Married Men	590	390	20
Unmarried Men	447	533	20
Married Women	500	450	50
Unmarried Women	378	572	50

Total number of people - $4000$
Probability of selecting a married person - $\frac{1000 + 1000}{4000} = 50 %$
Probability of selecting a democrat or a woman - $\frac{1945 + 2000 - 1022}{4000} = 73 %$
Republican given that the person is a man - $\frac{1037}{2000} = 51.85 %$
Democrat given the person is a woman - $\frac{1022}{2000} = 51.1 %$
Unmarried given the person is a Democrat - $\frac{1105}{1945} = 56.8 %$