Probability: Basics&Applications

Probability 101

Colloquial: Epistemic or Subjective	Decimal Expression	Expressed as a Ratio	Expressed as a Percentage	Expressed as Wagering Odds	Colloquial: Objective or Frequeny
Certainly True	1.0	1	100%	not applicable	Always so, or necessarily
Almost Certainly True	0.9	9/10	90%	1\|9	Almost Always
Very Probably True	0.67	2/3	67%	1\|2	Very Frequently
Probably True	0.51	51/100	51%	49\|51	Usually "most commonly"
Can't Tell	0.5	1/2	50%	1\|1 even odds	Equiprobably "it's a toss up"
Probably False	0.49	49/100	49%	51\|49	Usually not
Very Probably False	0.33	1/3	33%	2\|1	Very Seldom
Almost Certainly False	0.1	1/10	10%	9\|1	Almost Never
Certainly False	0.0	0	0%	not applicable	Never, or impossibly

Example: Assuming a Standard Deck of Cards {52 cards in four suits valued A to K}

p(I draw a club) = 1/4
p(I draw a king) = 1/13

Basic Theorems

Negation Theorem for not: p(not-P) = 1 - p(P)

example: p(I don't draw a club) = 3/4 {= 1 - 1/4}
example: p(I don't draw a king) = 12/13 { = 1 - 1/13}

Multiplication theorem for and: p(P&Q) = p(P) * p(Q,P)

p(I draw a king & it's a club) = 1/13 * 1/4 = 1/52
p(I draw a face card & it's a king) = 3/13 * 1/3 = 1/13
independence v dependence
- P and Q are independent iff p(Q,P) = p(Q)

p(I draw a king & it's a club) = 1/13 * 1/4 = 1/52
p(I toss heads the first time & heads again the next) = 1/2 * 1/2 = 1/4
drawing with replacement from a hat containing 2 red & 1 white chip
p(I draw red the first time & I draw red the second) = 2/3 * 2/3 = 4/9

P and Q are dependent iff p(Q,P) <> p(Q)
- p(I draw a face card & it's a king) = 3/13 * 1/3 = 1/13

p(I toss heads the first time and tails the first time) = 1/2 * 0 = 0
drawing without replacement from a hat containing 2 red & 1 white
p(I draw red the first time & I draw red the second) = 2/3 * 1/2 = 1/3

Addition theorem for or: p(P or Q) = p(P) + p(Q) - p(P&Q)
- example: P = I draw a face card; Q = I draw a spade
  p(P or Q) = (12/52 + 13/52 - 3/52 = 22/52)

example: P = the 1st coin toss is heads; Q = the 2nd toss is tails
p(P or Q) = .5 + .5 -.25 = .75
mutually exclusive v. compossible :
- P and Q are mutually exclusive iff p(P&Q) = 0
  - p(the first toss is heads or the first toss is tails) = 1/2 + 1/2 = 1
  - p(I draw a spade or I draw a heart) = 1/4 + 1/4 = 1/2
- P and Q are compossible iff p(P&Q) <> 0
  - p(the first toss is heads or the second is tails) = 1/2 + 1/2 - 1/4 = 3/4
  - p(I draw a spade or I draw a face card) = 12/52 + 13/52 - 3/52 = 22/52

Conditional Probability for if ... then: p(If P then Q) = p(Q,P)

theorem: p(Q,P) = p(P&Q)/p(P)
example: p(If I roll an odd number, it's prime) = (1/2*2/3)/1/2 = 2/3
example: p(If I draw a face card, it's a king) = (3/13*1/3)/3/13 = 1/3
example: p(If it rains the picnic will be cancelled) = (.6*.7)/.6 = .7
example: p(If a child gets an allowance they do chores) = (.4/.6) = .67 {p(A&C)=.4; p(A)=.6)}

Online Problems

Advanced Problem

Example: Problem 5 of More Probabilitiy Problems: According to the Consumer Reports website, Consumer Union did a nationwide survey of owners of manufactured (mobile) homes. The report determined that 6 out of 10 (or 60%) of the people had major problems with their homes. If 5 manufactured home owners are randomly selected, what is the probability that at least one of them had major problems with their homes?

let '&' signify and an '|' or
P = person 1 had major problems
Q = person 2 had major problems
R = person 3 had major problems
S = person 4 had major problems
T = person 5 had major problems

Solution by Multiplication plus Negation:

p(P|Q|R|S|T) = 1 - p(not-(P|Q|R|S|T)) {negation theorem}
p(not-(P|Q|R|S|T)) = p(not-P & not-Q & not-R & not-S & not-T) {equivalence}
p(not-P & not-Q & not-R & not-S & not-T) = (.4*.4*.4*.4*.4) = .01 {applying the multiplication theorem}
p(P|Q|R|S|T) = .99 {applying the negation theorem}