Colloquial: |
Decimal |
Expressed |
Expressed as a |
Expressed as |
Colloquial: |

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 |

Can't Tell |
0.5 |
1/2 |
50% |
1|1 |
Equiprobably |

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 |

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

**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 and Q are
- 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: p(P or Q) = p(P) + p(Q) - p(P&Q)**or**- 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 = I draw a face card; Q = I draw a spade
- 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

- P and Q are
**Conditional Probability**forp(If*if ... then:**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)}

**Math Goodies: Probability Basics | Certain & Impossible Events | Complements | &ing Independent Events | Conditional Probability****Addition & Multiplication Theorem Problems with Answers:**http://student.ccbcmd.edu/elmo/math141s/practice/and_or.htm- More Addition & Multiplication Theorem Problems: http://student.ccbcmd.edu/elmo/math141s/practice/and_or2.htm

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}