Probability Mechanics

Laws of probability: Addition

• The question of Or

• p(A or B) = p(A) + p(B)– Probability of getting a grape or lemon skittle in a bag

of 60 pieces where there are 15 strawberry, 13 grape, 12 orange, 8 lemon, 12 lime?

– p(G) = 13/60 p(L) = 8/60– 13/60 + 8/60 = 21/60 = .35 or a 35% chance we’ll get

one of those two flavors when we open the bag and pick one out

Laws of probability: Multiplication

• The question of And• If A & B are independent• p(A and B) = p(A)p(B)

• p(A and B and C) = p(A)p(B)p(C)– Probability of getting a grape and a lemon (after

putting the grape back) after two draws from the bag– p(Grape)*p(Lemon) = 13/60*8/60 = ~.0288

Conditional Probabilities and Joint Events

• Conditional probability – One where you are looking for the probability of some event with

some sort of information in hand– e.g. the odds of having a boy given that you had a girl already.1

• Joint probability– Probability of the co-occurrence of events– E.g. Would be the probability that you have a boy and a girl for

children i.e. a combination of events• In this case the conditional would be higher b/c if we

knew there was already a girl that means they’re of child-rearing age, able to have kids, possibly interested in having more etc.

Conditional probabilities

• If events are not independent then:

• p(X|Y) = probability that X happens given that Y happens– The probability of X “conditional on” Y

• p(A and B) = p(A)*p(B|A)

Conditional probability Example

• Example: once we grab one skittle we aren’t going to put it back (sampling without replacement) so:

– p(A and B and C) = p(A)*p(B|A)*p(C|A,B)

– Probability of getting grape and lemon = p(G)*p(L|G)

– (13/60)(8/59) = .0293

– Note: p(G)*p(L|G) = p(L)*p(G|L)

Joint Probability Example

• What is the probability of obtaining a Female who is Independent from this sample?

• In this case we’re looking for the joint condition of someone who is Female and Independent out of all possible outcomes:2/17 = 11.8

Lib Mod Cons Totals

M 4 4 2 10

F 12 9 3 24

Totals 16 13 5 34

Lib Mod Cons Totals

M 11.8 11.8 5.9 29.4

F 35.3 26.5 8.8 70.6

Totals 47.1 38.2 14.7 100%

Example: Political Party and Gender

Conditional probabilities

• Let’s do a conditional probability: If I have a male, what is the probability of him being in the ‘other’ category? Formally:

• p(A|B) = p(A,B)/p(B)= [p(A)*p(B|A)]/p(B)

 • p(O|M) = p(O,M)/p(M)= [p(O)* p(M|O)]/p(M) =

(.412*.714)/.588= ~.5

Conditional probabilities

• Easier way by looking at table- there are 10 males and of those 10 (i.e. given that we are dealing with males) how many are “Other”?

• p(O|M) = 5/10 or 50%.

Permutations

• Permutation is a sequence or ordering of events.

• Basic Question: if I have N objects, how many different orderings of them are there?

• Factorial: N!• Formula: N(N-1)(N-2)…(1)• Example: 5(5-1)(5-2)(5-3)(5-4)

– 5*4*3*2*1 = 120

Permutations

• General formula for finding the number of permutations of size k taken from n objects

PkN

N!

(N k)!

Example: 10 songs on the iPod and we only have time to hear 6. What is the number 6 song orderings that we can make?:

10! = 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

(10 - 6)! 4! 4 x 3 x 2 x 1

= 10 x 9 x 8 x 7 x 6 x 5 = 151200

PkN

N!

(N k)!

Example

Combinations

• General formula for finding the number of combinations of k objects you can choose from a set of n objects

CkN

N!

k!(N k)!

CkN

N!

k!(N k)!

e.g. How many sets of 6 song groupings (where their order is unimportant) can we make from 10 total (without repeating the same combinations)?

10! = 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 6!(10 - 6)! 6!(4!) (6 x 5 x 4 x 3 x 2 x 1)( 4 x 3 x 2 x 1)

= 10 x 9 x 8 x 7 = 5040 = 210(4 x 3 x 2 x 1) 24

Example

Binomial in action: Sign test

• Decision: bigger (+) or smaller (-) = binomial

• Makes no assumption about the distribution of means– Might be useful when data is highly skewed

Sign test example

• Memory span for digits and letters 13 subjects

• Digits:7.2, 6.2, 5.9, 8.1, 6.7, 7.0, 7.6, 8.0, 5.8, 6.5, 7.0, 6.9, 6.2

• Letters:6.8, 6.1, 5.9, 6.9, 6.0, 6.5, 7.2, 7.4, 6.0, 6.6, 7.4, 6.8, 6.5

• Sign+, +, =, +, +, +, +, +, -, -, -, +, -8 out of 12 [ignore the =]

Add probabilities

• What is the probability of at least 8 with more digit recall assuming .5 success rate typically, i.e. digit span = letter span?

• p(8) + p(9) + p (10) + p(11) + p(12)

Enter values

• Input the values into the equation

• Check your answer with:BINOMDIST(success,total N, prob., FALSE)

p(X )N!

X!(N X )!pXq(N X )

Equation with values

• End up with:

p(8) 12!

8!(12 8)!0.580.5(12 8)

Decision

• p(8) = .12

• Without even calculating p(9) & p(10) & p(11) & p(12), we can say that the probability of getting at least 8 people with more digit recall than letter recall >.05

• Thus by conventional standards we would fail to reject the null hypothesis– “There is no statistically significant difference found in

the recall span for digits versus letters”