1. Probability Mechanics

2. 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) = 13/60p(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

3. 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

4. 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.

5. 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)

6. 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)

7. 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

8. LibModConsTotals M44210 F129324 Totals1613534 LibMod ConsTotals M11.8 5.929.4 F35.326.58.870.6 Totals47.138.214.7100% Example: Political Party and Gender

9. 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

10. 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%.

11. 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

12. Permutations General formula for finding the number of permutations of size k taken from n objects

13. 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 Example

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

15. 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

16. 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

17. 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 =]

18. 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)

19. Enter values Input the values into the equation Check your answer with: BINOMDIST(success,total N, prob., FALSE)

20. Equation with values End up with:

21. 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”