Question

GRE Tutorial: Counting & Probability

Answer 1

\({\bf{Counting:}}\)
- when counting numbered items between m and n, subtract (m-n) then add 1 to get the total number of items there are

for example, if someone is selling tickets labelled 1-3, subtracting 3-1, then adding 1 back, yields the correct number of tickets (3 tickets). if you ever forget this rule, go back with an easy example like this to remind yourself you need to add 1 back

- if you only want to know how many items are between two endpoints, without including the endpoints themselves, subtract the two endpoints and subtract 1 more

so, for example, if three people numbered 1-3 are standing in a line, 3-1-1 = 1 gives the number of people in between them

this might seem inconsequential, but a major component of probability is counting the desired outcomes/total outcomes, so an error here could cost you an answer

Answer 2

\({\bf{Counting~Principle:}}\) if there are n ways to do one thing and m ways to do a second thing, there are m*n ways to do both tasks

easy example: if you have 3 types of bread and 2 toppings there are 2*3 = 6 sandwiches using 1 bread type and 1 topping type. prettyboringsandwichbuthwhatever

\({\bf{Systematic~List~Making:}}\) probably one of my least favorite types of counting questions - "how many ___ digit numbers have ___[insert some condition here], like "odd digits in the tens place, 5 or 4 in the hundreds place, etc.". you can systematically determine the number of possibilities by considering how many possibilities there are for each digit, then use the counting principle to determine the total # of possibilities

key point to remember: for counting purposes putting 0 in front of a number doesn't count as having another digit --> ex: if it asks for a 3 digit number don't consider 001, 002, etc. as candidiates; start from 100 and go up to 999

Answer 3

easy ex from my prep book: "total # of 3-digit integers that have only odd digits"

first digit possibilities: 1,3,5,7,9
second digit possibilities: 1,3,5,7,9
third digit possibilities: 1,3,5,7,9

so 5 possibilities for each digit --> 5*5*5 = 125

Answer 4

\({\bf{Venn~Diagrams}}\) if you are given something like "x individuals belong to category A, y individuals belong to category B, z individuals belong to both, etc." a good idea would be to make a venn diagram. you will most likely only need to consider 2-3 groups + any indivduals not in any of the groups

*** important note *** if it says "x individuals belong to category A" that includes
- individuals in category A only
- individuals in categories A + any combination of other categories

so pay attention to how the question is worded

Answer 5

\({\bf{Probability:}}\)
general formula: P(outcome) = # of possibilities for that outcome/# total possibilities
*** important note: when the problem involves removing items from a scenario, always consider replacement vs nonreplacement for consecutive draws ***

> P can only be between 0-1 inclusive so if you get something outside that range you messed up your calculation somewhere
> probability of 0 = event will *never* happen, ex: P(being a frequent rule-breaker and being modded) = 0
> probability of 1 = even will *always* happen, ex: P(somebody on QC spam-tags you at least once a week) = 1
> probability of all possible events sum up to 1
> if probability of an event occurring is P(E) then the probability of not occurring is 1 - P(E) assuming its a binary situation where it either will or won't occur
> for independent, consecutive events, multiply the probability of individual events *** must be independent for this to work ***

Answer 6

Anyway, that's the end of my tutorial, I hope it was a helpful resource. Source material is the19th edition Barron's prep book for the new GRE.