Stats CH. 5 EXAM 2
5.1 Consider the following terms and match to description: event A; sample space; and complement of event A (i) the event: that event A DOES NOT occur (ii) collection of one or more outcomes (iii) set of all outcomes
(i) complement A the event does not occur 1-P(a) (ii) event A (iii) sample space set of all outcomes
SALE NO SALE . TOTAL Agress 270 . 310 . 580 PAgress416 . 164 580 Column Total 686 . 474 1160 A. find P(A & S) B. find P(Pa & S)
A. Step #1 - aggressive sales 270 Step #2 - over the entire total (270/1160) B. 416/1160
5.3 3 step process to use counting techniques on calculator
1) MATH 2) -> right keypad arrow to PROB 3) select either: #1 nPr permutation #2 nCr combination #3 ! factorial
5.2 Review According to survey, of the women interviewed, 24% had asked for a raise, and of those women who asked for a raise, 45% received the raise. Find following probabilities 1) P(women asked for raise) 2) P(women received raise, given she asked for one) 3) P(women asked for raise AND received one)
1) P(women asked for raise) = 24% 2) P(women received raise, given she asked for one) 45% 3) P(women asked for raise AND received one) .24 x .45 = 10.8
5.2 Khan Academy Example On random day the probability that Rahul will eat a bagel for breakfast P(A)=0.6 and that he will eat pizaa fro lunch P(B)=0.5 and the conditional probability bagel given pizza P(A / B)=0.7 What is P(B / A) pizza for lunch given he ate a bagel for breakfast? Hint: Because P(A) is different from P(A / B) means they are dependent so use general rule of multiplication
1. P(A & B) = P(B) x P(A / B) or P(A) x P(B / A) use given info to solve P(A & B) because also mean P(B / A) = P(A & B) divided by P(A) 2. Solve P(A & B) = .5 x .7 = .35 3. P(B / A) = .35/.6 = .58
5.2 Review There are five multiple choice questions on an exam, each with 4 possible answers. Determine the number of possible answer sequences for the five questions. Only ONE set contains all five correct answers. What is the probability of getting all 5 answers correct?
4 answers 5 questions 4 x 4 x 4 x 4 x 4 = 1024 only one set all correct 1/1024 chance = .00098
5.3 Karen's Exam Review Problems #13 A sandwich is being ordered, which will include meat, cheese and bread. There are 6 meats, 3 cheeses, and 2 breads. Assuming you must only choose one of each how many different sandwiches are possible?
6x3x2= 36
5.2 Review You are given the information P(A) = .30 and P(B) = .40 a) Do you have enough information to compute P(A & B) b) enough for P( A OR B) c) BUT say that they are independent do you have enough for P( A & B)
A) NO becuase to do P(A & B) need to know whether independent or given P(A / B) or P( B / A) B) No not enough for P(A OR B) independent because its P(A) + P(B) - P(A & B) and cant figure A & B out but can do for utally exclusive P( A OR B ) C) Yes if theyre independent just multiply for (A & B)
5.2 Conditional Prob Define independent events
If one event happening has no affect on the other one happening i.e. coin flips, dice roll
5.3 Counting Techniques define n and r
N = # of objects' options r = amount of groups to be put into/ divided
SALE NO SALE . TOTAL Agress 270 . 310 . 580 PAgress416 . 164 580 Column Total 686 . 474 1160 A. find P(A or S) need calc
OR means: P(a) + p(s) - p(a&s) because they arent independent 580/1160 + 686/1160 - 270/1160 EQUALS 996/1160
5.2 Review Applying for two jobs JOB A is .70 while JOB B is .80 Assume the job offers are independent. a) compute the probability of either JOB A OR JOB B how does it compare to each individual?
P( A OR B) = P(A) + P(B) - P(A & B) .70 + .80 - .56 1.5 - .56 .94 Higher odds to get ATLEAST one job that each individual or getting both
5.2 If two events A and B are mutually exclusive, what is the value of P(A and B)
P(A and B) = 0 because they can't happen at the same time
5.2 Multiplication rule for independent events
P(A and B) = P(A) x P(B)
SALE NO SALE . TOTAL Agress 270 . 310 . 580 PAgress416 . 164 580 Column Total 686 . 474 1160 A. find P(Sale) B. P(S / A) Sale given Aggressive approach
P(sale) = 686/1160 B. Step #1 - number of aggressive sales (270) Step #2 - over the agressive row total (270/580)
5.3 Karen's Exam Review Problems #15 A "pick 3" lottery is entered by picking 3 digits (0-9). You only win if your three numbers match the exact order. If you play once what is the probability you win?
Step #1 - # winning / # in sample space #2 - 1 / 10 10 because 0-9 is 10 options #3 - 1/10P3 #4 - 10P3 calc function #5 - 1/720 = .001
5.3 Review #16 A state lottery is entered by choosing 5 numbers, each number being 1-35. ORDER does not matter how many ways can the winning numbers be chosen?
Step #1 - combination vs permutation; a combin because order doesnt matter Step #2 - n=35 r=5 Step #3 - 35C5 Step #4 - calc function = 324632
SALE NO SALE . TOTAL Agress 270 . 310 . 580 PAgress416 . 164 580 Column Total 686 . 474 1160 A. P(S / PA) Sale given Passive Aggressive approach B. Are these two events independent
Step #1 - number of passive aggressive sales (416) Step #2 - over the row total so pass sales + pass no sales (416/580) B. Independent means event A has an affect on event B, NO, NOT independent because not only passive agressive also agressive has affect on sales
5.2 Review There is money to send two of eight city council members to a conference in HNL. How many different combinations of two council members can be selected from the eight who want to go?
Use calc function 8! = 40320 2!(8-2)! = 1440 40320/1440 = 28
5.2 Review Applying for two jobs JOB A is .70 while JOB B is .80 Assume the job offers are independent. a) compute the probability of getting offers at both jobs. How does it compare to each individual offer
a) BOTH = P(A and B) = P(A) x P(B) .70 x .80 = .56 The probability of getting both job offers is significantly lower then each individual offer.
5.2 Coin flip independent and mutually exclusive because
getting a heads or tails doesn't affect the other outcome and you cant get both a heads AND a tails
5.2 Define mutually exclusive Drawing a king or an ace from a deck of cards because why?
if the events cannot happen together/ simultaneously
5.3 Counting Techniques The multiplication rule is the same as the...and what is the formula
permutations rule, same way to solve, use same formula
5.1 What is Probability Define sample space
set of all possible outcomes in brackets
5.3 Permutations definition
the number of ways to arrange IN ORDER "n" distinct objects taking "r" at a time
5.3 Combinations definition and formula
used for different groupings/combinations of "n" (ORDER doesn't matter) taken at "r" a time
