Statistics//Chapter 4.4, 4.5, 4.6
Using Multiplication Rule
A biologist experiments with a sample of to vascular plants (V) and four non-vascular plants (N). She wants to randomly select two of the plants for further experimentation. Find the probability that the fist selected plant is non-vascular (N) and the second Plant is also non-vascular (N). a) with replacement (independent) b) without replacement (dependent) Answer: sample space: V V N N N N a) P(A) = probability that the first one is non-vascular = #non-vascular/n = 4/6 = 0.667 Since this is with replacement, the first plant is put back, this means that the probability of choosing N is still 4/6 or 0.667 P(A and B) = P(A)*P(B) (4/6)*(4/6)=(16/36)= 0.444 The probability of choosing 2N out of 6 plants in 0.444 b) P(A) = probability that the first one is non-vascular = #non-vascular/n = 4/6 = 0.667 Since this is without replacement, the first plant is not put back, this means that the probability of choosing N is dependent on A (there will be one less N in the new sample space). V V N N N The probability that the second selection will be N is 3/5 P(A and B) = P(A)*P(B) (4/6)*(3/5)=12/30 = 2/5 or 0.400
Using Combinations Rule
Example: how many 5-card poker hands are possible? Answer: n=52 cards r=5 possible cards at once nCr = ((n!)/((n-r)!r!) 52C5 = ((52!)/((52-5)!5!) = 2,598,960
Fundamental Counting Rule
Fir a sequence of two events in which the first even can occur "m" different ways and the second can occur "n" different ways, the events together can occur a total of "m*n" different ways. ie. These are all the different orders or sequences of events that you can make "AB" is different from "BA" *Note when answering a question, always check whether it is a probability question or a counting possibility question
5% Guideline for Cumbersome Calulations
If a sample size is no more than 5% of the size of the population, treat it as being independent even if the selections are made without replacement. Example: Quality control randomly selects 3 different cars from a population of 200 including 5 that are defective. a) Does this involve independent events? b)can you use the 5% guideline for cumbersome events? c)calculate the probability that all 3 are not defective a) No; the selections are made without replacement b) Yes, since n = 3 represents 3/200 = 0.015 which is less than 0.05 of the population c) continue as if selections are independent. P(A and B) = P(A)*P(B) There are 195 good cars in the population of 200. The probability of selecting 3 good cars is: (195/200)*(195/200)*(195/200) = 0.927
Using Fundamental Counting Rule
The typical alarm system has a code that consists of 4 digits (0 through 9) that can be repeated and they must be entered in the correct order. Assume that you plan to gain access by trying codes until you get the right one. a) How many codes are possible? b) What if no numbers could be repeated? a) in this case order matters so 4433 and 3344 are considered different codes. 4 digits can be entered, numbers can be repeated, 10 possibilities each time 10*10*10*10 = 10,000 different possible sequences but only on is correct b) once you have one number, it will not get used again (without replacement) there for you would calculate 10*9*8*7 = 5040, this circumstance has fewer possibilities.
Intuitive Multiplication Rule
When finding the probability that event A occurs in one trial and event B occurs in the next trial, multiply the probability of event A by the probability of event B, but be sure that the probability of event B takes into account the previous occurrence of event A.
Factorial rule
A collection of n different items (items will not be replaced) can be arranged in order "n!" different ways ex. A newspaper contains a puzzle that has letters the reader must unscramble to form words. E T R L T E Answer: Different letters, can not be repeated, 6 in total 6! = 6*5*4*3*2*1 = 720 there are not 720 words but there are 720 different arrangements
Complement
A complement is all of the outcomes in which the event does not occur (opposite) The complement of "none" is "one or more" because the complement of "none" is any outcome except "none" *Note remember that the probability on an even + its complement = 1 A + Ac = 1
Conditional Probability
A probability obtained with the additional information that some other event has already occurred. P(B/A) denotes the conditional probability of event B occurring, given that event A has already occurred.
Multiplication Rule vs Addition Rule
Addition Rule: Helps to solve problems when one task is performed and we want to know the probability of two events happening during that task. ex. the probability of selecting someone who is female and a voter in one selection Multiplication Rule: Also deals with two events, but the events occur as a result of more than one task. ex. drawing one card, and then drawing the other (with or without replacement) *Note: The word "or" in P(A or B) suggests addition. Add P(A) and P(B) without counting any values twice The word "and" suggests multiplication. Multiply P(A) and P(B) but be sure that the probability of B takes event A into account
Formal Multiplication Rule
The probability that A happened is first multiplied by the probability that B happened given that A already occurred
Using Conditional Probability Formal Approach
Conditional Probability can be found by dividing the probability of events A and B both occurring by the probability of event A.
Factorial Symbol
Denoted "!" it is the product of decreasing positive whole numbers. ex: 5! = 5*4*3*2*1 = 120 0! = 1
Independent/Dependent Events vs Disjoint Events
Disjoint events cannot occur at the same time. If one happens the other does not. Independent events must be able to occur at the same time. If one happens, It has no influence on the other. *Disjoint events cannot be Independent but events which are not Disjoint may or may not be independent
Dependent Events
Events are dependent if the occurrence of one effects to probability of the occurrence of others. This does not mean that one of the events is a cause of the other.
Independent Events
Events are independent if the occurrence of one does not affect the probability of the occurrence of the other. aka. The probability of event B would not change even if event A occurs ex. Toss two coins. Getting heads with the first coin will not affect the probability of getting heads with the second coin. These events are independent.
Multiplication Rule for Independent Events
If A and B are independent events P(B/A) is really the same as P(B) since A happening doesn't affect the probability of B or vise versa
Finding the Probability of "At Least One"
It is easier to calculate the probability of none than it is to calculate the probability of at least one. Therefore to find the probability of at least one, calculate the probability of non and then subtract that result from 1. P(at lease one) = 1- P(none) Example: find the probability of a couple having at least 1 girl among 3 children assuming boys and girls are equally likely and the sex of a child is independent of the sex of their siblings. A= at least 1 of 3 children Ac = all children are boys P(Ac) = P(1/2^3) = 0.125 P(A) = 1 - 0.125 = 0.875 Probability of getting at least one girl is 0.875
Multiplication Rule
Multiplication of the probability of event A and the probability of event B, where the probability of event B is adjusted to reflect the fact that the first event A has already occurred. (conditional probability) To do this we must understand the difference between dependent and independent events
Permutations when some items are identical
Requirements 1. There are n items available and some items are identical to others 2. we select all of the n items (without replacement) 3. We consider rearrangements of distinct items to be different sequences If the preceding requirements are satisfied and if there are n1 alike, n2 alike,....nk alike, the number of permutations (or sequences) of all items selected without replacement is (n!/n1!*n2!*...nk!) ex. How many permutations (sequences) of 2A's, 1B, 1C? answer: (4!/(2!1!1!)) = 12 (AABC AACB ABAC ABCA ACAB ACBA BAAC BACA BCAA CAAB CABA CBAA)
Replacement
Sampling with replacement means selections are independent events. Sampling without replacement means selections are dependent events. ex. If you are asked to pick two cards from a deck, does the first card effect the probability of drawing a red second card? Answer. If you draw the first card and then put it back into the deck (replace it) before drawing the second, the probability remains the same (with replacement), If you do not replace the card, the probabilities of choosing a red second card change (without replacement)
Counting Techniques
Shortcuts which allow us to compute probabilities without listing all of the outcomes
Factorial rule continued
Sometimes order matters and sometimes there are repeated letters. ie. Selecting r objects from a group of n, without repetition. ex. How many different letter sequences are there if we select 4 letters (no repeats or replacement) from the alphabet? Answer: 26 letters in the alphabet, randomly selecting 4 without replacement. 26*25*24*23 = 358800 or ((26!)/(26-4)!)
Using Conditional Probability Intuitive Approach
The Conditional Probability of B given A, can be found by assuming that event A has occurred, and then calculating the probability that event B will occur. Example: If we randomly select someone who had inherited a mutation from this graph, what is the probability of getting someone who had it prior? Total people with mutations = 1/2+3/4 = 1.25 Total people who had mutations prior = 1/2 = 0.5 P(A|B) = P(AandB)/P(B) P(prior|mutation) = P(Prior and mutation)/P(mutation) = 0.5/1.25 = 0.4
Redundancy
The practice of using duplicate or backup systems to greatly decrease probability of failure and increase reliability (if one fails, the other should work) Example: A homeowner knows there there is a 0.1 probability that a new flashlight does not work when turned on. Find the probability that no flashlight works when she buys three. Answer 0.1^3 = 0.0001
Multiplication Rule for Several Events
The probability of any sequence of independent events is simply the product of their corresponding probabilities. Example: There is a new method of sex selection during pregnancy. In a test 12 couples try to have girls. a) If the test doesn't work, what is the likely hood of all babies being girls? If there are 12 girls in 12 births should we assume the method is effective? Answer: Normal probability that a baby will be a girl = 0.5 This is considered to be with replacement, because the 0.5 probability of one couple having girl doesn't effect the 0.05 probability of another couple having a girl. P(A,B,C...) = P(A)*P(B)*P(C)... (1/2)*(1/2)*(1/2)...=0.5^12 = 0.000244. The probability of this occurring is very small and it is unlikely of ever occurring naturally.
Using Permutations
The probability of winning the Maine lottery is 1/5,245,786. What is the probability of winning if you must select the correct 6 numbers from 1 to 42 in the same order in which they are drawn? the numbers are drawn without replacement (no repeats) Answer: Order matters ie. Permutations P(win) = # of ways to win/total # of possible sequences of 6 numbers. n = 42 r = 6 42P6 = ((6!)/(42-6)!)
Confusion of the Inverse
To incorrectly believe that P(A|B) and P(B|A) are the same. Example: confusing the conditional probability of (A) the patient has cancer given (B) a positive x-ray. With the inverse, (B) a positive x-ray given that (A) the patient has cancer In a test of 300 people out of 150 positive x-rays 130 had cancer and out of 150 negative x-rays 5 had cancer 1) P(A|B)/P(B)= P(has cancer|positive x-ray)/(all positive x-rays) 2) P(B|A)/P(B) P(positive x-ray|has cancer)/(all patients who have cancer) 1) 130/150 = 86.7% chance that a patient has cancer given that they had a positive x-ray 2) 130/135 = 96.3% chance that a patient had a positive x-ray given that they have cancer *Note notice that the numbers are very different!
Permutations Rule
Used when items are all different. Requirements: 1. There are n different items available ( this rule does not apply if some of the items are identical to others) 2. We select r of the n items (without replacement) 3. We consider rearrangements of the same items to be different sequences (ABC is different from CBA) If the preceding requirements are satisfied the number of permutations (sequences) of r items selected from n available items (without replacement) is: nPr = ((n!)/(n-r)!) n items taken r at a time nPr can be found using a calculator instead of using factorials
Combinations Rule
different ordering of the same items are not counted separately, (combinations are different from permutations). Used when order doesn't matter. Requirements: 1. There are n different items available 2. We select r of the n items (without replacement) 3. We consider rearrangements of the same items to be the same. (ABC is the same as CBA) If the preceding requirements are satisfied, the number of cominations of r items selected from n different items is: nCr = ((n!)/((n-r)!r!)
