# statistic fail

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8/2/2019 Statistic Fail

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http://www.mathsisfun.com/data/probability.html

Probability

How Likely

In the real world events can not be predicted with total certainty. The best we can do is

say how likely they are to happen, using the idea of probability.

Tossing a Coin

When a coin is tossed, there are two possible outcomes:

heads (H) or tails (T)

We say that the probability of the coin landing H is 1/2.

Similarly, the probability of the coin landing T is 1/2.

Throwing Dice

When a singledieis thrown, there are six possible outcomes: 1,

2, 3, 4, 5, 6.

The probability of throwing any one of these numbers is 1/6.

Probability

In general:

Probability of an event happening =Number of ways it can happen

Total number of outcomes

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Example: the chances of rolling a "4" with a die

Number of ways it can happen: 1 (there is only 1 face with a "4" on it)

Total number of outcomes: 6 (there are 6 faces altogether)

So the probability =1

6

Example: there are 5 marbles in a bag: 4 are blue, and 1 is red. What is the

probability that a blue marble will be picked?

Number of ways it can happen: 4 (there are 4 blues)

Total number of outcomes: 5 (there are 5 marbles in total)

So the probability =4

= 0.85

Probability Line

You can show probability on aProbability Line:

The probability is always between 0 and 1

Probability is Just a Guide

Probability does not tell us exactly what will happen, it is just a guide

Example: toss a coin 100 times, how many Heads will come up?

Probability says that heads have a 1/2 chance, so we would expect 50 Heads.

But when you actually try it out you might get 48 heads, or 55 heads ... or

anything really, but in most cases it will be a number near 50.

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Words

Some words have special meaning in Probability:

Experiment: an action where the result is uncertain.

Tossing a coin, throwing dice, seeing what pizza people choose are all examples of

experiments.

Sample Space: all the possible outcomes of an experiment

Example: choosing a card from a deck

There are 52 cards in a deck (not including Jokers)

So the Sample Space is all 52 possible cards: {Ace of Hearts, 2 of Hearts,

etc... }

The Sample Space is made up of Sample Points:

Sample Point: just one of the possible outcomes

Example: Deck of Cards

the 5 of Clubs is a sample point

the King of Hearts is a sample point

"King" is not a sample point. As there are 4 Kings that is 4 different sample

points.

Event: a single result of an experiment

Example Events:

Getting a Tail when tossing a coin is an event

Rolling a "5" is an event.

An event can include one or more possible outcomes:

Choosing a "King" from a deck of cards (any of the 4 Kings) is an event

Rolling an "even number" (2, 4 or 6) is also an event

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The Sample Space is all possible outcomes.

A Sample Point is just one possible outcome.

And an Event can be one or more of thepossible outcomes.

Hey, let's use those words, so you get used to them:

Example: Alex decide to see how many times a "double" wouldcome up when throwing 2 dice.

Each time Alex throws the 2 dice is an Experiment.

It is an Experiment because the result is uncertain.

The Event Alex is looking for is a "double", where both dice have the same

number. It is made up of these 6 Sample Points:

{1,1} {2,2} {3,3} {4,4} {5,5} and {6,6}

The Sample Space is all possible outcomes (36 Sample Points):

{1,1} {1,2} {1,3} {1,4} ... {6,3} {6,4} {6,5} {6,6}

These are Alex's Results:

Experiment Is it a Double?

{3,4} No

{5,1} No

{2,2} Yes

{6,3} No

... ...

After 100 Experiments, Alex had 19 "double" Events ... is that close to what

you would expect?

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http://rchsbowman.wordpress.com/2008/10/28/statistics-notes-probability-examples-using-the-

addition-rule-and-conditional-probability/

Statistics NotesProbability Examples using the Addition Rule and

Conditional Probability

Filed under: Statistics Tags: Statistics bowman @ 2:40 am

Addition Rule:

If events A and B are mutually exclusive (disjoint), then P(A or B) = P(A) + P(B). Otherwise, P(A or B) =

P(A) + P(B) P(A and B)

Example 1: mutually exclusive

In a group of 101 students 30 are freshmen and 41 are sophomores. Find the probability that a

student picked from this group at random is either a freshman or sophomore.

Note that P(freshman) = 30/101 and P(sophomore) = 41/101. Thus P(freshman or sophomore) =

30/101 + 41/101 = 71/101

Example 2: not mutually exclusive

In a group of 101 students 40 are juniors, 50 are female, and 22 are female juniors. Find the

probability that a student picked from this group at random is either a junior or female.

Note that P(junior) = 40/101 and P(female) = 50/101, and P(junior and female) = 22/101. Thus

P(junior or female) = 40/101 + 50/101 22/101 = 68/101

Not sure why? When we add 40 juniors to 50 females and get a total of 90, we have overcounted. The

22 female juniors were counted twice; 90 minus 22 makes 68 students who are juniors or female.

.

Conditional Probability

Recall that the probability of an event occurring given that another event has already occurred is called

a conditional probability.

http://rchsbowman.wordpress.com/2008/10/28/statistics-notes-probability-examples-using-the-addition-rule-and-conditional-probability/http://rchsbowman.wordpress.com/2008/10/28/statistics-notes-probability-examples-using-the-addition-rule-and-conditional-probability/http://rchsbowman.wordpress.com/2008/10/28/statistics-notes-probability-examples-using-the-addition-rule-and-conditional-probability/http://rchsbowman.wordpress.com/category/statistics/http://rchsbowman.wordpress.com/tag/statistics/http://rchsbowman.wordpress.com/tag/statistics/http://rchsbowman.wordpress.com/category/statistics/http://rchsbowman.wordpress.com/2008/10/28/statistics-notes-probability-examples-using-the-addition-rule-and-conditional-probability/http://rchsbowman.wordpress.com/2008/10/28/statistics-notes-probability-examples-using-the-addition-rule-and-conditional-probability/8/2/2019 Statistic Fail

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In a card game, suppose a player needs to draw two cards of the same suit in order to win. Of the 52

cards, there are 13 cards in each suit. Suppose first the player draws a heart. Now the player wishes to

draw a second heart. Since one heart has already been chosen, there are now 12 hearts remaining in a

deck of 51 cards. So the conditional probability P(Draw second heart|First card a heart) = 12/51.

.

Suppose an individual applying to a college determines that he has an 80% chance of being accepted,

and he knows that dormitory housing will only be provided for 60% of all of the accepted students.

The chance of the student being accepted and receiving dormitory housing is defined by

P(Accepted and Dormitory Housing) = P(Accepted) P(Dormitory Housing|Accepted) = (0.80)

(0.60) = 0.48

.

To calculate the probability of the intersection of more than two events, the conditional probabilities of

all of the preceding events must be considered. In the case of three events, A, B, and C, the probability

of the intersection P(A and B and C) = P(A) P(B|A) P(C|A and B).

Consider the college applicant who has determined that he has 0.80 probability of acceptance and

that only 60% of the accepted students will receive dormitory housing. Of the accepted students who

receive dormitory housing, 80% will have at least one room mate. The probability of being accepted

and receiving dormitory housing and having no room mates is calculated by:

P(Accepted and Dormitory Housing and No Roommates) = P(Accepted) P(Dormitory

Housing|Accepted) P(No Room mates|Dormitory Housing and Accepted) = (0.80) (0.60) (0.20)

= 0.096. The student has about a 10% chance of receiving a single room at the college. Not

happening.

.

The probability that event B occurs, given that event A has already occurred is

P(B|A) = P(A and B) / P(A)

This formula comes from the general multiplication principle and a little bit of algebra. I showed you

on an earlier post.

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Since we are given that event A has occurred, we have a reduced sample space. Instead of the entire

sample space S, we now have a sample space of A since we know A has occurred. So the old rule

about being the number in the event divided by the number in the sample space still applies. It is the

number in A and B (must be in A since A has occurred) divided by the number in A. If you then divided

numerator and denominator of the right hand side by the number in the sample space S, then you

have the probabilit