Wht is the probability of picking something twice
Probability is the likelihood or chance of an event occurring.
|Probability =||the number of ways of achieving success|
|the total number of possible outcomes|
For example, the probability of flipping a coin and it being heads is ½, because there is 1 way of getting a head and the total number of possible outcomes is 2 (a head or tail). We write P(heads) = ½ .
- The probability of something which is certain to happen is 1.
- The probability of something which is impossible to happen is 0.
- The probability of something not happening is 1 minus the probability that it will happen.
This video is a guide to probability. Expressing probability as fractions and percentages based on the ratio of the number ways an outcome can happen and the total number of outcomes is explained. Experimental probability and the importance of basing this on a large trial is also covered.
There are 6 beads in a bag, 3 are red, 2 are yellow and 1 is blue. What is the probability of picking a yellow?
The probability is the number of yellows in the bag divided by the total number of balls, i.e. 2/6 = 1/3.
There is a bag full of coloured balls, red, blue, green and orange. Balls are picked out and replaced. John did this 1000 times and obtained the following results:
- Number of blue balls picked out: 300
- Number of red balls: 200
- Number of green balls: 450
- Number of orange balls: 50
a) What is the probability of picking a green ball?
For every 1000 balls picked out, 450 are green. Therefore P(green) = 450/1000 = 0.45
b) If there are 100 balls in the bag, how many of them are likely to be green?
The experiment suggests that 450 out of 1000 balls are green. Therefore, out of 100 balls, 45 are green (using ratios).
Independent and Dependent Events
Suppose now we consider the probability of 2 events happening. For example, we might throw 2 dice and consider the probability that both are 6’s.
We call two events independent if the outcome of one of the events doesn’t affect the outcome of another. For example, if we throw two dice, the probability of getting a 6 on the second die is the same, no matter what we get with the first one- it’s still 1/6.
On the other hand, suppose we have a bag containing 2 red and 2 blue balls. If we pick 2 balls out of the bag, the probability that the second is blue depends upon what the colour of the first ball picked was. If the first ball was blue, there will be 1 blue and 2 red balls in the bag when we pick the second ball. So the probability of getting a blue is 1/3. However, if the first ball was red, there will be 1 red and 2 blue balls left so the probability the second ball is blue is 2/3. When the probability of one event depends on another, the events are dependent .
When working out what the probability of two things happening is, a probability/ possibility space can be drawn. For example, if you throw two dice, what is the probability that you will get: a) 8, b) 9, c) either 8 or 9?
a) The black blobs indicate the ways of getting 8 (a 2 and a 6, a 3 and a 5, . ). There are 5 different ways. The probability space shows us that when throwing 2 dice, there are 36 different possibilities (36 squares). With 5 of these possibilities, you will get 8. Therefore P(8) = 5/36 .
b) The red blobs indicate the ways of getting 9. There are four ways, therefore P(9) = 4/36 = 1/9.
c) You will get an 8 or 9 in any of the ‘blobbed’ squares. There are 9 altogether, so P(8 or 9) = 9/36 = 1/4 .
Another way of representing 2 or more events is on a probability tree.
There are 3 balls in a bag: red, yellow and blue. One ball is picked out, and not replaced, and then another ball is picked out.
The first ball can be red, yellow or blue. The probability is 1/3 for each of these. If a red ball is picked out, there will be two balls left, a yellow and blue. The probability the second ball will be yellow is 1/2 and the probability the second ball will be blue is 1/2. The same logic can be applied to the cases of when a yellow or blue ball is picked out first.
In this example, the question states that the ball is not replaced. If it was, the probability of picking a red ball (etc.) the second time will be the same as the first (i.e. 1/3).
This video shows examples of using probability trees to work out the overall probability of a series of events are shown. Both independent and conditional probability are covered.
The AND and OR rules (HIGHER TIER)
In the above example, the probability of picking a red first is 1/3 and a yellow second is 1/2. The probability that a red AND then a yellow will be picked is 1/3 × 1/2 = 1/6 (this is shown at the end of the branch). The rule is:
- If two events A and B are independent (this means that one event does not depend on the other), then the probability of both A and B occurring is found by multiplying the probability of A occurring by the probability of B occurring.
The probability of picking a red OR yellow first is 1/3 + 1/3 = 2/3. The rule is:
- If we have two events A and B and it isn’t possible for both events to occur, then the probability of A or B occuring is the probability of A occurring + the probability of B occurring.
On a probability tree, when moving from left to right we multiply and when moving down we add.
What is the probability of getting a yellow and a red in any order?
This is the same as: what is the probability of getting a yellow AND a red OR a red AND a yellow.
P(yellow and red) = 1/3 × 1/2 = 1/6
P(red and yellow) = 1/3 × 1/2 = 1/6
P(yellow and red or red and yellow) = 1/6 + 1/6 = 1/3
Probability GCSE Maths revision, covering probability single & multiple events, the rules of probability and probability trees, including examples and videos.
STATISTICS 230, FALL 2005
(Random House) a strong likelihood or chance of something. The relative possibility an event will occur . the ratio of the number of actual occurrences to the total number of possible occurrences.
Three Types of Probability
(equally probable outcomes) Let S=sample space (set of all possible distinct outcomes). Then the probability of an event =
2. Relative Frequency Definition
The probability of an event in an experiment is the proportion (or fraction) of times the event occurs in a very long (theoretically infinite) series of (independent) repetitions of experiment. (e.g. probability of heads=0.4992)
3. Subjective Probability
The probability of an event is a “best guess” by a person making the statement of the chances that the event will happen. (e.g. 30% chance of rain)
Definitions1 and 2 are consistent with one another if we are careful in constructing our model.
must be repeatable (at least in theory).
Has several possible distinct outcomes
A single repetition of the experiment is a “trial”.
The probability of an event A , denoted P(A) , is a number between 0 and 1.
(a) roll a fair die
(b) toss a fair coin until the first head appears
(c) Select a female student in Stat 230 and measure her height.
Sample Space S:
The set of all possible (indivisible) outcomes of a random experiment.
Discrete Sample space:
S has a finite or a countable number of points in it. (for example (a) and (b) above).
e.g. we write S= < >in this case. These “smallest” possible sets in S , are simple events .
Consists of a sample space S and a probability distribution (to be defined) defined on S
Any subset of the sample space (in the case that S is discrete).
Let S = < >be a discrete sample space. Then probabilities are numbers attached to the simple events for such that the following two conditions hold:
The set of values is called a probability distribution on S .
When the experiment is performed, some simple event in S must occur.
Definition: The probability P(A) of an general event A is the sum of the probabilities for all the simple events that make up A .
An event A made up of two or more simple events is called a compound event.
Properties of a Probability measure:
for any event A
Example: Toss a fair coin until the first head appears. S=
Toss a coin twice. Find the probability of getting 1 head.
Let S = < 0 heads, 1 head, 2 heads >and assume the simple events each have probability 1/3. Then P (1 head) = 1/3.
Roll a red die and a green die. Find the probability the total is 6.
Let (x,y) represent getting x on the red die and y on the green die.
All points in S have probability 1/36.
What if the dice are identical?
S has 21 points. P(A) =3/21?
What is the probability that a 5-card poker hand contains four of a kind? (four twos or four threes, . etc)
2.7 Machine Recognition of Handwritten Digits. An optical scanner determining which of the digits 0,1. 9 an individual has written in a square box. The system may of course be wrong
Describe a sample space S that includes points (x,y) , where x stands for the number actually written, and y stands for the number that the machine identifies.
Suppose that the machine is asked to identify very large numbers of digits, of which 0,1. 9 occur equally often, and suppose that the following probabilities apply to the points in your sample space:
|p(y,y)=0.100 for y=1,2,3,8|
Give a table with probabilities for each point (x,y) in S . What fraction of numbers is correctly identified?
Chapter 3: Probability — Counting Techniques
If sample space S= < >and each simple event has probability 1/n (i.e. is “equally likely”), then a compound event A consisting of r simple events, has probability
Example: Roll 3 fair dice. There are 6 6 6=216 possible outcomes, all equally likely. What is the probability of A = “three dice are different numbers”? Number of triples e.g. (1,2,3) with different numbers is 6 5 4 so
The Addition Rule:
Suppose we can either do job 1 in p ways or job 2 in q ways. Then we can do either job 1 or job 2, (not both) in p+q ways.
The Multiplication Rule:
Suppose we can do job 1 in p ways and an unrelated job 2 in q ways. Then we can do both job 1 and job 2 in p q ways.
n distinct objects are to be “drawn” or ordered from left to right (Order matters objects are drawn without replacement)
The number of ways to arrange n distinct objects in a row is n(n-1)(n-2) (2)(1)=n!
The number of ways to arrange r objects selected from n distinct objects is
n(n-1)(n-2) . (n-r+1)=n
The number of ways to choose r objects from n (a set, i.e. order doesn’t matter) is denoted by . For n and r both non-negative integers with
We form a 4 digit number by randomly selecting and arranging 4 digits from <1, 2, 3,…7>without replacement. Find the probability the number formed is (c) an even number over 3000.
|Number of choices||5||5||4||1 (ends in 2)|
|Number of choices||4||5||4||2 (ends in 4 or 6)|
Beware of Double Counting.
Find the probability a bridge hand (13 cards picked at random from a standard deck) has
A1: P (hand has 1 ace or more) = (is this correct?)
Suppose passengers get on an elevator at the basement floor. There are floors above (numbered . ) where passengers may get off.
Find the probability
no passenger gets off at floor 1
A =”passengers all get off at different floors”
B=”all passengers get off at the same floor”
C=”no-one gets of at the first two floors”
D=”all passengers get off at floors 1 or nine”
G=”exactly one person gets of at each of floors 3 and 4″
What assumption(s) underlies your answer to (a)? Comment briefly on how likely it is that the assumption(s) is valid.
e.g. S= <1111,1112. 9999>has points. Of these are such all passengers get off on different floors.
The Birthday Problem. Suppose there are persons in a room. Ignoring February 29 and assuming that every person is equally likely to have been born on any of the 365 other days in a year, find the probability that no two persons in the room have the same birthday. Find the numerical value of this probability for . A: =0.411,0.891,0.994 http://www.ship.edu/%7Edeensl/mathdl/stats/Birthday.html
Chapter 4: Probability Rules and Conditional Probability
An event A “occurs” if one of the simple events in A occur.
e.g. throw a die, S= <1,2,3,4,5,6>A=” Die is even”= <2,4,6>.
Combinations of events A,B.
Union: A B means A OR B (or possibly both) occurs.
Intersection: A B (usually written as AB in probability) is shaded.
AB means A and B both occur.
Complement: =points in S which are not in A
means ” A does not occur”
USING VENN DIAGRAMS:
EXAMPLE SECTION 4.1: Students finishing 2A Math:
22% have a math average 80%,
24% have a STAT 230 mark 80%,
20% have an overall average 80%,
14% have both a math average and STAT 230 80%,
13% have both an overall average and STAT 230 80%,
10% have all 3 of these averages 80%,
67% have none of these 3 averages 80%.
Find the probability a randomly chosen math student finishing 2A has math and overall averages both 80% and STAT 230 80%.
Events of interest.
Events are mutually exclusive (disjoint) if, for all (the empty event)
(at most one of these can “happen”)
A probability measure is a set function P( defined on subsets A of S (i.e. assigning a real number value P(A) to subsets A such that:
If are mutually exclusive events, ( a finite number or an infinite sequence)
Other rules governing probability models:
For any event A , 0 P(A) 1
If A and B are two events with A B , then P(A) P(B) .
For any two events A B
For any three events A B
Example: Roll a die 3 times. Find the probability of getting at least one 6.
Then = P(A) + P(B)+P(C)= (it this right?)
Example: A box contains 10 defective chips and 15 good ones. Twelve chips are drawn at random (without replacement). Find an expression for the probability that at most 5 defective chips are drawn.
Let =”Exactly x defective chips are drawn”.
Want P(A) where Then P(A)
Intersection of Events and Independence
Events A and B are independent if
Indiana Jones: Raiders of the Lost Ark
What is P(AW)?? P(A)=0.3, P(W)=0.4.
What is P(A\W) and when is it the same as P(A) ?
Definition: Events A and B are independent if
or if P(A \ B)=P(A) where provided denominator is greater than 0.
QUIZ UP TO & INCLUDING SECTION 4.3
Definition. Three or more events A A A are mutually independent if for every choice of
Example: Toss a die twice. Let A =
Why not call events independent if every pair of events is independent?
Two fair coins are tossed. Let A first coin is heads, B = second coin is heads, = we obtain exactly one head. Then A is independent of B and A is independent of Are A B mutually independent?
A B independent implies
A , are independent,
,B are independent
We are interested in the probability of the event A but we are given some relevant information, namely that another event B occurred.
Revise the probabilities assigned to points of S in view of this new information. If the information does not effect the relative probability of points in B then the new probabilities of points outside of B should be set to 0 and those within B rescaled to add to 1.
Definition: Conditional Probability:
For an event B with P(B) >0, define the conditional probability This is another probability measure on the same sample space S . Note that P(B\B) =1, and P( \B =0.
Multiplication Rule: P(AB)=P(B)P(A \ B)
Let A and B be events defined on the same sample space, with P(A)=0.3 , P(B)=0.4 and P(A \ B)=0.5 . Given that event B does not occur, what is the probability of event A ?
Three students Jane, Sue and Tom write an exam. Jane has a 90% chance of passing the exam, Sue has a 70% chance of passing and Tom has an 80% chance of passing. If exactly one of the students failed the exam, what is the probability it was Sue?
Note that we are given the event
Multiplication Rules (The sequel)
Let A,B,C,D be events in a sample space such that A , AB , ABC have positive probability. Then
Let be a partition of the sample space S into disjoint (mutually exclusive) events such that S . Let B be an arbitrary event in S . Then
Many methods of spam detection are based on features that appear more frequently in spam than in regular email. Conditional probability methods are then used to decide whether an email is spam or not.
Define the following events associated with a random email message.
|B||=||“Message is spam”|
|=||“Message is not spam|
|A||=||“Message contains the word Viagra”|
If we know the values of the probabilities P(B) , P(A \ B) and P(A \ ) , then we can find the probabilities P(B \ A) and P( \ A) .
From a study of email messages coming into a certain system it is estimated that P(B) =0.5, P(A \ B)=0.2 , and P(A\ )=0.001 . Find P(B\A and P( \A .
If you declared that any email containing the word Viagra was Spam, then find what fraction of regular (non-spam) emails would be incorrectly identified as Spam.
Review: “at least one of the events A , B ,C,D”=
“all of the events A , B,C,D “= ABCD
“none of the events A , B,C,D “=
How do we calculate probabilities of the above?
Choosing appropriate Sample spaces: e.g. 3.8, 3.10, 3.11
Example (p. 48):
In an insurance portfolio 10% of the policy holders are in Class (high risk), 40% are in Class (medium risk), and 50% are in Class (low risk). The probability a Class policy has a claim in a given year is .10; similar probabilities for Classes and are .05 and .02. Find the probability that if a claim is made, it is for a Class policy.
Tree- each path represents a sequence of events.
On a branch write the conditional probability of that event given preceding events
The probability at a node of the tree=product of probabilities on the branches leading to the node
= probability of the intersection of the events leading to it
STATISTICS 230, FALL 2005 Probability Probability: (Random House) a strong likelihood or chance of something. The relative possibility an event will occur . the ratio of the number of