Topic 6: Probability

Dr J Frost (jfrost@tiffin.kingston.sch.uk)

Last modified: 18th July 2013

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Frost A Frosty SpecialQuestions from the deep dark recesses of my head.

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Topic 6: ProbabilityPart 1 – Manipulating Probabilities

Part 2 – Random Variables

a. Random Variablesb. Discrete and Continuous Distributionsc. Mean and Expected Valued. Uniform Distributionse. Standard Deviation and Variance

a. Binomialb. Bernoullic. Poissond. Geometrice. Normal/Gaussian

Part 3 – Common Distributions

Some starting notesOnly some of those reading this will have done a Statistics module at A Level. Therefore only GCSE knowledge is assumed. There is some overlap with the field of Combinatorics. For probability problems relating to ‘arrangements’ of things, look there instead.

Probability questions frequently come up (although not technically requiring any more than GCSE theory). In my experience, applicants tend to do particularly bad at these questions.

In university interviews…Harder probability questions are quite rare (although one appeared towards the end of 2012’s paper)

In SMC

Used to be moderately common, but less so nowadays

In BMO

Two questions at the end of every paper. You could avoid these, but you broaden your choice if you prepare for these.

In STEP

Probability and Stats questions in…

But some basic probability/statistics will broaden your maths ‘general knowledge’. You’ll know for example what scientists at CERN mean when they refer in the news to the “ test” needed to verify that a new particle has been discovered!

ζPart 1: Manipulating ProbabilitiesTopic 6 – Probability

Events and Sets

An event is a set of outcomes.

“Even number thrown on a die”

Given that events can be represented as sets, we can use set operations. Suppose , say the event of throwing an even number, and , say the event of throwing a prime number. Then:

𝐸∩𝑃={2 } means set intersection. It gives the items which are members of both sets. It represents “numbers that are even AND prime”.

𝐸∪𝑃={2 ,3 , 4 ,5 ,6 } means set union. It gives the items which are members of either. It represents “numbers that are even OR prime”.

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GCSE Recap

𝑃 ( 𝐴∪𝐵 )=𝑃 ( 𝐴 )+𝑃 (𝐵)

When and are mutually exclusive (i.e. and can’t happen at the same time, or more formally , where is the ‘empty set’)…

𝑃 ( 𝐴∩𝐵 )=𝑃 ( 𝐴 )×𝑃 (𝐵)When and are independent (i.e. and don’t influence each other)…

𝑃 ( 𝐴∩𝐵 )=0When and are mutually exclusive…

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More useful identities

𝑃 ( 𝐴∪𝐵 )=𝑃 ( 𝐴 )+𝑃 (𝐵 )− 𝑃 (𝐴∩𝐵)When and are not mutually exclusive…

𝑃 ( 𝐴∪𝐵 )=𝑃 ( 𝐴 )+𝑃 (𝐵 )− 𝑃 ( 𝐴 ) 𝑃 (𝐵)When and are independent and mutually exclusive…

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Conditional probabilities

𝑃 ( 𝐴∨𝐵 )

We might want to express the probability of an event given that another occurred:

The probability that A occurred given B occurred.

To appreciate conditional probabilities, consider a probability tree:

1st pick 2nd pick

Red

Green

Red

Green

47

37

36

36

This represents the probability that a green counter was picked second GIVEN that a red counter was picked first.

Conditional probabilities

or

Using the tree, we can construct the following identity for condition probabilities:

A

B

𝑝 (𝐴)

𝑝 (𝐵∨𝐴) 𝑝 (𝐴∩𝐵)

Conditional probabilities

If and are independent, then what is ?

The Common Sense Method The Formal Method

If and are independent, then the probability of occurring is not affected by whether occurred, so:

Because and are independent.? ?

ExamplesThe events A and B are independent with and . Find:

𝑃 (𝐵 )

𝑃 (𝐴′∩𝐵 )

𝑃 (𝐵′∨𝐴)

(Source: Edexcel)

because and are independent.

So ?

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Bayes’ Rule

Bayes’ Rule relates causes and effects. It allows us find the probability of the cause given the effect, if we know the probability of the effect given the cause.

𝑃 (𝐶|𝐸 )= 𝑃 (𝐸|𝐶 )𝑃 (𝐶 )𝑃 (𝐸 )

Dr House is trying to find the cause of a disease. He suspects Lupus (as he always does) due to their kidney failure. The probability that someone has this symptom if they did have Lupus is 0.2. The probability that a random patient has kidney damage is 0.001, and the probability they have Lupus 0.0001. What is the probability they have Lupus given their observed symptom?

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Bayes’ RuleBut we don’t always need to know the probability of the effect.

𝑷 (𝑪|𝑬 )=𝒌𝑷 (𝑬|𝑪 ) 𝑷 (𝑪 )

Notice that in the distribution , is fixed, and the distribution is over different causes, where . This suggests we can write:

where is a normalising constant that is set to ensure our probabilities add up to 1, i.e.

𝑷 (𝑪|𝑬 )=𝑷 (𝑬|𝑪 )𝑷 (𝑪 )𝑷 (𝑬 )

Question: The probability that a game is called off if it’s raining is 0.7. The probability it’s called off if it didn’t rain (e.g. due to player illness) is 0.05. The probability that it rains on any given day is 0.2.Andy Murray’s game is called off. What’s the probability that rain was the cause?

Bayes’ RuleQuestion: The probability that a game is called off if it’s raining is 0.7. The probability it’s called off if it didn’t rain (e.g. due to player illness) is 0.05. The probability that it rains on any given day is 0.2.Andy Murray’s game is called off. What’s the probability that rain was the cause?

Write down information:

So

Then using Bayes’ Rule:

But . So , so .Then

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ζPart 2: Random VariablesTopic 6 – Probability

Random VariablesA random variable is a variable which can have multiple values, each with an associated probability.

The variable can be thought of as a ‘trial’ or ‘experiment’, representing something which can have a number of outcomes.

A random variable has 3 things associated with it:

1 The outcomes(formally known as the ‘support vector’)

The values the random variable can have (e.g. outcomes of the throw of a die)

2 A probability function The probability associated with each outcome.

3Parameters These are constants used in our probability

function that can be set (e.g. number of throws)

Example random variables

The single throw of a fair die.

{1, 2, 3, 4, 5, 6} None 𝑃 (𝑋=𝑥 )=16 ∀ 𝑥

This symbol means “for all”

Random variable (X) Outcomes Parameters? Probability Function

The single throw of an unfair die.

{1, 2, 3, 4, 5, 6} We can set the probability of each outcome: p1, p2, …, p6

𝑃 (𝑋=𝑥 𝑖 )=𝑝𝑖∀ 𝑖

Parameters are values we can control, but do not change across different outcomes.We’ll see plenty more examples.

For an outcome x and a random variable X, we express the probability as , meaning “the probability that the random variable X has the outcome x”.We sometimes write for short, with a lowercase p.In this example, we can use the probability associated with the particular outcome. We sometimes use to mean the ith outcome.

We use capital letters for random variables.

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Sketching the probability function

X

1 2 3 4 5 6

P(X)

It’s often helpful to show the probability function as a graph. Suppose a random variable X represents the single throw of a biased die:

Outcomes

Probabilities

Discrete vs Continuous DistributionsDiscrete distributions are ones where the outcomes are discrete, e.g. throw of a die, number of Heads seen in 10 throws, etc.In contrast continuous distributions allow us to model things like height, weight, etc.

Here’s two possible probability functions:

Number of times target hit (k)

P(X=

k)

0.4

0.3

0.2

0.1

1 2 3 4

Discrete Continuous

Height of randomly picked person (h)P(

X=h)

1.2m 1.4m 1.6m 1.8m 2.0m 2.2m 2.4m

Discrete vs Continuous Distributions

Number of times target hit (k)

P(X=

k)

0.4

0.3

0.2

0.1

1 2 3 4

Discrete

Probabilities add up to 1.i.e.

All probabilities must be between 0 and 1, i.e. .

We call the probability function the:Probability mass function (PMF for short)

Because our probability function is ultimately just a plain old function (provided it meets the above properties), we often see the function written as “” rather than .

Continuous Distributions

Height of randomly picked person (h)

P(X=

h)

1.0m 1.2m 1.4m 1.6m 1.8m 2.0m 2.2m 2.4m 2.6m

Does it make sense to talk about the probability of someone being exactly 2m?

Clearly not, but we could for example find the probability of a height being in a particular range.

𝒑 (𝟏 .𝟒≤𝒉<𝟏.𝟔 )=∫𝟏 .𝟒

𝟏 .𝟔

𝒑 (𝒉)𝒅𝒉?

i.e. We find the area under the graph. Note that the area under the entire graph will be 1:

The probability associated with a particular value is known as the probability density. It’s value alone is not particular meaningful (and can be greater than 1!), but finding the area in a range gives us a probability mass. This is similar to histograms, where the y-axis is the ‘frequency density’, and finding the area under the bars gives us the frequency.

Probability Density

Question: Archers fire arrows at a target. The probability of the arrow being a certain distance from the centre of the target is proportional to this distance. No archer is terrible enough that his arrow will be more than 1m from the centre.What’s the probability that an arrow is less than 0.5m from the centre?

Distance of arrow from centre (x)

P(X=

x)

Probability is proportional to distance.

Maximum distance is 1m.

1

Since area under graph must be 1, then maximum probability density must be 2, so that the area of triangle is ½ x 2 x 1 = 1.

2

0.5

We’re finding the probability of the arrow being between 0m and 0.5m, so find the area under the graph in this region. We can see this will be 0.25.

Answer: ?Source: Frosty Special

Probability Density

Question: Archers fire arrows at a target. The probability of the arrow being a certain distance from the centre of the target is proportional to this distance. No archer is terrible enough that his arrow will be more than 1m from the centre.What’s the probability that an arrow is less than 0.5m from the centre?

, so .

Source: Frosty Special

Step 1: Use the information to express the proportionality relationship:

Step 2: Determine constant by using the fact that

So and thus

Step 3: Finally, integrate desired range.

∫0

0.5

2 𝑥 𝑑𝑥=0.25

Alternatively, using a cleaner integration approach:

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Mean and Expected Value

The process of using a random variable to give us some values is known as sampling. For example, we might have measured the heights of a sample of people:

The mean of a sample you’ve known how to do since primary school:

Mean of a Sample Mean of a Random Variable

But what about the mean of a random variable X? This is known as the “expected value of X”, written E[X]. It can be calculated using:

or

depending on whether your variable is discrete or continuous.

Archery scores: 57, 94, 25, 42 X is “times target hit out of 3 shots”.

x 0 1 2 3

P(X=x) 0.25 0.5 0.05 0.2?

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Expected ValueQuestion: Two people randomly think of a real number between 0 and 100. What is the expected difference between their numbers? (i.e. the average range)(Source: Frosty Special) (Hint: Make your random variable the difference between the two numbers )

Step 1: Use the information to express the proportionality relationship:

As with many problems, it’s easier to consider a simpler scenario.Consider just say integers between 0 and 10. How many ways can the numbers be chosen if the range is 0? Or the range is 1? Or 2? What do you notice?

We can consider the two numbers (with range ), as a ‘window’ which we can ‘slide’ in the 0 to 100 region. The bigger the window, the less we can slide it. If they were to choose 0 and 100, we can’t slide at all.

Step 2: Determine constant by using the fact that

Integrating we get . If the limits are 0 and 100, we get , so

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Expected Value

(Hint: Make your random variable the difference between the two numbers )

𝐸 [ 𝑋 ]=∫0

100

𝑥𝑝 (𝑥 )𝑑𝑥

One of the harder problem sheet exercises is to consider what happens when we introduce a 3rd number!

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Step 3: Finally, given our known PDF, find E[X]

Question: Two people randomly think of a real number between 0 and 100. What is the expected difference between their numbers? (i.e. the average range)(Source: Frosty Special)

Modifying Random VariablesWe often modify the value of random variables.

Example: X = outcome of a single throw of a die, Y = outcome of another die

What does it mean? We add 1 to all the outcomes of the die (i.e. we now have 2 to 7)The probabilities remain unaffected.

How does the expected value change?

Clearly the mean value will also increase by one. i.e.:

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Consider X + 1

In general: ?

We often modify the value of random variables.

Example: X = outcome of a single throw of a die, Y = outcome of another die

What does it mean? We consider all possible outcomes of X and Y, and combine them by adding them. The new set of outcomes is 2 to 12. Clearly we need to recalculate the probabilities.

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Now consider X + Y

Modifying Random Variables

A uniform distribution is one where all outcomes are equally likely.

Uniform Distribution

Discrete Example Continuous Example

You throw a fair die. What’s the probability of each outcome?

(This ensures the probabilities add up to 1).

You’re generating a random triangle. You pick an angle in the range to use to construct your triangle, chosen from a uniform distribution.What is the probability (density) of picking a particular angle?

This ensures the area under your PDF graph (a rectangle with width and height ) is 1.

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Standard Deviation and Variance

We find the average of the squares of the displacements from the mean.

Example:1cm 4cm 7cm 12cm

Mean = 6Displacements are -5, -2, 1, 6So variance is:

Variance of a Sample Variance of a Random Variable

This is very similar to the sample variation. We’re finding the average of the squared displacement from the mean, i.e.:

Using the fact that :

i.e. We can find the “mean of the squares minus the square of the mean”.

Standard Deviation gives a measure of ‘spread’. It can roughly be thought of as the average distance of values from the mean. It’s often represented by the letter .

The variance is the standard deviation squared. i.e.

Since

The expected value of a value is just the value itself.

Standard Deviation and VarianceExample: Find the variance of this biased spinner (which just has the values 1 and 2), represented by the random variable X.

k 1 2

P(X=k) 0.6 0.4

So

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STEP Question

Fire extinguishers may become faulty at any time after manufacture and are tested annually on the anniversary of manufacture. The time T years after manufacture until a fire extinguisher becomes faulty is modelled by the continuous probability density function:

A faulty fire extinguisher will fail an annual test with probability p, in which case it is destroyed immediately. A non-faulty fire extinguisher will always pass the test. All of the annual tests are independent.

a) Show that the probability that a randomly chosen fire extinguisher will be destroyed exactly three years after its manufacture is

b) Find the probability that a randomly chosen fire extinguisher that was destroyed exactly three years after its manufacture was faulty 18 months after its manufacture.

“I need to consider each of the 3 cases.”

“I have a PDF. This requires me to use definite integration.”

What might be going through your head:

(We’ll do part (b) a bit later)

STEP Question

𝑓 (𝑡 )={ 2 𝑡(1+𝑡 2 )2

, 𝑓𝑜𝑟 𝑡≥0

¿ 0 , h𝑜𝑡 𝑒𝑟𝑤𝑖𝑠𝑒

Since we have a PDF, it makes sense to integrate it so we can find the probability of the extinguisher failing between some range of times.

The probability the extinguisher fails sometime in the first year is , during the second year and during the third year Let’s consider the three cases:

a) If it fails during the first year, it must survive the first two tests, before failing the third. This gives a probability of b) If it fails during the second year, it must survive the second test and fail on the third, giving (note that on the first

test, the probability of it surviving given it’s not faulty is 1)c) If it fails during the third year, then it fails during the third year. We get .

Adding these probabilities together gives us the desired probability.

Mean and Variance of Random VariablesA point P is chosen (with uniform distribution) on the circle . The random variable denotes the distance of from . Find the mean and variance of X. [Source: STEP1 1987]

An important first question is how we could chosen a random point on the circle with uniform distribution.Question: Could we for example choose the x coordinate randomly between -1 and 1, and use to determine ?

Click to choose points uniformly

across x.

No: We can see that because the lines are steeper either side of the circle, we’d likely have less points in these regions, and thus we haven’t chosen a point with uniform distribution around the circle. We’d have a similar problem if we were trying to pick a random point on a sphere, and picked a random latitude/longitude coordinate (we’d favour the poles)

(1,0)

Mean and Variance of Random VariablesA point P is chosen (with uniform distribution) on the circle . The random variable denotes the distance of from . Find the mean and variance of X. [Source: STEP1 1987]

In which case, how can we make sure we pick a point randomly?

Introduce a parameter for the angle anticlockwise from the x-axis say. Clearly this doesn’t give bias to certain regions of the arc (satisfying the ‘uniform distribution’ bit).

𝜃𝑋So what is the distance X?

(It’s an isosceles triangle, so split into 2)

So what is E[X]?

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Summary• Random variables have a number of possible outcomes, each with an associated

probability.• Random variables can be discrete or continuous.

• Discrete random variables have an associated probability mass function. We require that across the domain of the function (i.e. possible outcomes).

• Continuous random variables have an associated probability density function. Unlike ‘conventional’ probabilities, these can have a value greater than 1. We require that , i.e. the total area under the graph is 1.We can find a probability mass (i.e. the ‘conventional’ kind of probability) by finding the area under the graph in a particular range, using definite integration.

• While we have a ‘mean’ for a sample, we have an ‘expected value’ for a random variable, written . It can be calculated using for a discrete random variable, and for a continuous random variable. The expected value for a fair die for example is 3.5.

• The variance gives a measure of spread. For specifically it’s the average squared distance from the mean. We can calculate it using , which can be remembered using the mnemonic “mean of the square minus the square of the mean”, or “msmsm”.

• . i.e. Scaling our outcomes/adding has the same effect on the mean.

ζPart 3: Common DistributionsTopic 6 – Probability

Common DistributionsWe’ve seen so far that can build whatever random variable we like using two essential ingredients: specifying the outcomes, and specifying a PMF/PDF that associates a probability with each outcome. But there’s a number of well-known distributions for which we already have the outcomes and probability function defined: we just need to set some parameters.

Bernoulli

e.g. Throw of a (possible biased) coin.

Multivariate

e.g. Throw of a (possibly biased) die.

Binomial

e.g. Counts the number of heads and tails in 10 throws.

Multinomial

e.g. Counting the number of each face in 10 throws of a die.

Poisson

e.g. Number of cars which pass in the next hour given a known average rate.

Geometric

Exponential

e.g. The possible time before a volcano next erupts.

e.g. The number of times I have to flip a coin before I see a heads.

Dirichlete.g. The possible probability distributions for the throw of a die, given I threw a die 60 times and saw 10 ones, 10 twos, 10 threes, 10 fours, 10 fives and 10 sixes.

We won’t explore these.

Bernoulli Distribution

The Bernoulli Distribution is perhaps the most simple distribution. It models an experiment with just two outcomes, often referred to as ‘success’ and ‘failure’.

It might represent the single throw of a coin. (where ‘Heads’ could represent a ‘success’)

A single trial with two outcomes.

“Failure”/”Success”, or {0, 1}

p, the probability of success.

𝑃 (𝑋=𝑥 )={1−𝑝𝑥=0𝑝𝑥=1

Description Outcomes Parameters? Probability Function

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A trial with just two outcomes is known as a Bernoulli Trial.

A sequence of Bernoulli Trials (all independent of each other) is known as a Bernoulli Process.An example is repeatedly flipping a coin, and recording the result each time.

Binomial Distribution

Suppose I flip a biased coin. Let heads be a ‘success’ and tails be a ‘failure’. Let there be a probability that I have a success in each throw.

The Binomial Distribution allows us to determine the probability of a given number of successes in n (Bernoulli) trials, in this case, the number of heads in n throws.

Question: If I throw a biased coin (with probability of heads p) 8 times, what is the probability I see 3 heads?

H H H T T T T T

The probability of this particular sequence is: But there’s ways in which we could see 3 heads in 8 throws.Therefore

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Binomial DistributionTherefore, in general, the probability of k successes in n trials is:

Binomial DNumber of ‘successes’ in n trials.

{0, 1, 2, … , n}i.e. between 0 and n successes.

, the probability of a single success., the number of trials

𝑝 (𝑋=𝑘 )=(𝑛𝑘)  𝑝𝑘 (1−𝑝)𝑛−𝑘

Description Outcomes Parameters? Probability Function

? ? ?

We can write B(n,p) to represent the Binomial Distribution, where n and p are the parameters for the number of trials and probability of a single success.

If we want some random variable X to use this distribution, we can use . The means “has the distribution of”.

Frost Real-Life Example

While on holiday in Hawaii, I was having lunch with a family, where an unusually high number were left-handed: 5 out of the 8 of us (including myself). I was asked what the probability of this was. (Roughly 10% of the world population is left-handed.)

Suppose X is the random variable representing the number of left handed people.

Then

(This example points out one of the assumptions of the Binomial Distribution: that each trial is independent. But this was unlikely to be the case, since most on the table were related, and left-handedness is in part hereditary. Sometimes when we model a scenario using an ‘off-the-shelf’ distribution, we have to compromise by making simplifying assumptions.)

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Summary of Distributions so farSimilarly, a multivariate distribution represents a single trial with any number of outcomes. A multinomial distribution is a generalisation of the Binomial Distribution, which gives us the probability of counts when we have multiple outcomes.

Bernoulli

e.g. “What’s the probability of getting a Heads?”

Multivariatee.g. “What’s the probability of getting a 5?

Binomial

e.g. “What’s the probability of getting 3 Heads and 2 Tails?”

Multinomiale.g. “What’s the probability of rolling 3 sixes, 2 fours and a 1?

Generalise to n trials

Generalise to n trials

Generalise to k outcomes

(Use your combinatorics knowledge to try and work out the probability function for this!)

Poisson Distribution

Cars pass you on a road at an average rate of 5 cars a minute. What’s the probability that 3 cars will pass you in the next minute?

When you have a known average ‘rate’ of events occurring, we can use a Poisson Distribution to model the number of events that occur within that period.

We use to represent the average rate.

We can see that when the average rate is 10 (say per minute), we’re most likely to see 10 cars. But technically, we could see a million cars (even if the probability is very low!)

k is the number of events (e.g. seeing a car) that occur.

Poisson DistributionAssumptions that the Poisson Distribution makes:

1. All events occur independently (e.g. a car passing you doesn’t affect when the next car will pass you).

2. Events occur equally likely at any of time (e.g. we’re not any more likely to see cars at the beginning of the period than at the end)

Number of events occurring within a fixed period given an average rate.

{0, 1, 2, … } up to infinity.

, the average number of events in that period.

𝑃 (𝑋=𝑘 )= 𝜆𝑘

𝑘 ! 𝑒−𝜆

Description Outcomes Parameters? Probability Function

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is Euler’s Number, with the value 2.71…

i.e. The Poisson Distribution is a DISCRETE distribution.

Poisson Distribution

Example: An active volcano erupts on average 5 times each year. It’s equally likely to erupt at any time.

Q1) What’s the probability that it erupts 10 times next year?

𝑝 (𝑋=10 )=510

10 !𝑒−5=0.018

Q2) What’s the probability that it erupts at all next year?

1−𝑝 (𝑋=0 )=1− 50

0 !𝑒− 5=1−𝑒−5=0.993

Q3) What’s the probability that it next erupts between 2 and 3 years after the current date?

i.e. It erupts 0 times in the first year, 0 times in the second year, and at least once the third year.

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Relationship to the Binomial DistributionImagine that we segment this fixed period into a number of smaller chunks of time, in each of which an event can occur (which we’ll describe as a ‘success’), or not occur.

1 minute

A car passed in this period!

If we presumed that we only had at most one car passing in each of these smaller periods of time, then we could use a Binomial Distribution to model the total number of cars that pass across 1 minute, because it models the number of successes.

Of course, multiple cars could actually pass within each smaller segment of time. How would we fix this?

A car passed in this period!

Relationship to the Binomial DistributionWe could simply use smaller chunks of time – in the limit, we have tiny slivers of time, so instantaneous that we couldn’t possibly have two cars passing at exactly the same time.

1 minute

Now if we’d divided up our time into chunks where is large, and we expect an average of cars to pass, what then is the probability of a car passing in one chunk of time? (Only Year 8 probability needed!)

𝒑=𝝀𝒏?

Therefore, as n becomes infinitely large (so our slivers of time become instantaneous moments), we can use the Binomial Distribution to represent the number of events that occur within some period:

We need some fiddly maths to show this. tends to arise in maths when we have limits.

Uniform Distribution

We saw earlier that a uniform distribution is where each outcome is equally likely.

Each outcome is equally likely.

x1, x2, …, xn None.

Description Outcomes Parameters? Probability Function

? ?𝑝 (𝑋=𝑥 𝑖 )=

1𝑛 ∀ 𝑖

?

Examples: The throw of a fair die, the throw of a fair coin, the possible lottery numbers this week (presuming the ball machine isn’t biased!).

Geometric Distribution

You, Christopher Walken, are captured by the Viet Cong during the Vietnam War, and forced to play Russian Roulette. The gun has 6 slots on the barrel, one of which has a bullet, and the other slots empty. Before each shot, you rotate the barrel randomly, then shoot at your own head. If you survive, you repeat this ordeal.

Q1) What’s the probability that you die on the first shot?

Q2) What’s the probability that you die on the second shot?

You survive the first then die on the second:

Q3) What’s the probability that you die on the shot?

𝟏𝟔?

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Geometric Distribution

If you have a number of trials, where in each trial you can have a ‘success’ or ‘failure’, and you repeat the trial until you have a success (at which point you stop), then the geometric distribution gives you the probability of succeeding on the 1st trial, the 2nd trial, and so on.

Succeeding on the xth trial after previously failing.

{ 1, 2, 3, … }The trial on which you succeed.

The probability of success.

Description Outcomes Parameters? Probability Function

? ?𝑝 (𝑥 )= (1−𝑝)𝑥− 1𝑝

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Note that if , then For example, if we tossed a fair die until we saw a 1, we’d expect to have to throw the die times on average before we see a 1 (where the count includes the last throw).

Side Note: The distribution is called ‘geometric’ because if we were to list out the probabilities for , , and so on, we’d have a geometric series!

Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If one player hits the target and the other does not then the successful player wins. If both players hit the target, or if both players miss the target, then each has another attempt, with the same rules applying. If the probability of Tom hitting the target is always and the probability of Geri hitting the target is always , what is the probability that Tom wins the competition?

Level 2

Level 1

Level 4

Level 3

Level 5

SMC

A:

B: C:

D: E:

Geometric Distribution

The probability that they both hit or miss is .So Tom can win by either winning immediately , or initially drawing before winning: , or drawing twice and then winning: and so on. This gives us an infinite geometric series with and . Using , we get .

Frost Real-Life Example

My mum (who works at John Lewis), was selling London Olympics ‘trading cards’, of which there were about 200 different cards to collect, and could be bought in packs. Her manager was curious how many cards you would have to buy on average before you collected them all. The problem was passed on to me!(Note: Assume for simplicity that each card is equally likely to be acquired – unlike say ‘Pokemon cards’ [a childhood fad I never got into], where lower numbered cards are rarer)

Hint: Perhaps think of the trials needed to collect the next card as a geometric process? Then consider these processes all combined together.

Answer: cardsExplanation on next slide…

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Frost Real-Life Example

Answer: cards

To get the first card, we just need to buy 1 card.To get the second card, each time we buy a card, we have a chance of buying a new card (if not, we keep buying until we have a new one). Since the number of cards we need to buy to get this next card is geometrically distributed, we expected number of cards is Combined these expected number of cards we need to buy for each new card, we get . The bracketed expression is known as a ‘Harmonic Series’, which can be represented as . Typing “200 * H(200)” into www.wolframalpha.com got me the answer above.

This problem is more generally known as the “Coupon Collector’s Problem” http://en.wikipedia.org/wiki/Coupon_collector%27s_problem

Coin ConundrumsHow would you model a fair coin given you have just a fair die?Solution: Easy! Roll the die. If you get say an even number, declare ‘Heads’, else declare ‘Tails’.

How would you model a fair die given you have just a fair coin?Solution: A bit harder! Throw the coin 3 times, giving us 8 possible outcomes. Label the first 6 of these outcomes (e.g. HHH, HHT, …). If we get the last two outcomes, then reject these outcomes and repeat.An interesting side question is how many times on average we’d expect to have to throw the coin. If the probability of being able to stop is , the expected value of a geometric distribution is , i.e. . Since we throw the coin 3 times each time, then we expect an average of 4 throws.

How would you model a fair coin using an unfair coin?

Solution: Suppose the probability of Heads on the unfair coin is . Then throw this coin two times. We have four outcomes: HH, HT, TH and TT, with probabilities , , and respectively. Two of these outcomes have the same probability. So declare ‘Heads’ if you threw HT on the biased coin, ‘Tails’ if you threw HT, and repeat otherwise. (You’d expect to have to throw times on average).

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Gaussian/Normal Distribution

A Gaussian/Normal distribution is a continuous distribution which has a ‘bell-curve’ type shape. It’s useful for modelling variables where the values are clustered, about the mean, and spread out around it with probability dropping off.

IQ (“Intelligence Quotient”) x

P(X=

x)

70 85 100 115 130 145

Suppose there’s a known mean and a known standard deviation . We distribution is denoted as , parameterised by the mean and variance.Then if :

IQ is a good example. The mean is (by definition) 100, and the probability of having an IQ drops off symmetrically, with Standard Deviation 15 (by definition).

Z-values

IQ (“Intelligence Quotient”) x

P(X=

x)

70 85 100 115 130 145

We might be interested to know what percentage of the population have an IQ below 130.

The z-value is the number of Standard Deviations above the mean.

Some rather helpful mathematicians have compiled a table of values that give us , i.e. the probability of being below a particular z-value, for different values of z. This is unsurprisingly known as a z-table.

If and , how many Standard Deviations above the mean is 130?

Answer = 2?

Z-values

We look up the units and tenths digit of our z-value here…

…and our hundredths digit here.

So ?

Z-values

It’s useful to remember that 68% of values are within 1 s.d. of the mean, 95% within two, and 99.7% within 3 (when the variable is ‘normally distributed’)

When scientists referred to a “ test” needed to officially ‘discover’ the Higgs Boson, they mean that were the data observed to occur ‘by chance’ in the situation where the Higgs Boson didn’t exist (known as the null hypothesis), then the probability is less than that of being away from the mean of a randomly distributed variable: a 1 in 3.5m chance. A Level students studying S2 will encounter ‘hypothesis testing’.