Unit 4
Probability Guide
Mathematical Methods
How and when to use each type of probability
Binomial Distribution
Normal Distribution
Continuous Density Functions
Discrete Probability
Conditional Probability
Statistical Inference
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Binomial Distribution
How to Identify:
The first thing to realise is that Binomial Distribution is used with a series of Bernoulli
Trials, where a Bernoulli Trial is something that only has two outcomes. For example, flipping
a coin is considered a Bernoulli Trial as there are only two possible outcomes (heads or tails).
Getting an answer right in a test is a Bernoulli Trial as there are only two possible outcomes
(right answer or wrong answer) that are completely independent (one outcome has no effect on
the next outcome).
Be careful though, as there are some tricky ones to identify. Is "Rolling a 6 on a die" a Bernoulli
Trial?? Yes. There only two outcomes (rolling a 6, or not rolling a 6). Let’s look at an example.
Example:
Answer:
Let’s figure out how we can determine what type of probability to use. Like I said, we have to
first identify if there is a Bernoulli Trial here. Firstly, we know that the probability John hits
the bullseye is 1/4.
For each question you come across, ask yourself - "Are there more than two outcomes to this
trial?" If there is only two, it will most likely be a Bernoulli. This trial only has two outcomes hitting the bullseye or not hitting the bullseye. Therefore, we have a Bernoulli Trial and hence
a Binomial Distribution question on our hands.
Binomial Distribution is largely done in your calculator. Searching for this example question, I looked at Exam 1s first and couldn’t actually easily find one. They’ll always be on
CAS-enabled Exam 2.
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So, there are a few aspects we need to consider in this question.
1. The probability that John hits the bullseye at least once from four throws.
2. The probability that Rebecca hits the bullseye at least once from two throws.
3. The ratio between these two numbers.
Now that we’ve identified that it’s a Binomial Distribution question, we can work this out.
For Binomial Distribution, we always need p , n and x , where p = probability of success, n =
number of trials, and x = how many successes you have.
NOTE: x can be a range of values
John:
Rebecca:
n=4
p = 41
x = 1, 2, 3 and 4 (x > 0) - We know this be-
cause we’re looking for the probability that
he hits at least one bullseye
n=2
p = 21
x = 1 and 2 (x > 0) All this goes into our CAS
(Binomial CDF). I believe this is MENU →
5→5→E
All this goes into our CAS (Binomial CDF).
I believe this is MENU → 5 → 5 → E
n = 2, p = 12 , lower bound = 1, upper bound
=2
n = 4, p = 14 , lower bound = 1, upper bound
=4
Therefore, Pr(x > 0) = 0.75
Therefore, Pr(x > 0) = 0.6836
Ratio:
A and D are clearly not correct. Rebecca has a higher chance than John, therefore the first
number must be higher than the second. That counts out C. Then you just work out what the
ratios are in a decimal form.
32
= 1.1852
27
192
= 1.0971
175
Rebecca/John = 0.75/0.6836 = 1.0971
Therefore, the answer is E (192 : 175).
These questions can be quite simple once you get to know them. All you need to remember is
this: If the question refers to a trial that only has two outcomes, it is a Binomial Distribution
question.
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Normal Distribution
How to Identify:
The question will say "Let the random variable X be normally distributed with a mean of...
etc". It is very easy to identify a Normal Distribution question, because the question will state
that it is one.
NOTE: Normal Distribution follows the ’bell-curve’. Let’s look at an example.
Example:
Answer (6a):
µ = 2.5, σ = 0.3
X ∼ N (2.5, 0.3)
Z ∼ N (0, 1)
Standard normal distribution is represented by Z . This is used to find out the z-scores. A
z-score is "how many standard deviations away from the mean this value is".
P r (x > 3.1) = P r (Z < b)
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P r (X > 3.1) = P r (Z > 2) → P r (Z > 2) = P r (Z < −2)
P r (X > 3.1) = P r (Z < −2)
∴ b = −2
Answer (6b):
Find P r (X < 2.8|X > 2.5):
=
P r (X < 2.8 ∩ X > 2.5)
P r (X > 2.5)
Since µ = 2.5, 0.5 of the data lies on each side:
=
P r (X < 2.8 ∩ X > 2.5)
0.5
6
We know that P r (Z < −1) = 0.16, so:
P r (X < 2.2) = P r (X > 2.8) = 0.16
∴ P r (2.5 < X < 2.8) = 12 (1 − 2(0.16))
= 21 (1 − 0.32)
= 0.34
P r (X < 2.8|X > 2.5) =
0.34
0.5
∴ Pr(X < 2.8|X > 2.5) = 0.68
Identifying Normal Distribution questions is quite easy, because the question identifies it for
you. Try to develop a solid understanding of the bell curve and how the standard deviations/area work together, and this will make these questions much easier.
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Continuous Density Functions
How to Identify:
Like Normal Distribution, Continuous Density Functions are easy to identify. Many questions
will actually state "A continuous random variable, X , has a probability density function given
by..." so it is simple to tell when to use this type of probability. Whenever you see a function
set out as in the example below, use continuous density function probability.
The biggest thing to note with this type of probability is that the area under the graph equals
the probability of an outcome lying between two x-values. Integration is a large part of this
type of probability. The total area under the graph is equal to 1 and f (x) is always greater
than or equal to zero.
Example:
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Answer (8a):

 1 e −x
5
f (x) = 5
0
x ≥0
x ≤0
The median of X is m .
For median, 0.5 of the data lies on either side.
∴ Area under graph = 0.5 up to m .
R m 1 −x
e 5 d x = 0.5
0
5
Solve in CAS or by hand if Exam 1:
h
0.5 = −e
0.5 = −e
0.5 = −e
−x
5
e
−m
5
0
−m
5
−m
5
−0.5 = −e
im
− (−e 0 )
+1
−m
5
= 0.5
loge 0.5 =
−m
5
∴ m = −5loge 12
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Answer (8b):
P r (X < 1|X ≤ m) =
P r (X < 1 ∩ X ≤ m)
P r (X ≤ m)
Because m > 1, P r (X < 1 ∩ X ≤ m) = P r (X < 1)
∴ P r (X < 1)|X ≤ m) =
P r (X < 1)
P r (X ≤ m)
R1 1
P r (X < 1)|X ≤ m) =
0 5e
−x
5
dx
0.5
−1
∴ P r (X < 1)|X ≤ m) = 2(1 − e 5 )
Like I said above, make sure you understand that the area under the graph is the probability between those two points, so get comfortable using integration in this context.
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Discrete Probability
How to Identify:
Again, many questions tell you that the variable X is discrete. You can usually tell if it’s discrete if the question offers a probability table of some sort, or if it says " X is a discrete random
variable..."
If there is a probability table, we need to use discrete probability.
Example:
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Answer (7a):
We need to show that p = 23 OR p = 1.
All probabilities MUST equal to 1 when added together.
∴ 1 = 0.2 + 0.6p 2 + 0.1 + (1 − p) + 0.1
1 = 0.4 + 0.6p 2 + 1 − p
1 = 1.4 + 0.6p 2 − p
0 = 0.6p 2 − p + 0.4
Decimals are too hard to deal with, so let’s multiply everything by 10.
∴ 0 = 6p 2 − 10p + 4
0 = 3p 2 − 5p + 2
Use the cross method to factorise the above expression, giving 0 = (3p − 2)(p − 1).
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∴ 3p − 2 = 0, so p = 32
AND
∴ p − 1 = 0, so p = 1
⇒ p = 23 and p = 1 as required.
Answer (7bi)
Since p 6= 0 (because that would push the total probability over 1 which is not possible), we let
p = 32 to find E (x). Our new probability table looks a little like the following:
x
0
P r (X = x)
0.2
1
6
10
× ( 32 )2
2
3
4
0.1
1
3
0.1
∴ E (X ) = (0 × 0.2) + (1 × 53 × 49 ) + (2 × 0.1) + (3 × 13 ) + (4 × 0.1)
12
+ 0.2 + 1 + 0.4
E (X ) = 45
E (X ) = 12
45 + 1.6
12 8
+
45 5
12 72
E (X ) =
+
45 45
84
E (X ) =
45
28
∴ E (X ) =
15
E (X ) =
Answer (7bii)
We need to find P r (X ≥ E (X )).
E (X ) is slightly less than 2 (according to our answer for 7bi ).
∴ P r (X ≥ E (X )) = P r (X = 2) + P r (X = 3) + P r (X = 4)
13
P r (X ≥ E (X )) = 0.1 + 31 + 0.1
1 1 1
+ +
10 3 10
3 10 3
P r (X ≥ E (X )) =
+
+
30 30 10
8
∴ P r (X ≥ E (X )) =
15
P r (X ≥ E (X )) =
Keep in mind as well (for Discrete Probability):
E(X) = the sum of (x × P r (X = x)) of each section of the probabililty table
Var(X) = E (X 2 ) − [E (X )]2
p
SD(X) = V ar (X )
The sum of all probabilities in the table must be equal to 1.
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Conditional Probability
How to Identify:
Conditional probability is all about "If this happens, then this." If a previous outcome affects the next outcome, this is conditional probability.
Example:
Answer:
We can see that this question refers to conditional probability, because whether she is fit or not
in any given month has a direct effect on whether she is fit the next month.
We always want to construct a Tree Diagram for these types of questions as below. We know
that, as per the question, in the first month Paula is unfit. This sets the scene for the rest of
the tree diagram.
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The question asks us "What is the probability that she is fit in exactly two of the next three
months? " This means we need to add the ’paths’ of each relevant branch in the tree diagram
together to get our final answer as below.
P r (Two fit months) = P r (F F F 0 ) + P r (F F 0 F ) + P r (F 0 F F )
1 3 1
1 1 1
1 1 3
P r (Two fit months) = ( × × ) + ( × × ) + ( × × )
2 4 4
2 4 2
2 2 4
3
1
3
P r (Two fit months) =
+
+
32 16 16
2
6
3
+
+
P r (Two fit months) =
32 32 32
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∴ P r (Two fit months) =
32
To work out the probability of each ’path’ in the tree diagram, multiply the branches together.
When doing conditional probability, a tree diagram is the easiest way to actually understand
what is happening. It is much too difficult to try and visualise situations like these mentally.
So, if one outcome has a direct effect on the subsequent outcome, use a tree diagram to break
up the information.
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Statistical Inference
How to Identify:
Whenever a question mentions either proportions, distribution of p̂, margin of error, confidence
intervals, a sample statistic or a population parameter, you are dealing with statistical inference
(the new study design topic as of 2016). This is pretty much your guide for recognising those
types of questions.
Example:
Answer:
For proportion:
For margin of error:
15
100
3
∴ p̂ =
20
or 0.03.
p̂ =
M = 0.03 → Margin of error is equal to 3%,
∴M =
3
100
For a 95% confidence interval, find the inverse normal for area under the graph. A 95% confidence interval means that 95% of the data lies in the middle. Therefore, there is 2.5% on each
side.
To find z , use inverse normal: area less than z is 95 + 2.5% = 0.975
σ = 1, µ = 0, Area= 0.975
Put this in your CAS calculator (Inverse Normal menu item) to get z = 1.96
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r
We know that, according to our Margin of Error equation, M = z
p̂(1 − p̂)
, where
n
z = z -score for which the area corresponds to the confidence interval
p̂ = proportion of sample
n = sample size
We’re trying to find the sample size, n , so let’s input all of our other values in and solve for n .
r
p̂(1 − p̂)
n
r
0.15(1 − 0.15)
0.03 = 1.96
n
M =z
Solve for n using CAS, which gives:
n = 544 people
How to work out your z-score:
To find the z -score of a certain confidence interval, use Inverse Normal in CAS.
Work out all the area to the left of the z value. For a 99% confidence interval, area is 0.99 plus
0.005 (0.5%).
µ = 0, σ = 1, Area= 0.995
∴ z = 2.58
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Margin of Error:
A margin of error is very difficult to define. It is explained in textbooks as “The distance
between the endpoints of the confidence interval and the sample estimate”. It’s a bit ambiguous what this means, but you don’t really need to understand it. All you need to know is this:
r
M =z
That’s it for statistical inference!
p̂(1 − p̂)
n
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