01 · Inclusive interval
X is uniform on the integers 4 through 15. Find P(7≤X≤12).
Hint
List the endpoints and count between them.
Worked solution
7,8,9,10,11,12: six values, so 1/2.
Understand · explore · practise
Discrete uniform distributions
Calculate probabilities on equally likely finite supports, handle integer endpoints and combine an independent uniform variable with a non-uniform one.
Before you startProbability mass functions and independent events.
01 / Justify equal probability
Uniform means equal mass at every supported value.
For N equally likely supported values, each probability is 1/N.
An event has probability: number of qualifying values ÷ N.
The selection process must justify equal likelihood. “Chosen at random” does not by itself specify every possible mechanism. State a uniform choice when using simple counts.
Count the eligible integersExplore
X is uniform on the twelve integers 4 through 15, inclusive. Each has probability 1/12.
All twelve integers qualify: probability 1.
02 / Count both ends when they are included
Consecutive integers from a through b number b−a+1.
Uniform integers 4 through 15:Worked example
Number of values: 15−4+1=12
Subtracting endpoints alone counts gaps, not values.
For 5≤X<11, the values are 5,6,7,8,9,10
Six values give probability 6/12=1/2.
For 5<X<11, exclude 5 as well
Five values give 5/12.
02 · Beyond the support
Find P(0≤X≤6).
Hint
Only supported values contribute.
Worked solution
Only 4,5,6 qualify, giving 3/12=1/4.
03 / A property also defines a subset
Count values, not the number of words in the event.
03 · Prime
Find P(X is prime) for the model.
Hint
Which primes lie from 4 through 15?
Worked solution
5,7,11,13: four values, probability 1/3.
04 · Multiple
Find P(X is a multiple of 3).
Hint
Include all eligible multiples.
Worked solution
6,9,12,15: probability 4/12=1/3.
05 · Complement
Find P(X is not 8).
Hint
Complement one supported value.
Worked solution
1−1/12=11/12.
06 · Decimal boundary
Find P(6.2<X<10.8).
Hint
X still only takes integers.
Worked solution
7,8,9,10 qualify. Probability 4/12=1/3.
04 / Uniform support need not be consecutive
Count the listed values rather than the width of their range.
A uniform variable Y takes values {−4,−1,2,9}. It has four possible values, each of probability 1/4. The gaps between labels do not create additional outcomes.
07 · Gapped event
Find P(Y>0).
Hint
Use the supported positive values.
Worked solution
2 and 9 qualify, so 2/4=1/2.
08 · Impossible value
Find P(Y=0).
Hint
Zero is between two labels but is not supported.
Worked solution
0.
09 · Uniform selection versus repeated labels
A bag has three tokens labelled 1 and one labelled 8. A token is selected uniformly. Is its recorded label uniform on {1,8}?
Hint
Count the physical tokens contributing to each label.
Worked solution
No. The label probabilities are 3/4 and 1/4. The four tokens are equally likely; the two labels are not.
05 / A new variable can lose uniformity
Different source values can lead to the same result.
Watch: equal source masses combine after squaring
Pause, replay or seek freely. The notes explain the same idea and stay in view.
U is uniform on {−2,−1,0,1,2}. Define V=U².Worked example
V=0 comes only from U=0
Probability1/5.
V=1 comes from U=−1 or 1
Probability 2/5.
V=4 comes from U=−2 or 2
Probability 2/5. V is not uniform on its three values.
10 · Absolute value
For the same U, find the distribution of W=|U|.
Hint
Combine opposite values.
Worked solution
W=0,1,2 have respective probabilities 1/5,2/5,2/5.
06 / Weight each case of a second variable
Independence allows the original X distribution within each Y case.
X is uniform on 1 through 10. Independently, Y is 2,5,9 with probabilities 0.5,0.3,0.2. Find P(X>Y).Worked example
Given Y=2, eight X values are larger
Contribution 0.5×8/10=0.40.
Given Y=5, five X values are larger
Contribution 0.3×5/10=0.15.
Given Y=9, one X value is larger
Contribution 0.2×1/10=0.02.
Add the three disjoint Y cases
P(X>Y)=0.40+0.15+0.02=0.57.
The 30 listed (X,Y) pairs are not equally likely: their probabilities depend on Y. Counting qualifying pairs and dividing by 30 would ignore those weights.
11 · Equality
For these independent X,Y, find P(X=Y).
Hint
Each possible Y value is one of the ten X values.
Worked solution
0.5×1/10 + 0.3×1/10 + 0.2×1/10 = 0.10.
12 · Inclusive comparison
Find P(X≥Y).
Hint
Add equality to the strict comparison.
Worked solution
0.57+0.10=0.67.
07 / Check the joint model
Uniform marginals alone do not imply independent selections.
13 · Reverse comparison
For the independent model, find P(X<Y).
Hint
The three comparisons are disjoint and exhaustive.
Worked solution
1−0.57−0.10=0.33.
14 · Shared selection
X is uniform on 1 through 10 and Y is set equal to X. Can P(X=Y) be found by multiplying ten pairs of 0.1 probabilities?
Hint
The two variables are fully linked.
Worked solution
No. P(X=Y)=1. Multiplication of the marginal probabilities would require independence, which fails here.
08 / Count only after checking the model
Equal mass is an assumption with consequences.
List the support, check equal likelihood and count the values satisfying the event. Respect strict endpoints and gaps. When comparing independent variables with unequal masses, weight the conditional counts. A transformed uniform variable may not remain uniform.
Section 1 of 8 · Justify equal probability