01 · Find the constant
X takes values 1,3,5 with probabilities k,2k,3k respectively. Find k and P(X≥3).
Hint
Sum all three masses before selecting the event.
Worked solution
6k=1, so k=1/6. P(X≥3)=2/6+3/6=5/6.
Understand · explore · practise
Statistical distributions: mixed practice
Practise choosing discrete and binomial models, constructing capped distributions, calculating tails, proving cutoffs and combining repeated or shifted counts.
Before you startThe Statistical distributions lessons; attempt questions independently before opening hints.
01 / Choose a model before a formula
Name the random variable and its possible values.
First ask: what quantity is being recorded?
A success count in fixed trials, a stopping time, a uniform label and a count of qualifying groups need different models.
Use paper to attempt the questions. Hints and complete solutions open only when requested. State assumptions, show event translations and keep enough precision for boundary decisions.
Choose a method before calculatingExplore
Check the method
02 / Normalise a probability table
A valid table has nonnegative masses summing to one.
02 · Strict interval
For that X, find P(1<X<5).
Hint
Only one supported value lies strictly inside.
Worked solution
Only X=3 qualifies, so the probability is 2k=1/3.
03 / Count supported integers
Inclusive endpoint counting comes before division.
03 · Uniform labels
U is uniform on the integers −3 through 8 inclusive. Find P(U>4).
Hint
There are 8−(−3)+1 labels.
Worked solution
Twelve labels are equally likely. Values 5,6,7,8 qualify, so probability 4/12=1/3.
04 · Absolute value
Find P(|U|≤2). Is U² uniform on its distinct values?
Hint
Count −2 through 2, then compare how many labels give each square.
Worked solution
P(|U|≤2)=5/12. U² is not uniform: 0 comes only from U=0, but 1 comes from U=−1 or 1, so those masses differ.
04 / Construct a capped stopping distribution
The last mass includes success or failure at the cap.
Watch: fixed trial count or stopping count?
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 · Three-attempt cap
Independent attempts succeed with probability 0.4. Stop at the first success or after three attempts. Give the distribution of T, the attempts used.
Hint
For T=3, only the first two attempts must fail.
Worked solution
P(T=1)=0.4; P(T=2)=0.6×0.4=0.24; P(T=3)=0.6²=0.36. The total is 1.
06 · Final success
Find the probability that the first success occurs on attempt 3, and explain why it differs from P(T=3).
Hint
Add a success requirement on the third attempt.
Worked solution
0.6²×0.4=0.144. P(T=3)=0.36 also includes three failures, of probability 0.6³=0.216.
05 / Justify a binomial model
A familiar-looking count may fail a key assumption.
07 · Without replacement
Four objects are sampled without replacement from a small box of three marked and five unmarked objects. Is the marked count exactly B(4,3/8)?
Hint
A selected marked object changes the next chance.
Worked solution
No. The draws are dependent and their conditional success chances change. Use the actual sampling model.
08 · More than two colours
A fixed number of independent spins have red, blue and green outcomes with constant probabilities. Can the number of green results be binomial?
Hint
Classify the outcomes for this question.
Worked solution
Yes: green is success and red-or-blue is failure. Use n equal to the number of spins and p equal to the green probability.
06 / Translate before using a cumulative function
Use X~B(8,0.25) for the next three questions.
09 · Point probability
Find P(X=3).
Hint
Use C(8,3) and five failures.
Worked solution
56×(1/4)³×(3/4)⁵ = 1701/8192 ≈ 0.207641602.
10 · Inclusive interval
Find P(2≤X≤4).
Hint
Use F(4)−F(1), or add three point masses.
Worked solution
19845/32768 ≈ 0.605621338.
11 · Upper tail
Find P(X≥2).
Hint
Complement counts 0 and 1.
Worked solution
1−F(1) = 41479/65536 ≈ 0.632919312.
07 / Prove a boundary with adjacent integers
A working value alone need not be the required extremum.
12 · Upper-tail cutoff
For X~B(8,0.25), find the smallest r with P(X≥r)≤0.05.
Hint
Compare the tails starting at 4 and 5.
Worked solution
At r=4 the tail is approximately 0.11381531; at r=5 it is 0.02729797. Thus r=5 is the smallest valid threshold.
13 · Strict equality
For Y~B(3,0.5), find the smallest r with P(Y≥r)<1/8, allowing r from 0 through 4.
Hint
P(Y≥3)=1/8 exactly.
Worked solution
r=4. The strict target excludes r=3; r=4 gives the impossible event with probability 0. If thresholds were restricted to 0..3, no solution would exist.
08 / Solve for a required trial count
Use the complement of no successes and check the target.
14 · Minimum attempts
Independent trials succeed with probability 0.3. Find the minimum n for at least a 90% chance of one or more successes.
Hint
Require 0.7ⁿ≤0.1.
Worked solution
At n=6, success probability=0.882351<0.9. At n=7, it is 0.9176457≥0.9. Thus n=7; the log calculation brackets the same two integers.
15 · Certain target
Can the same model attain exactly 100% success probability with finitely many trials?
Hint
Is 0.7ⁿ ever zero for finite n?
Worked solution
No. The chance of no success remains positive for every finite n.
09 / Distinguish individuals from qualifying groups
Find the group probability before using the outer model.
16 · One group
A group contains three independent fair trials. It qualifies if at least two succeed. Find its qualification probability q.
Hint
Add the masses at 2 and 3.
Worked solution
q=(3+1)/8=1/2.
17 · Four groups
Four such groups are independent. Find the probability that exactly two qualify.
Hint
The outer count is B(4,1/2).
Worked solution
6×(1/2)²×(1/2)²=3/8. The outer count has four group trials, not twelve individual trials.
10 / Subtract a certain contribution from the target
Only the random part needs the binomial model.
18 · Known answers plus guesses
An eight-question quiz has two certainly correct known answers and six independent random guesses, each with success chance 1/4. Find P(total score≥4), one mark per correct answer.
Hint
Write T=2+G with G~B(6,1/4).
Worked solution
T≥4 means G≥2. P(G≥2) = 1 − P(G=0) − P(G=1) = 1909/4096 ≈ 0.466064453. Modelling all eight answers with p=1/4 would ignore the known marks.
11 / Check the model, event and boundary
Use mistakes to choose the next lesson to revisit.
If a question failed, identify whether the issue was the model assumptions, the possible values, the event inequality or the calculation. Revisit that specific lesson and then attempt the question again without viewing the solution. A correct answer should also explain what the probability refers to.
Section 1 of 11 · Choose a model before a formula