01 · Complete the space
Find P(A∪B) and the probability of neither.
Hint
Add the three disjoint interior regions.
Worked solution
Union=0.2+0.3+0.1=0.6. Neither=1−0.6=0.4.
Understand · explore · practise
Statistics review: probability and distributions
Cumulative original practice linking Venn regions, conditional trees, finite distributions, binomial tails and repeated independent samples.
Before you startProbability and statistical distributions.
01 / Define the event before choosing a formula
Similar words can describe different random variables.
Fixed-trial success counts and waiting times are different.
State the trial process, the event and any independence assumption. These numerical examples are constructed for practice.
Count or waiting time?Explore
Reveal the model and calculation
02 / Recover totals from disjoint regions
Add regions first, then check the requested relationship.
A constructed model has P(A only)=0.2, P(B only)=0.3 and P(A∩B)=0.1.
02 · Marginals
Find P(A) and P(B).
Hint
Include the intersection in each event.
Worked solution
P(A)=0.3 and P(B)=0.4.
03 · Independence
Are A and B independent?
Hint
Compare joint probability with the product.
Worked solution
No. P(A∩B)=0.1, but P(A)P(B)=0.3×0.4=0.12.
04 · Mutual exclusion
Are A and B mutually exclusive?
Hint
Check whether their intersection can occur.
Worked solution
No. The intersection has probability 0.1, not zero.
03 / Use the probability for the branch you are on
Without replacement changes the remaining population.
A bag has three green and two purple counters. Two are chosen randomly without replacement.
05 · Different colours
Find the probability of different colours.
Hint
Add green–purple and purple–green paths.
Worked solution
(3/5)(2/4)+(2/5)(3/4)=3/5=0.6.
06 · Second green
Find the probability the second counter is green.
Hint
Sum both paths ending in green.
Worked solution
(3/5)(2/4)+(2/5)(3/4)=3/5. The equal numerical answer to question 05 is specific to these counts, not a general identity.
07 · Conditional branch
Given that the first counter is purple, what is the probability the second is green?
Hint
Update the bag after the first draw.
Worked solution
3/4: three green counters remain among four counters.
08 · Supplier mixture
Supplier A provides 70% of parts, with fault probability 0.02. B provides the rest, with fault probability 0.06. Find the overall probability of no fault.
Hint
Find the weighted fault probability and complement it.
Worked solution
Fault probability=0.7×0.02+0.3×0.06=0.032. No fault=0.968.
04 / Check that a proposed distribution is valid
Probabilities describe mutually exclusive values.
Define P(X=x)=x/10 for x=1,2,3,4 and zero otherwise.
09 · Validity
Check this probability distribution.
Hint
Check signs and the total.
Worked solution
All probabilities are nonnegative and (1+2+3+4)/10=1.
10 · Interval event
Find P(1<X≤3).
Hint
List included values before adding.
Worked solution
X=2 or 3, so the probability is (2+3)/10=0.5.
11 · Complement
Find P(X≠4).
Hint
Complement the mass at four.
Worked solution
1−4/10=0.6.
12 · Uniform or not
Is this a discrete uniform distribution?
Hint
Compare probabilities of the supported values.
Worked solution
No. The probabilities increase with x; a discrete uniform distribution would give each of these four values probability 1/4.
05 / Distinguish an exact count from a tail
The random count includes all eight planned trials.
Watch: three successes differs from waiting until trial three
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Each of eight independent trials succeeds with probability 0.3. Let Y be the success count.Worked example
Y~B(8,0.3)
Fixed number of trials, common success probability and independence.
P(Y=2)=C(8,2)(0.3)²(0.7)⁶=0.29647548
This includes all possible positions of the two successes.
P(Y>2)=1−P(Y≤2)=0.44822619
Strictly more than two means 3 through 8.
13 · Off-by-one
Is P(Y>2) the same as 1−P(Y≤1)?
Hint
What count does the second expression include?
Worked solution
No. The second expression is P(Y≥2), so it includes exactly two.
14 · Sample model
Why would sampling eight counters without replacement from a small mixed bag generally not give B(8,0.3)?
Hint
Consider changing probabilities and dependence.
Worked solution
The remaining composition changes after each draw and outcomes are dependent. A fixed common success probability with independent trials is not justified.
06 / A group event can become one new binary trial
Use the group-event probability, not the original trial probability.
Take three independent groups of eight independent trials from the same model. Call a group qualifying if it has more than two successes. The group success probability is q=0.44822619.
15 · Second binomial
Give the distribution of the number Z of qualifying groups.
Hint
There are three groups and one qualifying probability per group.
Worked solution
Z~B(3,q), with q=0.44822619, provided the groups are independent.
16 · Two qualifying groups
Find P(Z=2).
Hint
Include the three possible positions of the nonqualifying group.
Worked solution
3q²(1−q)≈0.33256519. Using 0.3 for q would confuse individual trials with group events.
07 / A first-success event specifies earlier failures
Do not insert a binomial coefficient automatically.
In independent trials with success probability 0.3, “first success on trial three” requires failure, failure, success. Its probability is 0.7²×0.3=0.147. By contrast, “exactly three successes in four trials” allows four arrangements and has probability 4×0.3³×0.7=0.0756.
17 · Later trials
If four trials are performed, does “first success on trial three” constrain the fourth outcome?
Hint
The first-success event has already happened.
Worked solution
No. The fourth can be either success or failure. Summing those possibilities leaves probability 0.147.
18 · Model distinction
Explain why the number of successes in four trials is not the number of trials until first success.
Hint
Their values describe different things.
Worked solution
The first is a fixed-trial count with support 0,…,4. The second is a waiting time requiring a stopping rule and may exceed four if trials continue. Define the variable explicitly.
08 / Match each calculation to its event
Check independence, support and boundaries.
Add disjoint regions and alternative tree paths; multiply conditional probabilities along a path. Validate finite distributions, translate strict inequalities carefully and identify when a group count needs a second binomial model. Keep waiting events distinct from fixed-trial success counts.
Section 1 of 8 · Define the event before choosing a formula