01 · Complement of qualification
Find the chance that one group does not qualify.
Hint
Complement the whole two-success event.
Worked solution
1−27/128=101/128. Nonqualification includes 0,1,3 or4 individual successes.
Understand · explore · practise
Repeated binomial experiments
Calculate a within-group binomial probability, use it as the success chance for independent repeated groups and distinguish a binomial count from a fixed offset plus a random count.
Before you startBinomial point probabilities, complements and independence.
01 / Define two different random variables
A successful group is a new event.
First calculate q=P(a group qualifies).
Then, for m independent identical groups, the number Y of qualifying groups is B(m,q).
The inner distribution counts individual successes within one group. The outer distribution counts groups satisfying a chosen rule. Their trial totals and success probabilities usually differ.
Count groups, not individual successesExplore
Each group contains four independent trials with success chance 1/4. A group qualifies if it has exactly two successes. Five groups are independent. Y counts qualifying groups.
Group probability q=27/128; Y~B(5,q). The twenty individual trials form a different count.
02 / Calculate the probability of one qualifying group
The group rule determines q.
Watch: individual trials become one group outcome
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Four independent trials each have success probability 1/4. A group qualifies exactly when two succeed.Worked example
Let X count successes within one group
X~B(4,1/4).
q=P(X=2)=6×(1/4)²×(3/4)²
The six orders are disjoint.
q=27/128=0.2109375
Keep this exact value for the outer calculation.
02 · Change the rule
If a group instead qualifies when at least one trial succeeds, what is q?
Hint
Complement four failures.
Worked solution
1−(3/4)⁴=175/256. Changing the group rule changes the outer probability.
03 / Build the outer binomial distribution
There are five group trials, each with probability q.
Five independent groups follow the original exactly-two rule. Find P(exactly two groups qualify).Worked example
Y~B(5,27/128)
The outer success is a qualifying group, not one individual success.
P(Y=2) = 10 × (27/128)² × (101/128)³
Choose which two of the five groups qualify.
P(Y=2)≈0.218595794
Do not replace q with the original individual chance 1/4.
03 · Outer inputs
State the n and p for the outer distribution.
Hint
Count the groups and use their qualification chance.
Worked solution
n=5 and p=27/128.
04 · All groups
Find the probability that all five groups qualify.
Hint
Every outer trial succeeds.
Worked solution
(27/128)⁵≈0.000417608.
05 · At least one group
Find P(Y≥1).
Hint
Complement five nonqualifying groups.
Worked solution
1−(101/128)⁵≈0.694115817.
04 / Do not replace the group count with a total count
Two qualifying groups does not specify the total number of individual successes.
06 · Twenty trials
Let Z count all successes among the twenty independent individual trials. State its distribution.
Hint
All individual probabilities are 1/4.
Worked solution
Z~B(20,1/4). This is not Y, which counts five groups meeting the exactly-two rule.
07 · Same total, different groups
Compare within-group counts (2,2,0,0,0) and (4,0,0,0,0). What are Z and Y in each?
Hint
Both have four individual successes.
Worked solution
First: Z=4,Y=2. Second: Z=4,Y=0. Knowing Z does not determine Y.
08 · Stop on qualification
If groups are tried until the first qualifying group, is the number of groups tried B(5,q)?
Hint
Is the number of group trials fixed at five?
Worked solution
No. A stopping count is a different variable. Define the stopping rule and any cap before constructing its distribution.
05 / Check assumptions at both levels
A correct inner calculation does not establish outer independence.
09 · Shared conditions
Groups share an uncertain environmental condition that affects all their results. Is the outer binomial automatically justified?
Hint
The qualification events may be linked.
Worked solution
No. Independence between groups must be justified separately. Common marginal qualification probabilities are not sufficient.
10 · Different group sizes
Some groups have four trials and others have six, with the same exactly-two rule. Can one common q be assumed?
Hint
The inner trial count changes.
Worked solution
No. Calculate each group probability; a single binomial outer model requires a common q as well as independence.
Retain full precision for q. Rounding q before raising it to powers changes the final answer; round only the final reported result.
06 / Separate certain marks from guessed marks
A fixed contribution shifts the support.
A ten-question quiz awards one mark per correct answer. Three answers are certainly known; the remaining seven are guessed independently from four equally likely choices.Worked example
Let G count correct guesses
G~B(7,1/4).
Total score T=3+G
T takes integer values 3 through 10. T is not B(10,1/4).
T≥6 means G≥3
P(T≥6) = P(G≥3) = 3991/16384 ≈ 0.243591309.
11 · Exactly five marks
Find P(T=5).
Hint
Translate to G=2.
Worked solution
21 × (1/4)² × (3/4)⁵ = 5103/16384 ≈ 0.311462402.
12 · Impossible score
Under the stated assumptions, what is P(T=2)?
Hint
Three marks are already certain.
Worked solution
0. The minimum total is 3.
07 / Name the event at each stage
Check whether you are counting successes, groups or a shifted total.
13 · Incorrect outer p
A solution uses Y~B(5,1/4) for qualifying groups. Identify the error.
Hint
One group contains four individual trials.
Worked solution
The outer success chance is P(X=2)=27/128, not the chance of one individual success.
14 · Incorrect quiz model
Why is T~B(10,1/4) inappropriate in the quiz example?
Hint
The three known answers do not have success probability 1/4.
Worked solution
There is no common success probability across all ten answers. Model the seven random guesses and add the three certain marks.
08 / Inner event, outer count, separate assumptions
Keep the variable definitions beside the calculations.
Find q for a single group, justify independent identical repetitions and then count qualifying groups with B(m,q). If a result includes a fixed contribution, subtract it from the target before using the random-count distribution.
Section 1 of 8 · Define two different random variables