01 · Another target
For p=0.25, express a target of at least 90% as a bound on no successes.
Hint
Complement both the event and target.
Worked solution
0.75ⁿ≤0.10.
Understand · explore · practise
Finding a required number of trials
Find the minimum number of independent trials needed for a target chance of at least one success, using complements, logarithms and adjacent-integer checks.
Before you startBinomial zero-success probabilities, logarithms and inequalities.
01 / Make the trial number the unknown
At least one success is the complement of no successes.
P(X≥1)=1−(1−p)ⁿ for X~B(n,p).
This assumes n independent trials with the same success chance p.
The unknown is n, not the success count. The complement avoids adding probabilities for one, two, three and all the other possible success counts.
How many independent attempts?Explore
Each attempt succeeds with probability 0.2. Find the smallest positive integer n giving at least a 95% chance of one or more successes.
02 / Convert the target into a failure bound
A high success target leaves a small probability for no successes.
Each attempt succeeds independently with probability 0.2. Require P(at least one success)≥0.95.Worked example
No success in n attempts has probability 0.8ⁿ
All n attempts fail.
1−0.8ⁿ≥0.95
This is the required success probability.
0.8ⁿ≤0.05
The chance of complete failure must be at most 5%.
02 · One attempt
For p=0.2 and n=1, what is P(at least one success)?
Hint
Use the complement or the stated success chance.
Worked solution
1−0.8=0.2.
03 / Divide by a negative logarithm
The inequality reverses because log(1−p) is negative.
Watch: the first integer crossing the target
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Solve 0.8ⁿ≤0.05.Worked example
n log(0.8)≤log(0.05)
Log is increasing, so this step preserves the inequality.
n≥log(0.05)/log(0.8)≈13.425
Dividing by the negative number log(0.8) reverses it.
The first possible integer is 14
Now verify the original probabilities at 13 and 14.
03 · Wrong direction
A student writes n≤13.425 after dividing by log(0.8). Explain the error.
Hint
What sign does the divisor have?
Worked solution
log(0.8)<0, so division reverses ≤ to ≥.
04 · Log base
Must these be natural logarithms?
Hint
Use the same base in the numerator and denominator.
Worked solution
No. Any standard logarithm base greater than 1 gives the same ratio. Natural or base-10 logarithms both work.
04 / Check the adjacent trial counts
The rounded-up candidate still needs a boundary check.
Use unrounded probabilities.Worked example
At n=13: 1−0.8¹³≈0.94502442
This is below 0.95, so thirteen fails.
At n=14: 1−0.8¹⁴≈0.95601953
This reaches the target.
The minimum is n=14
As n increases, 0.8ⁿ decreases and the success probability increases.
05 · A 90% target
For p=0.25, show that the minimum n for P(X≥1)≥0.90 is 9.
Hint
Compare n=8 and n=9.
Worked solution
At 8, 1−0.75⁸≈0.89988708<0.90. At 9, 1−0.75⁹≈0.92491531≥0.90. Thus 9 is the minimum.
06 · Reliability target
For p=0.1, find the minimum n for P(X≥1)≥0.99.
Hint
Solve 0.9ⁿ≤0.01 and check neighbours.
Worked solution
The log ratio is approximately 43.709. At 43 the success probability is approximately 0.98922474; at 44 it is 0.99030226. The minimum is 44.
05 / An exact equality may be excluded
Distinguish at least the target from more than the target.
For independent fair trials, P(at least one success in three trials)=1−(1/2)³=7/8 exactly.
07 · Inclusive target
Find the minimum positive n giving a probability of at least 7/8 when p=1/2.
Hint
Equality is permitted.
Worked solution
n=3. Two trials give 3/4, which fails.
08 · Strict target
Find the minimum positive n giving a probability greater than 7/8.
Hint
Three gives equality only.
Worked solution
n=4, giving 15/16. For a strict inequality n>a, an integer value a itself is excluded.
09 · Rounding rule
Is “always take the ceiling of the log ratio” valid for a strict probability target?
Hint
What if the exact ratio is an integer?
Worked solution
No. For n>a, use the smallest integer strictly greater than a. If a is an integer, that is a+1. Always check the original condition.
06 / More trials help only under the model
The formula relies on independence and a common success chance.
10 · Shared obstruction
All attempts fail whenever one shared obstruction is present. Can 1−(1−p)ⁿ be used from the marginal p alone?
Hint
The failure events may be linked.
Worked solution
No. Independence needs justification; a shared cause can stop extra attempts improving the chance as predicted.
11 · Changing chances
Independent attempts have success chances p₁,p₂,…,pₙ. Write the chance of at least one success.
Hint
Multiply their individual failure probabilities.
Worked solution
1−(1−p₁)(1−p₂)…(1−pₙ). A single power applies only when all p values are equal.
This calculation chooses a fixed number of trials to perform. It does not say that the first success will occur on the last trial, and it does not guarantee success.
07 / Treat impossible and certain cases separately
The logarithmic rearrangement assumes 0<p<1.
12 · Impossible success
If p=0, can any finite n reach a positive success target?
Hint
Each trial always fails.
Worked solution
No. The chance of at least one success is zero.
13 · Certain success
If p=1 and at least one trial is allowed, what minimum positive n reaches a target of 0.95?
Hint
One trial succeeds certainly.
Worked solution
n=1. The logarithmic formula is not needed.
14 · A 100% target
For 0<p<1, can a finite n give P(at least one success)=1 exactly?
Hint
The all-failure probability remains positive.
Worked solution
No. It approaches 1 as n grows but stays strictly below 1 for every finite n.
08 / Complement, solve, then check integers
Report the minimum with evidence from both sides.
Write the target probability, use the complement of no successes and take logs only when 0<p<1. Reverse the inequality when dividing by log(1−p). Test the candidate and its predecessor without premature rounding, and state the independence assumption.
Section 1 of 8 · Make the trial number the unknown