01 · More working values
Does r=8 also meet the 0.05 condition? Why is it not the answer?
Hint
P(X≥8)≈0.00278151.
Worked solution
Yes, but 7 is smaller and already works. The question asks for the smallest threshold.
Understand · explore · practise
Finding binomial cutoffs
Find binomial thresholds that meet a target probability, distinguish smallest and largest cutoffs and prove the boundary using an adjacent integer.
Before you startBinomial cumulative probabilities, tails and integer inequalities.
01 / Read what is being minimised or maximised
A probability condition can be true for several thresholds.
Find a threshold that works, then check its neighbour.
A smallest valid r requires r to work and r−1 to fail; a largest valid k requires k to work and k+1 to fail.
The adjacent check proves the boundary when the relevant probability is monotonic. Upper-tail probabilities P(X≥r) decrease as r increases; lower-tail probabilities F(k)=P(X≤k) increase as k increases.
Find the first upper-tail cutoff that worksExplore
X~B(12,0.25). Find the smallest integer r with P(X≥r)≤0.05. Choose r and test the boundary.
02 / Find a smallest upper-tail threshold
Raising the threshold makes the event smaller.
Watch: a working threshold and its failing neighbour
Pause, replay or seek freely. The notes explain the same idea and stay in view.
X~B(12,0.25). Find the smallest integer r for which P(X≥r)≤0.05.Worked example
At r=6, P(X≥6)≈0.05440223
This is greater than 0.05, so r=6 fails.
At r=7, P(X≥7)≈0.01425278
This is at most 0.05, so r=7 works.
The smallest threshold is r=7
Every lower threshold has an upper-tail probability at least as large as the failing value at 6.
02 · Cumulative conversion
Express the test P(X≥r)≤0.05 using F.
Hint
The excluded counts run through r−1.
Worked solution
1−F(r−1)≤0.05, equivalently F(r−1)≥0.95.
03 / Find a largest lower-tail cutoff
Raising a lower-tail cutoff makes its event larger.
For X~B(12,0.25), find the largest integer k with P(X≤k)≤0.20.Worked example
F(1)≈0.15838176≤0.20
k=1 works.
F(2)≈0.39067501>0.20
k=2 fails.
The largest cutoff is k=1
Higher values also fail because F cannot decrease.
03 · A smaller target
Find the largest k with F(k)≤0.05, given F(0)≈0.03167635 and F(1)≈0.15838176.
Hint
Check the two adjacent cutoffs.
Worked solution
k=0: F(0) works and F(1) fails.
04 · Wrong neighbour
A student checks k=1 works and k=0 works. Does that prove k=1 is the largest?
Hint
Which direction could contain a larger valid cutoff?
Worked solution
No. Check k=2 fails. A smaller working value does not establish maximality.
04 / Find a smallest cumulative cutoff
A cumulative target asks when the total first reaches the target.
For the same X, find the smallest k with F(k)≥0.95.Worked example
F(5)≈0.94559777<0.95
Five is too small.
F(6)≈0.98574722≥0.95
Six reaches the target.
The required cutoff is k=6
Compare this with the upper-tail threshold r=7: the complement relation uses r−1.
05 · Related cutoffs
Why are k=6 and r=7 consistent rather than contradictory?
Hint
Compare F(k) with P(X≥r).
Worked solution
P(X≥7)=1−F(6). The two conditions refer to complementary events with adjacent integer boundaries.
06 · Interpretation
What does F(6)≥0.95 say in words?
Hint
Include the endpoint six.
Worked solution
The chance of no more than six successes is at least 95%. It does not guarantee the count will be at most six.
05 / Check whether equality is allowed
Strict targets can change the answer at an exact equality.
Let Y~B(5,0.5). P(Y≥4)=6/32=0.1875 and P(Y≥5)=1/32=0.03125. These exact fractions let us distinguish a strict target without rounding.
07 · Inclusive target
Find the smallest r with P(Y≥r)≤0.1875.
Hint
At r=4 equality is allowed; P(Y≥3)=0.5.
Worked solution
r=4. It works exactly, while r=3 fails.
08 · Strict target
Find the smallest r with P(Y≥r)<0.1875.
Hint
The equality at r=4 no longer works.
Worked solution
r=5. The probability is 0.03125; r=4 fails the strict condition.
09 · Lower-tail equality
For this Y, F(0)=1/32 and F(1)=6/32. Find the largest k with F(k)<0.1875.
Hint
Equality is excluded.
Worked solution
k=0. F(0)=0.03125 works and F(1)=0.1875 fails.
06 / Search systematically and retain precision
Round the reported probability after deciding the inequality.
Use a table or calculator to bracket the boundary, then check the adjacent integers with enough precision to decide the target. For a search, record each integer and its probability. Do not round a probability to the target and assume equality.
10 · Rounded ambiguity
A table prints 0.0500. Is that enough to decide whether the underlying probability is ≤0.05?
Hint
Different values can round to the same four decimals.
Worked solution
Not always. Both 0.04996 and 0.05004 round to 0.0500, but lie on opposite sides of the target. Use more precision.
11 · Boundary proof
For an upper-tail search, r=9 works and r=8 fails. Which condition does this establish?
Hint
The tail decreases with r.
Worked solution
Nine is the smallest working threshold, provided the tested inequalities match the question.
07 / Check the allowed threshold range
Some probability targets have no solution in the specified range.
12 · Zero target
For X~B(12,0.25), can P(X≥r)=0 occur with 0≤r≤12?
Hint
Every supported count has positive probability.
Worked solution
No. At r=12 the probability is (1/4)¹²>0. If r=13 is permitted, the event is impossible and its probability is 0.
13 · Certain lower tail
What is the smallest k with F(k)=1 for that X?
Hint
The top supported count has positive mass.
Worked solution
k=12. F(11)<1, while F(12)=1.
14 · Strict impossibility
Can any threshold have P(X≥r)<0?
Hint
Probability is nonnegative.
Worked solution
No. A strict negative or zero upper bound of this form cannot be met.
08 / Working value plus failing neighbour
State the inequality, probability and integer conclusion.
Decide whether you seek a smallest or largest value. Use monotonicity to choose the correct neighbour, keep strict and inclusive targets distinct and check any stated threshold range. Give both boundary probabilities so the conclusion is justified.
Section 1 of 8 · Read what is being minimised or maximised