01 · No more than
Find P(no more than three successes).
Hint
No more than includes three.
Worked solution
F(3)=0.99328.
Understand · explore · practise
Binomial tails and intervals
Translate at least, at most and strict inequalities into binomial events, then use complements and differences of cumulative probabilities without off-by-one errors.
Before you startCumulative binomial probabilities F(k)=P(X≤k).
01 / Translate words into included integers
Write the qualifying counts before choosing a formula.
Let F(k)=P(X≤k).
The endpoint is included. Every complement or subtraction must respect that definition.
Throughout the examples, X~B(5,0.2). Its cumulative totals are F(0)=0.32768, F(1)=0.73728, F(2)=0.94208, F(3)=0.99328, F(4)=0.99968 and F(5)=1.
Which counts qualify?Explore
X~B(5,0.2). Select an event and compare its included counts with its cumulative formula.
02 / At most versus fewer than
A strict upper bound excludes its endpoint.
Compare at most two with fewer than two successes.Worked example
At most two means X≤2
Use F(2)=0.94208.
Fewer than two means X<2, or X≤1
Use F(1)=0.73728.
The difference is the mass at 2
P(X=2)=0.2048.
02 · Less than
Find P(fewer than three successes).
Hint
Include only 0,1,2.
Worked solution
F(2)=0.94208.
03 / Complement the excluded lower counts
At least r starts at r, so exclude counts through r−1.
Watch: the complement stops one count earlier
Pause, replay or seek freely. The notes explain the same idea and stay in view.
P(X≥r)=1−F(r−1); P(X>r)=1−F(r).
These expressions assume integer r. First translate noninteger wording into included integer counts.
Find P(at least two successes).Worked example
At least two includes 2,3,4,5
Its complement is 0 or 1.
P(X≥2)=1−F(1)
Subtract 0.73728 from 1.
P(X≥2)=0.26272
Using 1−F(2) would wrongly exclude the mass at 2.
03 · More than two
Find P(X>2).
Hint
The complement is X≤2.
Worked solution
1−F(2)=0.05792.
04 · At least one
Find P(X≥1).
Hint
Only zero successes is excluded.
Worked solution
1−F(0)=0.67232.
05 · Explain the error
A student uses 1−F(3) for P(X≥3). What event did they calculate?
Hint
The complement of X≤3 is X>3.
Worked solution
P(X>3)=P(X≥4). The correct result is 1−F(2)=0.05792.
04 / Subtract the part below the interval
An inclusive lower endpoint stays in the answer.
P(a≤X≤b)=F(b)−F(a−1).
For integers a≤b, the first total includes 0 through b; the second removes 0 through a−1.
Find P(2≤X≤4).Worked example
F(4) includes 0,1,2,3,4
We need 2,3,4 only.
Subtract F(1), which includes 0,1
Do not subtract F(2): that would also remove 2.
0.99968−0.73728=0.26240
Direct check: 0.2048+0.0512+0.0064.
06 · Inclusive interval
Find P(1≤X≤3).
Hint
Use F(3)−F(0).
Worked solution
0.99328−0.32768=0.66560.
07 · One-point interval
Find P(2≤X≤2) using cumulative totals.
Hint
Subtract neighbouring totals.
Worked solution
F(2)−F(1)=0.2048.
05 / Translate both interval endpoints
The two signs can require different integer adjustments.
08 · Mixed signs
Find P(2<X≤4).
Hint
The qualifying counts are 3 and 4.
Worked solution
F(4) − F(2) = 0.99968 − 0.94208 = 0.05760.
09 · Both strict
Find P(1<X<4).
Hint
Only 2 and 3 qualify.
Worked solution
F(3) − F(1) = 0.99328 − 0.73728 = 0.25600.
10 · Noninteger bounds
Find P(1.4<X≤3.8).
Hint
List the possible integer counts.
Worked solution
Only 2 and 3 qualify, so F(3)−F(1)=0.25600.
06 / Check empty and certain events
A formula should agree with the support.
11 · At least zero
Find P(X≥0).
Hint
Every possible count qualifies.
Worked solution
1. Equivalently 1−F(−1)=1−0.
12 · Beyond the maximum
Find P(X>5).
Hint
At most five successes can occur.
Worked solution
0. Equivalently 1−F(5)=0.
If an interval contains no possible integer values, its probability is zero. Avoid applying an ordered-interval subtraction to reversed endpoints without first checking the event.
07 / Match the count to the wording
The counted event may be an unwanted outcome.
13 · Defect limit
Five independent items each have defect probability 0.2. A pack is acceptable if it contains fewer than two defective items. Find its acceptance probability.
Hint
Let X count defects. Accept when X≤1.
Worked solution
F(1)=0.73728, under the stated assumptions.
14 · Rejection
Under the same rule, find the rejection probability.
Hint
Reject when there are at least two defects.
Worked solution
1−F(1)=0.26272. Acceptance and rejection sum to 1.
08 / Draw the boundary before calculating
Use a short count list to prevent off-by-one mistakes.
For lower tails, identify the last included integer. For upper tails, complement the excluded lower counts. For intervals, take the cumulative total through the upper endpoint and subtract only the counts below the lower endpoint. Check the result is between zero and one.
Section 1 of 8 · Translate words into included integers