Calculate binomial tails and intervals with exact boundaries
Translate at least, at most and strict inequalities into integer events before summing or taking a complement.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.4. A probability table needs non-negative probabilities summing to one. E(X) is a probability-weighted mean; Var(X) = E(X²) − [E(X)]², with squared units. A binomial model B(n,p) counts successes in a fixed non-negative integer number n of independent two-outcome trials with a constant success probability p. A geometric model Geo(p) counts the trial of the first success, starting at 1, with independent trials and a constant 0 < p ≤ 1. Formula proofs are not required. The expectation of the geometric distribution is required; its variance is not required by section 5.4. Normal approximations belong to section 5.5 and are not used here. Exact model probabilities do not establish real-world independence or constant rates without evidence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate exact binomial probabilities.
- Use complementary events and disjoint sums.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which values belong to 1 ≤ X ≤ 3?
What you will learn
- Calculate binomial tail and interval probabilities.
- Preserve inclusive and strict integer boundaries.
Translate the words first
For a success count, ‘at least two’ means X ≥ 2, ‘at most two’ means X ≤ 2, and ‘more than one’ also means X ≥ 2 because X is integer-valued. An interval 1 ≤ X ≤ 3 includes all three values 1, 2 and 3. List the allowed integers before selecting a calculation.
For X ~ B(4,1/3), P(X ≥ 2) = 1 − P(X = 0) − P(X = 1). The excluded probabilities are 16/81 and 32/81, so the answer is 33/81 = 11/27. Subtracting only P(X = 0) would calculate at least one instead. A complement must exclude exactly the unwanted event.
Choose the shorter valid sum
For X ~ B(5,0.40), P(1 ≤ X ≤ 3) is the sum of P(X = 1), P(X = 2) and P(X = 3): 0.2592 + 0.3456 + 0.2304 = 0.8352. These are disjoint count values, so their probabilities add. Multiplying them would describe no relevant event in this single experiment.
For X ~ B(3,1/4), P(X > 1) = P(X = 2) + P(X = 3) = 9/64 + 1/64 = 5/32. The complementary P(X ≤ 1) is 27/32. Keep exact fractions or unrounded probabilities until the end, especially if subtracting a small tail from one. Calculator commands must use the same endpoint convention as the written event.
Pause and explain
Which complement calculates P(X ≥ 2)?
Worked example
X ~ B(4,1/3). Find the probability of at least two successes.
Show the worked solution
- At least two means X ≥ 2; the complement is X = 0 or 1.
- P(X = 0) = (2/3)⁴ = 16/81 and P(X = 1) = 4(1/3)(2/3)³ = 32/81.
- Subtract both from one to obtain 33/81 = 11/27.
Answer P(X ≥ 2) = 11/27.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
X ~ B(4,1/3). Find P(X ≥ 2).
Give me a hint
Subtract the probabilities of zero and one success.
Compare my reasoning
- P(X = 0) = 16/81.
- P(X = 1) = 4(1/3)(2/3)³ = 32/81.
- P(X ≥ 2) = 1 − 48/81 = 11/27.
11/27.
Look for these in your work
- Include both excluded count values.
- Retain exact fractions until the final simplification.
X ~ B(5,0.40). Find P(1 ≤ X ≤ 3).
Give me a hint
Calculate and add the three inclusive count probabilities.
Compare my reasoning
- P(X = 1) = 5(0.4)(0.6)⁴ = 0.2592.
- P(X = 2) = 10(0.4)²(0.6)³ = 0.3456; P(X = 3) = 10(0.4)³(0.6)² = 0.2304.
- Their disjoint sum is 0.8352 = 522/625.
0.8352 = 522/625.
Look for these in your work
- Include both endpoints.
- Add mutually exclusive values rather than multiplying them.
X ~ B(3,1/4). Find P(X > 1) and its complementary probability P(X ≤ 1).
Give me a hint
A strict lower boundary at one means the counts two and three.
Compare my reasoning
- P(X = 2) = 9/64 and P(X = 3) = 1/64.
- P(X > 1) = 10/64 = 5/32.
- P(X ≤ 1) = 1 − 5/32 = 27/32.
5/32 and 27/32.
Look for these in your work
- Convert the strict boundary using the integer support.
- Check that complementary probabilities sum to one.
Common mistakes
- At least and at most include their stated value.
- Strict boundaries move to the next admissible integer.
- Subtract the complete complementary event rather than one convenient term.
What can you explain now?
For an integer count, does X > 1 mean X ≥ 2?
Compare with the explanation
Yes; there are no possible values strictly between 1 and 2.
Use the support of the variable when translating inequalities.
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Make it stick
Write the allowed integer values first, then choose a disjoint sum or its exact complement.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.