IB Maths AA HL
Topic 4 ā Statistics & Probability
Paper 2
~7 min read
Calculating Binomial Probabilities
Once X ~ B(n, p) is set up, every probability question is a one-line GDC call. The skill is translating exam wording into the right inequality before pressing the button. Strict inequalities convert to weak ones using P(X < k) = P(X ā¤ k ā 1).
š What you need to know
- Single value: BinomPdf(n, p, x) ā P(X = x).
- Cumulative range: BinomCdf(n, p, lower, upper) ā P(lower ā¤ X ā¤ upper).
- Upper tail: P(X ā„ k) = 1 ā P(X ā¤ k ā 1).
- Strict to weak: P(X < k) = P(X ā¤ k ā 1); P(X > k) = P(X ā„ k + 1).
- Open range: P(a < X < b) = P(a + 1 ā¤ X ā¤ b ā 1).
- Formula (booklet): P(X = r) = nCr Ā· pr Ā· (1 ā p)n ā r ā but GDC is faster.
- Always state the inequality before computing ā earns method marks even if you mistype.
- Round to 3 sf unless the question says otherwise.
Single-value probabilities
Probability of exactly r successes (booklet)
P(X = r) = nCr Ā· pr Ā· (1 ā p)n ā r
For a single value, use BinomPdf on the GDC. Different calculators name it differently ā Binomial PD, BPD, Binomial Pdf ā but it always wants n, p, and x.
Cumulative and range probabilities
For a range, use BinomCdf ā it sums P(X = i) from a lower limit to an upper limit. Most calculators take both endpoints; some take only an upper limit, in which case use the identities below.
| You want | Computation | Why |
|---|
| P(X ā¤ k) | BinomCdf(n, p, 0, k) | lower bound is 0 (no negative successes) |
| P(X ā„ k) | 1 ā P(X ā¤ k ā 1) | complement of “less than k“ |
| P(X > k) | 1 ā P(X ā¤ k) | complement of “at most k“ |
| P(a ā¤ X ā¤ b) | BinomCdf(n, p, a, b) | both endpoints included |
Strict inequalities ā weak inequalities
Because X only takes integer values, you can rewrite any strict inequality (< or >) as a weak one (ā¤ or ā„) by shifting the bound by 1.
P(X < k)
= P(X ā¤ k ā 1)
drop down by 1
P(X > k)
= P(X ā„ k + 1)
step up by 1
Open range trick: P(a < X < b) = P(a + 1 ā¤ X ā¤ b ā 1). Bump both endpoints inwards by 1 and use BinomCdf normally. Example: 5 < X < 12 becomes 6 ā¤ X ā¤ 11.
š§ Recipe ā calculate any binomial probability
- Confirm X ~ B(n, p) and write it down.
- Translate the wording into an inequality on integer values.
- Convert strict (<, >) to weak (ā¤, ā„) by shifting bounds by 1.
- Choose the GDC function: BinomPdf for “= k“, BinomCdf for ranges; complement for upper-tail.
- Round to 3 sf and check 0 ā¤ answer ā¤ 1.
Worked examples
WE 1Single value ā P(X = k)
Let X ~ B(15, 0.4). Find P(X = 7).
Identify n = 15, p = 0.4, x = 7
Use BinomPdf on the GDC
P(X = 7) = BinomPdf(15, 0.4, 7)
= 0.17708…
P(X = 7) ā 0.177 (3 sf)
by formula: Ā¹āµCā Ā· 0.4ā· Ā· 0.6āø ā but the GDC does it in one keystroke
WE 2Cumulative ā P(Y ā¤ k)
Let Y ~ B(25, 0.6). Find P(Y ā¤ 12).
Lower bound is 0, upper bound is 12
Use BinomCdf on the GDC
P(Y ā¤ 12) = BinomCdf(25, 0.6, 0, 12)
= 0.15376…
P(Y ā¤ 12) ā 0.154 (3 sf)
“at most 12” includes 12 ā boundary IS in the count
WE 3Upper tail using complement ā P(X ā„ k)
Let X ~ B(30, 0.2). Find P(X ā„ 8).
Complement: P(X ā„ 8) = 1 ā P(X ā¤ 7)
Compute P(X ā¤ 7) on the GDC
P(X ā¤ 7) = BinomCdf(30, 0.2, 0, 7) = 0.76079…
Subtract from 1
P(X ā„ 8) = 1 ā 0.76079… = 0.23920…
P(X ā„ 8) ā 0.239 (3 sf)
“at least 8” ā the bound is INCLUDED, so subtract P(X ā¤ 7), not P(X ā¤ 8)
WE 4Inclusive range ā P(a ā¤ W ā¤ b)
Let W ~ B(50, 0.45). Find P(20 ā¤ W ā¤ 28).
Both endpoints already inclusive ā just use BinomCdf
P(20 ā¤ W ā¤ 28) = BinomCdf(50, 0.45, 20, 28)
= 0.75825…
P(20 ā¤ W ā¤ 28) ā 0.758 (3 sf)
lower = 20, upper = 28 ā both included since both inequalities use ā¤
WE 5Strict-strict range ā P(a < Z < b)
Let Z ~ B(18, 0.55). Find P(5 < Z < 12).
Convert strict-strict to weak-weak: bump both inwards by 1
5 < Z < 12 ā 6 ā¤ Z ā¤ 11
Apply BinomCdf with the new bounds
P(6 ā¤ Z ā¤ 11) = BinomCdf(18, 0.55, 6, 11)
= 0.75590…
P(5 < Z < 12) ā 0.756 (3 sf)
always write the conversion line ā earns the setup mark even if calc slips
WE 6Real-world ā multiple-choice quiz with guessing
A multiple-choice quiz has 24 questions, each with 4 options. A student guesses every answer. Let G be the number of correct guesses.
(a) State the distribution of G. (b) Find P(G = 6). (c) Find P(G ā¤ 5). (d) Find P(G ā„ 10). (e) Find P(4 ā¤ G ā¤ 9).
(a) Set up ā n = 24, p = 1/4 = 0.25
G ~ B(24, 0.25)
(b) Single value
P(G = 6) = BinomPdf(24, 0.25, 6) = 0.18525… ā 0.185
(c) Cumulative
P(G ā¤ 5) = BinomCdf(24, 0.25, 0, 5) = 0.42215… ā 0.422
(d) Upper tail via complement
P(G ā„ 10) = 1 ā P(G ā¤ 9) = 1 ā 0.94533…
= 0.05466… ā 0.055
(e) Inclusive range
P(4 ā¤ G ā¤ 9) = BinomCdf(24, 0.25, 4, 9) = 0.83031… ā 0.830
(b) 0.185; (c) 0.422; (d) 0.055; (e) 0.830
scoring 10+ by guessing alone is unlikely (ā 5.5%) ā sanity check passes
š” Top tips
- Always write the inequality in clean form before pressing buttons ā earns method marks even if you mistype.
- Strict-to-weak shortcut: P(X < 5) = P(X ā¤ 4); P(X > 5) = P(X ā„ 6).
- Use the complement for upper-tail: P(X ā„ k) = 1 ā P(X ā¤ k ā 1).
- Round to 3 sf by default ā booklet convention.
- For “between a and b inclusive”, BinomCdf takes lower = a, upper = b directly.
ā Common mistakes
- Using BinomPdf when “at most” or “at least” is asked ā those need BinomCdf, not the single-value function.
- Off-by-one: P(X < 5) is NOT P(X ā¤ 5); it’s P(X ā¤ 4).
- Writing P(X ā„ k) = 1 ā P(X ā¤ k) ā the correct complement is 1 ā P(X ā¤ k ā 1).
- Mixing up “lower/upper” in the GDC’s BinomCdf ā lower goes first, not the other way round.
- Reading “no more than 5” as “fewer than 5” ā they’re different (ā¤ 5 vs < 5).
That closes the Binomial Distribution sub-topic. Up next: the Normal Distribution ā a continuous distribution where the GDC handles the heavy lifting in much the same way, but the underlying mechanics are quite different. Expect to use it for height, weight, exam scores, and any phenomenon where values cluster around a mean with diminishing tails.
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