Exactly → PDF: Want exactly 5? Include only the bar at 5.
Choose the wording and move the count to see which probabilities to include.
Interactive diagram
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Exactly 5
Use the highlighted bars above.
Step by step
- X ~ B(20, 0.3). Find P(X = 5).
- Use binompdf(20,0.3,5).
Final answer
P(X = 5) ≈ 0.179.
What do I enter on the GDC?
binompdf(n,p,r)
n = attempts; p = success chance; r = exact count.
Example: (20, 0.3, 5). Show this setup in your workings.
Do I need the combinations formula?
The formula combines all possible success/failure arrangements.
For the exam, use binompdf on your GDC. Manual factorial calculations are not needed.
GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
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At most includes; fewer than excludes: At most 3: keep 0–3. Fewer than 3: keep 0–2.
Choose the wording and move the count to see which probabilities to include.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
At most 3
Use the highlighted bars above.
Step by step
- X ~ B(20, 0.3). Keep 0, 1, 2 and 3.
- Use binomcdf(20,0.3,3).
Final answer
P(X ≤ 3) ≈ 0.107.
Fewer than 7: which boundary?
Exclude 7, so the largest allowed count is 6.
P(X < 7) = binomcdf(20,0.3,6) ≈ 0.608.
What does CDF do?
Adds probabilities from 0 up to your boundary, including it.
Example: CDF(3) includes 0, 1, 2 and 3.
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Keep the right-hand end: At least 7: keep 7–20. More than 7: keep 8–20.
Choose the wording and move the count to see which probabilities to include.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
At least 7
Use the highlighted bars above.
Step by step
- X ~ B(20, 0.3). Remove 0–6 to keep 7–20.
- Use 1 − binomcdf(20,0.3,6).
Final answer
P(X ≥ 7) ≈ 0.392.
More than 7: what changes?
Exclude 7, so remove 0–7.
1 − binomcdf(20,0.3,7) ≈ 0.228.
More than 5 example
Keep 6–20; remove 0–5.
1 − binomcdf(20,0.3,5) ≈ 0.584.
Why “1 minus”?
All counts together have probability 1.
Subtract the counts you do not want; the right-hand tail remains.
Between: keep the middle: Between 3 and 8 inclusive means keep 3, 4, 5, 6, 7 and 8.
Choose the wording and move the count to see which probabilities to include.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Between 3 and 8
Use the highlighted bars above.
Step by step
- X ~ B(20, 0.3). CDF(8) includes 0–8.
- Remove 0–2: binomcdf(20,0.3,8) − binomcdf(20,0.3,2).
Final answer
P(3 ≤ X ≤ 8) ≈ 0.851.
Why subtract CDF(2), not CDF(3)?
You want to keep 3. Remove only counts below 3: 0, 1 and 2.
General rule
For a ≤ X ≤ b:
CDF(b) − CDF(a − 1).
Example: 3–8 → CDF(8) − CDF(2).
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Choose the wording before the command: PDF = one exact count. CDF = counts from 0 up to a boundary.
| Exam words | Means | TI-84 command |
|---|---|---|
| Exactly 7 | X = 7 | binompdf(n,p,7) |
| At most 7 | X ≤ 7 | binomcdf(n,p,7) |
| Fewer than 7 | X ≤ 6 | binomcdf(n,p,6) |
| At least 7 | X ≥ 7 | 1 − binomcdf(n,p,6) |
| More than 7 | X ≥ 8 | 1 − binomcdf(n,p,7) |
| Between 3 and 7 inclusive | 3 ≤ X ≤ 7 | binomcdf(n,p,7) − binomcdf(n,p,2) |
At least one?
Keep 1 and above; remove 0.
1 − binomcdf(n,p,0).
How can I remember the boundary?
Includes the number: at most / at least.
Excludes the number: fewer than / more than.
A game with two targets: There are two targets: A and B. In each round, take one shot at each. Hit both to win the round. Chance of hitting A: 80%. Chance of hitting B: 90%. The shots are independent.
One round: hit target A AND target B
You take one shot at each target. You win only if both shots hit.
- Chance of winning: 0.8 × 0.9 = 0.72 (72%).
- Hitting just one target is not a win.
Play 12 rounds
Each round is independent, with the same 72% chance of winning.
- X = number of rounds you win.
- X ~ B(12, 0.72): 12 rounds, 0.72 chance of winning each round.
What is the chance of winning at least 10 rounds?
“At least 10” means 10, 11 or 12 wins.
- Keep: 10, 11, 12 wins.
- Remove: 0, 1, 2, …, 9 wins.
Find the chance on your calculator
You play 12 rounds. Find the chance of winning at least 10.
Step by step
- binomcdf(12,0.72,9) gives the chance of 0–9 wins. These are the counts we do not want.
- Subtract from 1 to keep 10–12 wins: 1 − binomcdf(12,0.72,9).
Final answer
About 0.304, or a 30.4% chance.
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A coin game: Play 40 rounds. Each round has 10 independent tosses, with a 65% chance of heads each toss. Win with 6 or more heads. How many rounds do we expect to win?
One round = 10 tosses
5 heads + 5 tails is a tie. 6–10 heads wins.
- H counts heads: H ~ B(10, 0.65).
Find the chance of winning one round
Remove 0–5 heads to keep 6–10. That is why the calculator uses 5.
- 1 − binomcdf(10,0.65,5) ≈ 0.7515.
- So one round has about a 75.15% chance of winning.
40 rounds × chance of winning
40 × 0.751495… ≈ 30.1. Use the unrounded calculator value.
- About 30 wins on average. Your actual number of wins can vary.
Why not 40 × 0.65?
0.65 is the chance of one head. We need the chance of winning a whole round, about 0.7515.
GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Find the smallest number, step by step
Read the question
Take 60 independent shots, each with a 70% chance of scoring. Find the smallest number to fill the blank:
“There is less than a 30% chance of scoring ___ or more goals.”
Choose a starting point
Average = 60 × 0.7 = 42 goals. Start by trying 42. It is a starting point, not the answer.
Try the simulator below
Select At least. Slide 42 → 43 → 44 → 45.
42+ → 56.3%
43+ → 45.1%
44+ → 34.2%
45+ → 24.4% — stop!
Answer: 45 goals
45 is the first number that works. 44 is still above 30%, so 45 is the smallest answer.
Start → calculate → increase → stop below 30%
Start at 42. Move the slider to 43, 44, then 45. Watch the chance decrease.
Interactive diagram
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How is 56.3% calculated?
1 − binomcdf(60,0.7,41) ≈ 0.563 = 56.3%.
CDF(41) adds the chance of 0–41 goals. Subtract from 1 to keep 42–60 goals.
Largest goal count: use At most
Find the largest number to fill this blank:
“Scoring ___ goals or fewer has less than a 70% chance.”
In the simulator above, select At most. Try 43, then 44.
43 or fewer → 65.8% ✅ below 70%.
44 or fewer → 75.6% ❌ above 70%.
Answer: r = 43 goals. The probability is 65.8%.
Calculator: binomcdf(60,0.7,43) adds the chances of 0–43 goals.
GDC walkthrough
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Continue the target game from section 6
Same game, one new question
Two targets: A (80% hit chance) and B (90%). One shot at each; hit both to win a round. The shots are independent.
12 independent rounds = one game. Win the game with 10 or more round wins.
Section 6 found the chance of winning one game. Now play 25 games: how many games do we expect to win?
Recap: chance of winning one round
Hit both: 0.8 × 0.9 = 0.72.
One round has a 72% chance of a win.
Recap: chance of winning one game
Count round wins: X ~ B(12, 0.72).
Keep 10, 11, 12 wins; remove 0–9.
1 − binomcdf(12,0.72,9) ≈ 0.304.
One game has about a 30.4% chance of a win.
New step: expected wins across 25 games
25 × 0.303706… ≈ 7.59 games.
About 7.6 games won on average. Use the unrounded calculator value.
One round → one game → 25 games
Why not 25 × 0.72?
0.72 is the chance of winning a round.
We repeat whole games, so use the chance of winning a game: about 0.304.
Extra: given at least 8 round wins
Chance of 8–10 wins, given at least 8:
[binomcdf(12,0.72,10) − binomcdf(12,0.72,7)] ÷ [1 − binomcdf(12,0.72,7)] ≈ 0.858.
Extra: wins in a row
Exactly 9 wins can be in any order. 9 consecutive wins means 9 in a row.
In 10 rounds with a 60% win chance, exactly 9 wins in a row has two orders: loss first or loss last.
2 × 0.6⁹ × 0.4 ≈ 0.00806.
AA exam practice: a success on the final trial
“The third success occurs on trial five” fixes the last outcome. Count two successes in the first four trials, then multiply by the chance of success on trial five.
Worked example [4 marks]
Independent trials each have success probability 0.4. Find the probability that the third success occurs on the fifth trial.
Step by step
- The fifth trial must be a success, and exactly two of the preceding four must succeed.
- There are six possible orders for those first four trials.
- On a TI-84: binompdf(4,0.4,2) × 0.4. Keep the unrounded value until the final answer.
Final answer
0.138 (3 s.f.).
Try the next question before opening its solution.
Your turn [4 marks]
Independent trials each have success probability 0.3. Find the probability that the second success occurs on the fourth trial.
Step by step
- Fix the final success, then count one success in the preceding three trials.
- Or use binompdf(3,0.3,1) × 0.3.
Final answer
0.132 (3 s.f.).
Exam tip: This is not just “three successes in five trials”: that would include orders whose third success occurs earlier. The GDC calculates the binomial part.
Additional AA exam practice
IB-style question — set it up
A biased coin lands heads with probability 0.4.
It is tossed 12 times.
State the distribution of the number of heads X.
Step by step
- Fixed trials n, constant p, two outcomes.
- Write the model.
Final answer
X ~ B(12, 0.4).
IB-style question — exactly k
X ~ B(10, 0.3).
Find P(X = 4).
Step by step
- Exactly 4 successes → binompdf.
- Evaluate on the GDC.
Final answer
P(X = 4) ≈ 0.200.
IB-style question — at most / at least
X ~ B(10, 0.3).
Find P(X ≤ 2) and P(X ≥ 3).
Step by step
- At most 2 → binomcdf.
- At least 3 is the complement.
Final answer
P(X ≤ 2) ≈ 0.383; P(X ≥ 3) ≈ 0.617.
IB-style question — conditional in a binomial
X ~ B(10, 0.4). Find P(X ≥ 5 | X ≥ 3).
Step by step
- Conditional probability: the X ≥ 5 case sits inside the X ≥ 3 case, so the numerator is just P(X ≥ 5).
- Each from 1 − binomialcdf on the GDC.
Final answer
≈ 0.441.
IB-style question — find the number of trials
A machine produces an item that is faulty with probability 0.1.
How many items must be inspected so that the probability of finding at least one faulty item is at least 0.9?
Step by step
- P(at least one) = 1 − P(none) ≥ 0.9.
- Take logs and solve.
Final answer
At least 22 items must be inspected.
IB-style question — exactly k
X ~ B(8, 0.25).
Find P(X = 2), giving your answer to 3 significant figures.
Step by step
- Exactly 2 successes → binompdf(n, p, k).
- Evaluate on the GDC.
Final answer
P(X = 2) ≈ 0.311.