Exactly → PDF: Want exactly 5? Include only the bar at 5.
Choose the wording and move the count to see which probabilities to include.
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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.
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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
Step through the exact calculator keystrokes, screen by screen, in study mode.
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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.
Conditional binomial counts and consecutive outcomes
Given means restrict the outcomes: For P(A | B), divide the probability of A and B by P(B). When A is already inside B, the numerator is just P(A).
A count, given a lower bound
X ~ B(8, 0.35). Given at least two successes, find the probability of at least four successes.
Step by step
- Every outcome X ≥ 4 is also an outcome X ≥ 2.
- Use two upper tails on the GDC.
Final answer
About 0.353. The denominator changes because we have been told at least two successes occurred.
Consecutive is a different instruction: If each trial succeeds independently with probability 0.35, the chance that the next four specified trials all succeed is 0.35⁴ = 0.01500625. Do not use the binomial “any four positions” factor when the positions are already fixed. A run somewhere in a longer sequence needs a separate analysis of possible overlaps.
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Find an unknown group size from a binomial probability
Find the number of counters
A bag contains N counters, four of them gold. A counter is drawn and replaced five times. The probability of no gold results is 0.32768. Find N.
Step by step
- Replacement keeps the gold probability constant.
- No gold means all five draws are not gold.
- Solve with the GDC, or take the positive fifth root.
Final answer
N = 20. Check with binompdf(5,4/20,0) = 0.32768. The answer must be a whole-number population size of at least 4.
A Poisson probability becomes a binomial success chance
Cross-topic HL connection: The Poisson model is taught in topic 4.17. Here use it only to find the probability for one interval, then count independent intervals with a binomial model.
One hour: a Poisson count
Y ~ Poisson(2). A quiet hour has Y = 0.
Turn the event into a chance
p = P(Y = 0) = e⁻² ≈ 0.1353. Keep full calculator precision.
Five independent hours: a binomial count
X counts quiet hours: X ~ B(5, p).
At least one quiet hour
Calls arrive at a uniform average rate of two per hour and counts in disjoint hours are independent. Find the chance that at least one of the next five hours has no calls.
Step by step
- A success for X means zero calls in one hour.
- Subtract the chance of no quiet hours.
Final answer
About 0.517. The Poisson mean is 2 calls per hour; the binomial success chance is e⁻², not 2.
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A binomial tail used in an HL hypothesis test
Cross-topic HL connection: Formal hypothesis testing is taught in topic 4.18. The calculation here reuses the binomial upper tail; the alternative hypothesis determines which tail is needed.
Evidence of improvement?
A method previously succeeded with probability 0.5. An independent trial of 20 attempts gives 15 successes. Test H₀: p = 0.5 against H₁: p > 0.5 at the 5% significance level.
Step by step
- Under H₀, count successes with the null probability.
- Results at least as high as 15 form the upper tail.
- Compare the p-value with the significance level.
Final answer
Reject H₀. There is evidence at the 5% level that the success probability has increased. This does not prove the new method always works.