Binomial probabilities and cumulative probabilities | Year 12

01 / LEARN

The idea, step by step

One exact count

For $X \sim Bin (n, p)$ , $P (X = k) = (\begin{matrix} n \\ k \end{matrix}) p^{k} {(1 - p)}^{n - k}$ for k = 0, 1, …, n. One specified sequence with k successes has probability $p^{k} {(1 - p)}^{n - k}$ ; the combination counts the different positions for those successes.

The power on failure probability is n − k, not n. All valid counts together form a full probability distribution. When p is 0 or 1, handle the certain outcome directly rather than relying on a calculator convention for 0⁰.

Cumulative probability and careful endpoints

“At most r” means $X \leq r$ ; “at least r” means $X \geq r$ , whose complement is $X \leq r - 1$ . A probability between a and b inclusive is $P (X \leq b) - P (X \leq a - 1)$ .

On technology, use the binomial probability function for an exact value and the cumulative function for a range beginning at zero. State the model and the expression as well as the numerical answer so that the endpoint choice can be checked.

A common trap

At least 3 is 1 − P(X ≤ 3).

Instead: That removes X = 3 as well. At least 3 is 1 − P(X ≤ 2).

02 / WORKED EXAMPLES

See the thinking

Example 1

Exactly two successes

$X \sim Bin (8, 0.3)$ . Find P(X = 2).

$P (X = 2) = (\begin{matrix} 8 \\ 2 \end{matrix}) {(0.3)}^{2} {(0.7)}^{6}$ .
The combination is 28, so the product is 0.29647548.
Round only the final result to four decimal places.

P(X = 2) ≈ 0.2965.

Example 2

At least one is easier by complement

$X \sim Bin (5, 0.2)$ . Find P(X ≥ 1).

The only excluded value is X = 0.
P(X = 0) = 0.8⁵ = 0.32768.
Subtract from 1: 1 − 0.32768 = 0.67232.

P(X ≥ 1) = 0.67232.

Check your understanding

For integer-valued X, how do you express P(X > 4) using a cumulative probability?

Show answer

$1 - P (X \leq 4)$ .

Strictly above 4 is the complement of at most 4.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

$X \sim Bin (4, 1 / 2)$ . Find P(X = 2).
2 marks
$X \sim Bin (3, 0.2)$ . Find P(X = 0).
2 marks
$X \sim Bin (4, 0.1)$ . Find P(X ≥ 1).
2 marks
$X \sim Bin (5, 1 / 2)$ . Find P(X ≤ 1).
2 marks
$X \sim Bin (4, 1 / 2)$ . Find P(1 ≤ X ≤ 3).
2 marks
$X \sim Bin (10, 0.3)$ . Write a cumulative-probability expression for P(3 ≤ X ≤ 7).
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

$X \sim Bin (4, 1 / 2)$ . Find P(X = 2).
2 marks
$(\begin{matrix} 4 \\ 2 \end{matrix}) {(1 / 2)}^{4} = 6 / 16 = 3 / 8$ .
$3 / 8$ .
$X \sim Bin (3, 0.2)$ . Find P(X = 0).
2 marks
All three trials fail: 0.8³ = 0.512.
0.512.
$X \sim Bin (4, 0.1)$ . Find P(X ≥ 1).
2 marks
Use 1 − P(X = 0).
1 − 0.9⁴ = 0.3439.
0.3439.
$X \sim Bin (5, 1 / 2)$ . Find P(X ≤ 1).
2 marks
$((\begin{matrix} 5 \\ 0 \end{matrix}) + (\begin{matrix} 5 \\ 1 \end{matrix})) / 2^{5} = 6 / 32 = 3 / 16$ .
$3 / 16$ .
$X \sim Bin (4, 1 / 2)$ . Find P(1 ≤ X ≤ 3).
2 marks
The excluded values are 0 and 4.
1 − 1/16 − 1/16 = 14/16 = 7/8.
$7 / 8$ .
$X \sim Bin (10, 0.3)$ . Write a cumulative-probability expression for P(3 ≤ X ≤ 7).
2 marks
The total to 7 includes all desired values.
Remove the values 0, 1 and 2, not the total to 3.
$P (X \leq 7) - P (X \leq 2)$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.

Official curriculum source ↗

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