Binomial probabilities and cumulative probabilities | Year 12

01 / LEARN

Understand the idea.

One exact count

For $X \sim Bin (n, p)$ , an exact count $k$ has probability $(\begin{matrix} n \\ k \end{matrix}) p^{k} {(1 - p)}^{n - k}$ . The powers describe one sequence with $k$ successes and $n - k$ failures; the combination counts all sequences with that count.

Check that $k$ is an integer from $0$ to $n$ . The exponents must add to the total number of trials. For $p = 0$ or $p = 1$ , describe the certain count directly instead of relying on a convention for $0^{0}$ .

Cumulative probability and careful endpoints

Translate wording before using cumulative technology. At most $r$ means $X \leq r$ ; at least $r$ means $X \geq r$ . Its complement is $X \leq r - 1$ , which leaves the required endpoint $r$ included.

For inclusive integer bounds, use $P (a \leq X \leq b) = P (X \leq b) - P (X \leq a - 1)$ . Record this expression with the model and numerical result so another reader can check both the input values and the chosen endpoints.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Exactly two successes

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

Worked solution

1
Substitute the exact count into the binomial formula.
$P (X = 2) = (\begin{matrix} 8 \\ 2 \end{matrix}) {(0.3)}^{2} {(0.7)}^{6}$
Two successes leave six failures, and their positions can vary.
2
Evaluate the combination and product.
$\begin{aligned} (\begin{matrix} 8 \\ 2 \end{matrix}) & = \frac{8!}{2! 6!} = 28 \\ P (X = 2) & = 28 (0.09) (0.117649) \\ = 0.29647548 \end{aligned}$
All twenty-eight disjoint placements have the same sequence probability.
3
Round the final probability.
$P (X = 2) \approx 0.2965$
Keeping the full product avoids intermediate rounding error.

Answer

P(X = 2) ≈ 0.2965.

EXAMPLE 02

At least one is easier by complement

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

Worked solution

1
Use the complement of no successes.
$P (X \geq 1) = 1 - P (X = 0)$
Only the count zero is excluded by at least one.
2
Find the no-success probability.
$P (X = 0) = (\begin{matrix} 5 \\ 0 \end{matrix}) {(0.2)}^{0} {(0.8)}^{5} = 0.32768$
All five independent trials must fail.
3
Subtract from one.
$P (X \geq 1) = 1 - 0.32768 = 0.67232$
The complementary events together have total probability one.

Answer

P(X ≥ 1) = 0.67232.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

The event $X > 4$ excludes exactly $X \leq 4$ , so $P (X > 4) = 1 - P (X \leq 4)$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

$X \sim Bin (4, \frac{1}{2})$ . 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 $P (X \geq 1)$ .
2 marks
$X \sim Bin (5, \frac{1}{2})$ . Find $P (X \leq 1)$ .
2 marks
$X \sim Bin (4, \frac{1}{2})$ . Find $P (1 \leq X \leq 3)$ .
2 marks
$X \sim Bin (10, 0.3)$ . Write a cumulative-probability expression for $P (3 \leq X \leq 7)$ .
2 marks

Read the worked solutions

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

$X \sim Bin (4, \frac{1}{2})$ . Find $P (X = 2)$ .
2 marks
$(\begin{matrix} 4 \\ 2 \end{matrix}) {(\frac{1}{2})}^{4} = \frac{6}{16} = \frac{3}{8}$ .
$\frac{3}{8}$ .
$X \sim Bin (3, 0.2)$ . Find $P (X = 0)$ .
2 marks
All three trials fail: ${0.8}^{3} = 0.512$ .
0.512.
$X \sim Bin (4, 0.1)$ . Find $P (X \geq 1)$ .
2 marks
Use $1 - P (X = 0)$ .
$1 - {0.9}^{4} = 0.3439$ .
0.3439.
$X \sim Bin (5, \frac{1}{2})$ . Find $P (X \leq 1)$ .
2 marks
$\frac{(\begin{matrix} 5 \\ 0 \end{matrix}) + (\begin{matrix} 5 \\ 1 \end{matrix})}{2^{5}} = \frac{6}{32} = \frac{3}{16}$ .
$\frac{3}{16}$ .
$X \sim Bin (4, \frac{1}{2})$ . Find $P (1 \leq X \leq 3)$ .
2 marks
The excluded values are 0 and 4.
$1 - \frac{1}{16} - \frac{1}{16} = \frac{14}{16} = \frac{7}{8}$ .
$\frac{7}{8}$ .
$X \sim Bin (10, 0.3)$ . Write a cumulative-probability expression for $P (3 \leq X \leq 7)$ .
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 ↗

In this lesson

Understand the idea.

One exact count

Cumulative probability and careful endpoints

See the method in action.

Exactly two successes

Substitute the exact count into the binomial formula.

Evaluate the combination and product.

Round the final probability.

At least one is easier by complement

Use the complement of no successes.

Find the no-success probability.

Subtract from one.

Try it yourself.

Check your understanding

Keep practising