Model repeated independent trials with the binomial distribution

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Statistics and Probability

IB Mathematics: Analysis and Approaches SL — Study topic SL 4.8

A multiple-choice question, a quality check, or a repeated game can be described as a sequence of trials. When each trial has two possible outcomes and the conditions remain consistent, the binomial distribution provides a model for counting successes. The key is to check the situation before choosing a formula: a calculation can be accurate but still use the wrong model if the trials are not independent or the success probability changes.

What you will learn

Decide when a repeated-trial situation can be modelled by a binomial distribution.
Interpret the parameters and calculate probabilities for a specified number of successes.
Find probabilities such as “at least” or “between” by combining outcomes or using a complement.
Use a graphing calculator to check binomial probabilities and interpret a distribution graph.

1. Prior knowledge: describing one trial

A trial is one repetition of an activity. For a binomial model, choose one outcome to call a success; the other outcome is a failure. These names are labels, not judgements. For example, if a seed germinates, that may be the success. Let

p

be the probability of success on one trial. Then the probability of failure is

1-p

, because the two outcomes cover all possibilities.

A sequence of trials is independent when the result of one trial does not change the probabilities on another. The success probability must also be the same on every trial. These conditions are part of the model, not details to ignore. A fixed number of trials, two outcomes per trial, independence, and constant probability are the checks for using a binomial distribution.

P(\text{failure})=1-p

Define what counts as a success before calculating.
Check that the number of trials is fixed.
Check independence and a constant success probability.

2. The binomial model and its formula

If there are

n

independent trials, each with success probability

p

, let

X

be the number of successes. We write

X\sim B(n,p)

to identify a binomial random variable. Here,

n

is a positive whole number and

0\leq p\leq1

.

To have exactly

k

successes, first consider one particular arrangement: its probability is

p^k(1-p)^{n-k}

. There are

{n\choose k}

arrangements of

k

successes among

n

trials. Multiplying gives the probability of exactly

k

successes. The combination

{n\choose k}

counts selections without regard to order; it can be evaluated with a calculator’s combination function.

The possible values of

X

are the whole numbers from

0

to

n

. “At least” and “at most” probabilities collect several of these values. For instance, “at least one” can often be found more efficiently by subtracting the probability of zero successes from

1

. This works because “zero successes” and “at least one success” are complementary events.

P(X=k)={n\choose k}p^k(1-p)^{n-k}

In $X\sim B(n,p)$ , $n$ is the number of trials and $p$ is the probability of success per trial.
Use the exact-success formula for a single value of $X$ .
For a range, add the probabilities of the included whole-number outcomes or use a complement when simpler.

3. Numerical, graphical, and contextual representations

A probability table lists

P(X=k)

for each possible number of successes. Its entries are non-negative and add to

1

, since one of the possible counts must occur. A bar graph of the same distribution places the values of

k

along the horizontal axis and their probabilities on the vertical axis. Each bar represents one exact count, not a range.

A graphing calculator can evaluate a single binomial probability using a binomial probability function, often called a probability mass function or `binompdf`. A cumulative function, often called `binomcdf`, gives the probability of a count up to a chosen value. Names and menu paths vary by calculator, so check what the function returns before using it. For example, a cumulative value at

k

means

P(X\leq k)

; subtracting the cumulative value at

a-1

from that at

b

gives

P(a\leq X\leq b)

.

Technology is useful for accurate evaluation and for viewing how probability is distributed over possible counts. It does not decide whether the binomial assumptions are appropriate. State the model, identify the event, and explain whether the calculator output is exact-count or cumulative.

P(a≤ X≤ b)=P(X≤ b)-P(X≤ a-1)

The horizontal axis of a binomial bar graph shows success counts; the vertical axis shows probabilities.
Use a single-value function for $P(X=k)$ and a cumulative function for $P(X\leq k)$ .
Confirm calculator function conventions and report a suitable accuracy.

4. A reliable problem-solving routine

Begin by defining success and writing the values of

n

and

p

. Check the model conditions in the context. Translate the wording into an event such as

X=3

,

X\geq1

, or

2\leq X\leq5

. Then choose the exact-probability formula, a sum, or a complement. Keep unrounded values during intermediate calculations and round the final answer as requested.

A useful technology check is to calculate a probability both from the formula and from a calculator function on a manageable example. Agreement helps catch a mistaken parameter or event. In an exam-style response, include enough working to show what the output represents; a calculator number alone may not communicate the reasoning.

Define the random variable in context.
Translate words carefully into inequalities or an exact value.
Show the model and event, then state the probability with units or context where relevant.

Worked example

Exactly two successes

A player makes each penalty shot with probability

0.3

. Assume shot outcomes are independent. In

8

shots, find the probability of exactly

2

successful shots.

Set up the model
A success is a made shot. There are a fixed $8$ shots, the success probability is $0.3$ each time, and the outcomes are assumed independent, so the binomial model applies.
$X\sim B(8,0.3)$
Substitute the exact count
Exactly two successes can occur in ${8\choose2}$ arrangements. Each such arrangement has two successes and six failures.
$P(X=2)={8\choose2}(0.3)^2(0.7)^6$
Evaluate
Evaluating the expression gives the probability. A calculator can check the arithmetic using a single-value binomial function.
$P(X=2)=0.29647548$

Answer: The probability is approximately

0.2965

, or 29.65% to two decimal places as a percentage.

Check: The answer is between

0

and

1

. The factor

(0.7)^6

represents the six missed shots.

Worked example

At least one success

An electronic component passes an inspection with probability

0.8

. For a batch of

6

independently inspected components, find the probability that at least one fails.

Define the count
Let $X$ count the components that fail. The failure probability on one inspection is $1-0.8=0.2$ , so the number of failures follows a binomial model.
$X\sim B(6,0.2)$
Use the complement
At least one failure is the complement of no failures. No failures means that all six components pass, which has probability $(0.8)^6$ by independence.
$P(X\geq1)=1-P(X=0)=1-(0.8)^6$
Calculate
Subtracting the probability of no failures from $1$ gives the required probability.
$P(X\geq1)=0.737856$

Answer: The probability that at least one component fails is approximately

0.7379

.

Check: The complementary event is exactly zero failures, not exactly one failure.

Worked example

A range of success counts

A plant produces a flower with probability

0.4

in each of

10

independent growing trials. Find the probability of between

3

and

5

flowers, inclusive.

Model and translate
Let $X$ be the number of flowers. “Between $3$ and $5$ , inclusive” includes the three counts $3$ , $4$ , and $5$ .
$X\sim B(10,0.4), P(3≤ X\leq5)$
Add the exact-count probabilities
These possible counts do not overlap, so their probabilities can be added. Use the binomial formula once for each count.
$P(3≤ X\leq5)=P(X=3)+P(X=4)+P(X=5)$
Evaluate or check with technology
The three terms evaluate to $0.214990848$ , $0.250822656$ , and $0.2006581248$ . A cumulative calculator function can also find the range by subtracting $P(X\leq2)$ from $P(X\leq5)$ .
$P(3≤ X\leq5)=0.6664716288$

Answer: The probability is approximately

0.6665

to four decimal places.

Check: The inclusive endpoints are handled by including both

P(X=3)

and

P(X=5)

.

Common mistakes and how to avoid them

Using

p

as the probability of failure in the formula.

Correction: Define success first. Use

p

for success and

1-p

for failure throughout.

Applying a binomial model when the success probability changes or trials affect one another.

Correction: Check independence and constant probability in the context before calculating.

Interpreting “at most

k

” as “less than

k

.”

Correction: At most

k

means

X\leq k

, so the count

k

is included.

Using a cumulative calculator output as though it were the probability of exactly one count.

Correction: Check the function’s meaning: cumulative output includes all counts up to its input.

Lesson summary

A binomial model requires a fixed number of independent trials, two outcomes per trial, and the same success probability each time.
For $X\sim B(n,p)$ , the probability of exactly $k$ successes is ${n\choose k}p^k(1-p)^{n-k}$ .
For ranges, add the included exact-count probabilities or use cumulative probabilities; complements can simplify “at least” questions.
Use technology to evaluate and check probabilities, while showing what event and model the output represents.

Check your understanding

Question 1

A fair coin is tossed

5

times independently. If

X

is the number of heads, what is

P(X=0)

?

$\frac{1}{32}$
$\frac{5}{32}$
$\frac{1}{2}$
$\frac{31}{32}$

Show answer and explanation

\frac{1}{32}

Zero heads means five tails, so the probability is

(0.5)^5=\frac{1}{32}

.

Question 2

For a binomial random variable with

n=7

, what does “at most

2

successes” mean?

$X<2$
$X\leq2$
$X\geq2$
$X=2$

Show answer and explanation

X\leq2

At most

2

includes

0

,

1

, and

2

successes, so the event is

X\leq2

.

Question 3

A binomial model has

n=4

and success probability

p=0.25

. What is the probability of exactly one success?

$0.1055$
$0.3164$
$0.4219$
$0.7500$

Show answer and explanation

0.3164

Use

{4\choose1}(0.25)(0.75)^3=0.31640625

, which rounds to

0.3164

.

Key terms

Trial: One repetition of an activity with two outcomes in the binomial model.
Success: The chosen outcome being counted; it is a label for one of the two outcomes.
Independent trials: Trials for which one result does not change the probabilities on the others.
Binomial distribution: A probability model for the number of successes in a fixed number of independent trials with a constant success probability.
Cumulative probability: The probability of a count being at or below a specified value, such as $P(X\leq k)$ .

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