IB Applications and Interpretation SL · Discrete random variables and the binomial distribution · Section 3 of 3

The binomial distribution: probabilities, mean and variance

3 worked examples · By Álvaro López

Key idea

The binomial distribution models the number of successes in a series of repeated trials, such as a marksman's hits, the defective screws in a batch or the correct answers in a test answered at random. In the IB exam it is almost always solved with the graphic display calculator (GDC), but you must recognise when it is binomial, identify $n$ and $p$ , and translate the question into $P (X = k)$ , $P (X \leq k)$ or $P (X \geq k)$ .

Bernoulli trials

A Bernoulli trial is a trial with only two possible outcomes, which we call success and failure. If the probability of success is $p$ , that of failure is $1 - p$ . "Success" does not mean something good: it is simply the outcome we are counting (for example, that a part is faulty).

Tossing a coin and checking whether it lands heads, or checking whether a randomly chosen student wears glasses, are Bernoulli trials.

The binomial distribution

Conditions 3 and 4 are the ones most often broken. If you draw balls from an urn without returning them, the composition changes at each draw: the probability stops being constant and the trials stop being independent, so it is not binomial. With replacement, it is.

For example, if you toss a balanced coin 4 times, the probability of obtaining exactly 1 head is $(1 4) (\frac{1}{2})^{1} (\frac{1}{2})^{3} = \frac{4}{16} = \frac{1}{4}$ . The four results "one head and three tails" (HTTT, THTT, TTHT, TTTH) each have probability $\frac{1}{16}$ .

Mean and variance

These formulas also allow the reverse: if the question gives the mean and the variance, $n$ and $p$ can be found (see the method below).

Cumulative probabilities

Many questions ask not for a specific value but for a range ("at most 3", "at least 5", "between 2 and 4"). The calculator has two functions for the binomial: one gives $P (X = k)$ (point probability) and the other gives $P (X \leq k)$ (cumulative probability). With the second and the complement, everything can be solved.

Method · Translating a range question into the calculator

Write $X \sim B (n, p)$ , identifying correctly what a success is.
Translate the question into an inequality about $X$ and convert it to the form " $\leq$ ", using that $X$ only takes whole-number values:

Question	Calculated as
$P (X = k)$	point function with $x = k$
$P (X \leq k)$	cumulative function with $x = k$
$P (X < k)$	$P (X \leq k - 1)$
$P (X \geq k)$	$1 - P (X \leq k - 1)$
$P (X > k)$	$1 - P (X \leq k)$
$P (a \leq X \leq b)$	$P (X \leq b) - P (X \leq a - 1)$

Enter $n$ , $p$ and the corresponding value of $k$ into the calculator, and read off the result.
Round to 3 significant figures, unless the question says otherwise, and answer with a sentence.

Finding n or p

If only $p$ is known and the smallest $n$ that satisfies a condition is asked for (for example, that the probability of at least one success exceeds a value), set up an inequality with the complement, $1 - (1 - p)^{n}$ , and test values of $n$ on the calculator or use logarithms.

Worked examples

Example 1 Basic

Free throws

A basketball player scores 70% of her free throws, and each throw is independent of the others. She takes 6 free throws. Let $X$ be the number of throws she scores.

a) State the distribution of $X$ and calculate the probability that she scores exactly 4 throws.

b) Calculate the mean number of throws she is expected to score.

c) Calculate the standard deviation of $X$ .

d) Calculate the probability that she scores at most 3 throws.

Solution

a) There are $n = 6$ independent trials with two outcomes and $p = 0.7$ constant, so $X \sim B (6; 0.7)$ . $P (X = 4) = (4 6) 0. 7^{4} 0. 3^{2} = 15 \cdot 0.2401 \cdot 0.09 = 0.324$

b) $E (X) = n p = 6 \cdot 0.7 = 4.2$ throws.

c) $σ = n p (1 - p) = 6 \cdot 0.7 \cdot 0.3 = 1.26 = 1.12$ .

d) Using the cumulative function of the GDC, with $n = 6$ , $p = 0.7$ and $k = 3$ : $P (X \leq 3) = 0.256$

Example 2 Medium

A test answered at random

A test has 12 questions and each one has 4 options, of which only one is correct. A student answers all of them at random. Let $X$ be the number of correct answers.

a) Show that $X$ is binomial and state its parameters. Calculate the probability of getting exactly 3 questions right.

b) Calculate the probability of getting at most 2 questions right.

c) Calculate the probability of getting at least 5 questions right.

d) Calculate the mean and the standard deviation of the number of correct answers.

Solution

a) Each question is a trial with two outcomes (correct or wrong), the trials are independent because they are answered at random, and the probability of a correct answer is always $p = \frac{1}{4}$ . There are $n = 12$ trials, so $X \sim B (12; \frac{1}{4})$ . $P (X = 3) = (3 12) (\frac{1}{4})^{3} (\frac{3}{4})^{9} = 0.258$

b) With the cumulative function of the GDC ( $n = 12$ , $p = 0.25$ , $k = 2$ ): $P (X \leq 2) = 0.391$

c) "At least 5" is the complement of "at most 4": $P (X \geq 5) = 1 - P (X \leq 4) = 1 - 0.842 = 0.158$

d) $E (X) = 12 \cdot \frac{1}{4} = 3$ and $σ = 12 \cdot \frac{1}{4} \cdot \frac{3}{4} = 2.25 = 1.5$ .

That is, guessing at random, one expects 3 correct answers out of 12.

Example 3 Hard

Finding the parameters of a binomial

In a bottling plant, the number $X$ of bottles with a defective cap in a box follows a binomial distribution. It is known that, per box, the mean of $X$ is 12 and its variance is 7.2.

a) Find the number $n$ of bottles per box and the probability $p$ that a bottle has a defective cap.

b) Calculate the probability that a box has exactly 10 defective bottles.

c) Calculate the probability that a box has more than 12 defective bottles.

d) Calculate the probability that the number of defective bottles is between 10 and 14, both included.

Solution

a) The two equations are set up: $n p = 12 n p (1 - p) = 7.2$ Dividing the second by the first: $1 - p = \frac{7.2}{12} = 0.6$ , so $p = 0.4$ . Then $n = \frac{12}{0.4} = 30$ . It satisfies $0 < p < 1$ and $n$ is an integer, so $X \sim B (30; 0.4)$ .

b) With the point function of the GDC ( $n = 30$ , $p = 0.4$ , $k = 10$ ): $P (X = 10) = 0.115$

c) $P (X > 12) = 1 - P (X \leq 12)$ . With the cumulative function, $P (X \leq 12) = 0.578$ , so $P (X > 12) = 1 - 0.578 = 0.422$

d) As $X$ is a whole number, $10 \leq X \leq 14$ is equivalent to $X \leq 14$ minus $X \leq 9$ : $P (10 \leq X \leq 14) = P (X \leq 14) - P (X \leq 9) = 0.648$

In the exam: marks are given for writing the two equations and the quotient that gives $p$ , and then for stating the model $X \sim B (30; 0.4)$ before using the calculator. In part d) the usual mistake is to subtract $P (X \leq 10)$ instead of $P (X \leq 9)$ , which leaves out the value 10. Use the GDC values without rounding, and round only the final answer.

Common mistakes

Summary

$X \sim B (n, p)$ when the successes are counted in $n$ independent trials, each with two outcomes and a constant probability of success $p$ .
$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$ , for $k = 0, 1, \dots, n$ .
$E (X) = n p$ , $Var (X) = n p (1 - p)$ and $σ = n p (1 - p)$ .
The GDC gives $P (X = k)$ (point) and $P (X \leq k)$ (cumulative). For the rest: $P (X < k) = P (X \leq k - 1)$ , $P (X \geq k) = 1 - P (X \leq k - 1)$ , $P (X > k) = 1 - P (X \leq k)$ and $P (a \leq X \leq b) = P (X \leq b) - P (X \leq a - 1)$ .
To find $n$ and $p$ from the mean and the variance: $1 - p = σ^{2} / μ$ and $n = μ / p$ .
In the exam, write the model $X \sim B (n, p)$ and the probability you calculate, and round to 3 significant figures.

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