Binomial probabilities: recognise B(n, p), then translate the words: IB Maths AA SL

Spot it in the exam

Exam phrases to spot (6)

“find the probability that exactly $r$ …”
“at least one”
“fewer than $k$ ”
“more than $k$ ”
“between $a$ and $b$ inclusive”
“ $n$ items are chosen at random”

In disguise (5)

"a random sample of $n$ is taken" from a large population with a known proportion: each member is a trial with the same $p$
$p$ is not a number in the question: it comes from an earlier part (a normal probability, a table, an earlier binomial)
"no more than one of the remaining days": count the successes (or the failures) in the trials that are left
$\mathrm{E}(X)$ and $\mathrm{Var}(X)$ are given, or linked, and $p$ (or $n$ ) is asked for: $np$ and $np(1 - p)$ give the equations
$n$ is the unknown, and a cumulative probability such as $\mathrm{P}(X \le k)$ is stated (rounded) instead of a target to pass: tabulate it against $n$ , as for a least $n$ , and give the $n$ whose value is closest. There is no inequality and no rounding up (a markscheme may accept the neighbouring $n$ as well)

Variants (5)

asked = exactly, at most, fewer than, at least, more than, between (inclusive), at least one
$p$ from = given, a normal part, a table, an earlier binomial; $n$ large (hundreds)
count = successes or failures (either works)
$p$ from a relation between the mean and the variance
$n$ unknown, with at least $k$ successes: a table over $n$

Looks like it, but isn't (6)

draws without replacement from a small group: Tree diagrams and repeated draws
, a few repeats of an experiment: Build the distribution from the experiment (count the orders)
$n$ unknown, and "at least one": Find n
"given that": Given that X is in a range
only the expected number, $np$ : the expected number of occurrences
a bracket of probability-like numbers raised to a power, where one term's number is asked and nothing is random: Expand, or find one term or coefficient (the same number as a binomial probability:

Worked example

Several words in one question

Paper 2 (calculator)

Question

An archer hits the target with probability $0.65$ on each shot, independently of her other shots. She shoots $16$ arrows. Find the probability that she hits the target
(a) exactly $11$ times; [2]
(b) fewer than $9$ times; [2]
(c) between $8$ and $12$ times inclusive. [2]

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See it move: Which bars count?

From11

To11

P(exactly 11) $= \mathrm{P}(X = 11) = 0.201$ : one bar.
$X \sim \mathrm{B}(16, 0.65)$ : 17 bars, from 0 to 16, with mean $np = 10.4$ .
Fewer than 9 leaves 9 out: the bars up to 8.
Between 8 and 12 inclusive keeps both ends: up to 12 minus up to 7, not up to 8.
The calculator's cdf adds the bars from 0, so every range is up to minus up to.

Which method?

$X$ counts whole numbers, so every phrase becomes a $\le$ form before the GDC (binomial probabilities):

exactly $r$
Translation
$\mathrm{P}(X = r)$
On the GDC
pdf
at most $k$
Translation
$\mathrm{P}(X \le k)$
On the GDC
cdf up to $k$
fewer than $k$
Translation
$\mathrm{P}(X \le k - 1)$
On the GDC
cdf up to $k - 1$
at least $k$
Translation
$\mathrm{P}(X \ge k) = 1 - \mathrm{P}(X \le k - 1)$
On the GDC
$1 -$ cdf up to $k - 1$
more than $k$
Translation
$\mathrm{P}(X \ge k + 1) = 1 - \mathrm{P}(X \le k)$
On the GDC
$1 -$ cdf up to $k$
$a$ to $b$ inclusive
Translation
$\mathrm{P}(a \le X \le b) = \mathrm{P}(X \le b) - \mathrm{P}(X \le a - 1)$
On the GDC
cdf up to $b$ minus cdf up to $a - 1$
at least one
Translation
$1 - \mathrm{P}(X = 0) = 1 - (1 - p)^n$
On the GDC
pdf at $0$

The words	Translation	On the GDC
exactly $r$	$\mathrm{P}(X = r)$	pdf
at most $k$	$\mathrm{P}(X \le k)$	cdf up to $k$
fewer than $k$	$\mathrm{P}(X \le k - 1)$	cdf up to $k - 1$
at least $k$	$\mathrm{P}(X \ge k) = 1 - \mathrm{P}(X \le k - 1)$	$1 -$ cdf up to $k - 1$
more than $k$	$\mathrm{P}(X \ge k + 1) = 1 - \mathrm{P}(X \le k)$	$1 -$ cdf up to $k$
$a$ to $b$ inclusive	$\mathrm{P}(a \le X \le b) = \mathrm{P}(X \le b) - \mathrm{P}(X \le a - 1)$	cdf up to $b$ minus cdf up to $a - 1$
at least one	$1 - \mathrm{P}(X = 0) = 1 - (1 - p)^n$	pdf at $0$

With $n$ unknown: for at least one, $1 - (1 - p)^n >$ target is Find n. For at least $k$ with $k \ge 2$ there is no formula to solve: tabulate $\mathrm{P}(X \ge k) = 1 - \mathrm{P}(X \le k - 1)$ against $n$ (table of values), and give the least $n$ that works, with the values for $n - 1$ and $n$ written down.

Steps

Check the model:
a fixed number of trials, two outcomes, the same $p$ , independent trials. Say what a success is.
State $X \sim \mathrm{B}(n, p)$ with the numbers.
This is the recognition mark: the word "binomial" can earn it (one markscheme accepts it even with wrong parameters), but $n$ and $p$ often carry a mark of their own and every later mark uses them, so write them; $np$ on its own doesn't. A $p$ from an earlier part goes in unrounded (markschemes also accept the answer that its 3 s.f. value gives, but the stored value is safe).
Translate the words
(table above) and write the probability statement.
GDC:
binomial probabilities. Round to 3 s.f., or to the accuracy asked.
Check:
the answer is between $0$ and $1$ , and its size fits the mean $np$ (a count far from $np$ has a small probability).

Common mistakes

Strict words read as non-strict
"fewer than $9$ " is $X \le 8$ , and "more than $20$ " is $1 - \mathrm{P}(X \le 20)$ , not $1 - \mathrm{P}(X \le 19)$ .
"At least $k$ " as $1 - \mathrm{P}(X \le k)$
that leaves out $X = k$ itself. "At least $8$ " is $1 - \mathrm{P}(X \le 7)$ ; the complement form has its own method mark.
"Between … inclusive" cut at the wrong place
$\mathrm{P}(8 \le X \le 12) = \mathrm{P}(X \le 12) - \mathrm{P}(X \le 7)$ .

More mistakes (5)

No recognition
write " $X \sim \mathrm{B}(16, 0.65)$ " (or "binomial, $n = 16$ , $p = 0.65$ "). $np$ alone doesn't show the model.
A binomial for draws without replacement
$p$ changes after each draw. Markschemes then give only the mark for writing the first draw's probability, and nothing more (Tree diagrams and repeated draws).
Remaining trials
when some trials are over, only the ones left are random, and the successes so far reduce the number still needed.
Keeping an impossible $p$
a negative $p$ , or $p > 1$ , must be rejected with the reason.
Calculator notation as the answer
write $\mathrm{P}(X \le 8) = 0.159$ , never the GDC's command.

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