Mathematics 9709 · Probability & Statistics 1

The binomial distribution

A binomial is $n$ repeated yes/no trials.
Each trial is success with probability $p$ or failure with probability $q=1-p$ .
The trials must not affect each other.
$X$ counts how many successes happen.
The ${}^nC_r$ counts how many orders give exactly $r$ successes.

X ~ B(n, p): the formula and the tails

Conditions: fixed $n$ , independent, two outcomes, constant $p$ . Then for $r=0,1,\dots,n$ :

First see what “exactly $r$ successes” means: the successes can sit in many different positions.

success (S) failure (F)

One possible arrangement of 10 trials

12345678910

order 1 of 120

successes r = 3success probability p = 0.40failure probability q = 0.60

1. one arrangement
0.40³ × 0.60⁷ = 0.00179

2. how many arrangements
10 choose 3 = 120

3. add them together
120 × 0.00179 = 0.21499

Below are a few other arrangements with exactly 3 successes. Only the positions move. Each row has the same probability; there are 120 rows like this in total.

order 1

order 2

order 61

order 120

P(X = 3) = C(10, 3) × 0.40³ × 0.60⁷ = 0.21499

The formula has two pieces: count the possible orders, then multiply by the probability of one order.

$P(X=r)={}^nC_r\,p^r q^{\,n-r}$

B(10, 0.4): how many successes in 10 trials. The bars are discrete (gaps), peak near np = 4, and sum to 1.

Worked example

Start simple: a single term

Spinner lands on A with prob 0.4; in 10 spins P(exactly 3 A's): ${}^{10}C_3\,(0.4)^3(0.6)^7=120\times0.064\times0.0279936$ .

Answer

\approx0.215

Translate the words into an r-list FIRST. “At least one” = 1 − P(0); for a tail near the top, sum the few top terms (or do 1 − the few terms on the other side).

Worked example

Upper tail: sum the top terms

$X\sim B(8,0.7)$ , P(more than 6) $=P(7)+P(8)$ : $P(7)=8(0.7)^7(0.3)=0.19765$ , $P(8)=0.7^8=0.05765$ .

Answer

0.19765+0.05765\approx0.255

Worked example

The '1 − top terms' shortcut

85% are rhesus-positive; in 40, P(fewer than 39) $=1-[P(39)+P(40)]$ : $P(39)=40(0.85)^{39}(0.15)=0.01060$ , $P(40)=0.85^{40}=0.00150$ .

Answer

1-0.01210=0.988

Worked example

A range: turn the words into an r-list

$X\sim B(11,0.3)$ , “more than 4 but fewer than 8” = $\{5,6,7\}$ : $P(5)+P(6)+P(7)=0.132+0.0566+0.0173$ . (9709/52/O/N/23 Q2)

Answer

\approx0.206

Name it: $X\sim B(n,p)$ with the numbers (the method-mark line).
Translate words → an exact r-list (“fewer than k” = $X\le k-1$ ).
Short list → sum $P(r)$ ; big chunk → $1-(\text{few terms on the other side})$ .
Each term ${}^nC_r p^r q^{n-r}$ ; add unrounded, round last.

Wording	Event	Best method
exactly k	$X=k$	${}^nC_k p^k q^{n-k}$
at least one	$X\ge1$	$1-q^n$
fewer than k	$X\le k-1$	sum if small, else 1−top terms
more than k	$X\ge k+1$	$1-P(X\le k)$
between a and b	r-list	sum the listed terms

Examiner note

Write at least one full

{}^nC_r p^r q^{n-r}

term explicitly (with

0<p<1

r\ne0,n

) before the arithmetic — a bare decimal scores 0; for a range, no terms omitted. (9709/52/M/J/24 Q6)

Why the ⁿCᵣ?

${}^nC_r$ counts the orders in which $r$ successes can fall among $n$ trials; each order has the same probability $p^r q^{n-r}$ , so multiply. Same arrangement-counting as Ch5.

Common mistake

Dropping the

{}^nC_r

(the most common slip); mis-reading the inequality (“fewer than 5” is

X\le4

, “more than 5” is

X\ge6

) so the r-list shifts by one; and giving failures

p

instead of

q

(the exponents must sum to

n

Is it really binomial?

The model breaks if the four conditions fail: no fixed $n$ (you repeat until something happens — that's geometric), non-constant $p$ or dependent trials (drawing without replacement from a small group changes the odds each pick), or more than two outcomes. Picking 2 children from “1 girl, 3 boys” is not $B(2,\tfrac14)$ — the second pick depends on the first. (9709/52/O/N/23 Q2)

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The binomial distribution

A binomial is $n$ repeated yes/no trials.
Each trial is success with probability $p$ or failure with probability $q=1-p$ .
The trials must not affect each other.
$X$ counts how many successes happen.
The ${}^nC_r$ counts how many orders give exactly $r$ successes.

X ~ B(n, p): the formula and the tails

Conditions: fixed $n$ , independent, two outcomes, constant $p$ . Then for $r=0,1,\dots,n$ :

First see what “exactly $r$ successes” means: the successes can sit in many different positions.

success (S) failure (F)

One possible arrangement of 10 trials

12345678910

order 1 of 120

successes r = 3success probability p = 0.40failure probability q = 0.60

1. one arrangement
0.40³ × 0.60⁷ = 0.00179

2. how many arrangements
10 choose 3 = 120

3. add them together
120 × 0.00179 = 0.21499

Below are a few other arrangements with exactly 3 successes. Only the positions move. Each row has the same probability; there are 120 rows like this in total.

order 1

order 2

order 61

order 120

P(X = 3) = C(10, 3) × 0.40³ × 0.60⁷ = 0.21499

The formula has two pieces: count the possible orders, then multiply by the probability of one order.

$P(X=r)={}^nC_r\,p^r q^{\,n-r}$

B(10, 0.4): how many successes in 10 trials. The bars are discrete (gaps), peak near np = 4, and sum to 1.

Worked example

Start simple: a single term

Spinner lands on A with prob 0.4; in 10 spins P(exactly 3 A's): ${}^{10}C_3\,(0.4)^3(0.6)^7=120\times0.064\times0.0279936$ .

Answer

\approx0.215

Translate the words into an r-list FIRST. “At least one” = 1 − P(0); for a tail near the top, sum the few top terms (or do 1 − the few terms on the other side).

Worked example

Upper tail: sum the top terms

$X\sim B(8,0.7)$ , P(more than 6) $=P(7)+P(8)$ : $P(7)=8(0.7)^7(0.3)=0.19765$ , $P(8)=0.7^8=0.05765$ .

Answer

0.19765+0.05765\approx0.255

Worked example

The '1 − top terms' shortcut

85% are rhesus-positive; in 40, P(fewer than 39) $=1-[P(39)+P(40)]$ : $P(39)=40(0.85)^{39}(0.15)=0.01060$ , $P(40)=0.85^{40}=0.00150$ .

Answer

1-0.01210=0.988

Worked example

A range: turn the words into an r-list

$X\sim B(11,0.3)$ , “more than 4 but fewer than 8” = $\{5,6,7\}$ : $P(5)+P(6)+P(7)=0.132+0.0566+0.0173$ . (9709/52/O/N/23 Q2)

Answer

\approx0.206

Name it: $X\sim B(n,p)$ with the numbers (the method-mark line).
Translate words → an exact r-list (“fewer than k” = $X\le k-1$ ).
Short list → sum $P(r)$ ; big chunk → $1-(\text{few terms on the other side})$ .
Each term ${}^nC_r p^r q^{n-r}$ ; add unrounded, round last.

Wording	Event	Best method
exactly k	$X=k$	${}^nC_k p^k q^{n-k}$
at least one	$X\ge1$	$1-q^n$
fewer than k	$X\le k-1$	sum if small, else 1−top terms
more than k	$X\ge k+1$	$1-P(X\le k)$
between a and b	r-list	sum the listed terms

Examiner note

Write at least one full

{}^nC_r p^r q^{n-r}

term explicitly (with

0<p<1

r\ne0,n

) before the arithmetic — a bare decimal scores 0; for a range, no terms omitted. (9709/52/M/J/24 Q6)

Why the ⁿCᵣ?

${}^nC_r$ counts the orders in which $r$ successes can fall among $n$ trials; each order has the same probability $p^r q^{n-r}$ , so multiply. Same arrangement-counting as Ch5.

Common mistake

Dropping the

{}^nC_r

(the most common slip); mis-reading the inequality (“fewer than 5” is

X\le4

, “more than 5” is

X\ge6

) so the r-list shifts by one; and giving failures

p

instead of

q

(the exponents must sum to

n

Is it really binomial?