Mathematics 9709 · Probability & Statistics 1

The geometric distribution

Geometric questions stop counting how many successes and start counting how long you wait.
You keep trying until the first success.
$X$ is the trial number where the first success finally happens.
To succeed exactly on trial $r$ , you fail $r-1$ times, then succeed.
So $P(X=r)=q^{\,r-1}p$ , with no ${}^nC_r$ .

X ~ Geo(p): the formula and the q^r tails

	Binomial B(n, p)	Geometric Geo(p)
what's fixed	n trials, fixed	trials open-ended
X counts	# successes	trial of the 1st success
coefficient	${}^nC_r$	none (one order)
P(X = r)	${}^nC_r p^r q^{n-r}$	$q^{\,r-1}p$
mean	$np$	$1/p$
range of X	$0,1,\dots,n$	$1,2,3,\dots$

$P(X=r)=q^{\,r-1}p,\qquad P(X>r)=q^{\,r},\qquad P(X\le r)=1-q^{\,r}$

Geo(0.2): the trial of the first success. It always peaks at r = 1 and decays by a factor of q each step — never use the infinite sum, use the q^r tails.

Worked example

Start simple: a single value

Success prob 0.66 ( $q=0.34$ ), first success on trial 3: $P(X=3)=0.34^2(0.66)$ .

Answer

\approx0.0763

Worked example

On or before: sum the small cases

Same $p=0.66$ , “on or before the 2nd trial”: $P(X\le2)=P(1)+P(2)=p+pq=0.66+0.66(0.34)$ . For small $r$ you can just add the terms — that is why the $1-q^r$ shortcut works.

Answer

\approx0.884

The tail shortcuts P(X > r) = qʳ and P(X ≤ r) = 1 − qʳ package the whole infinite sum — never add up an endless GP.

Worked example

Where the tail shortcut shines

18% wear contact lenses ( $q=0.82$ ). First wearer among the first 15: $P(X\le15)=1-0.82^{15}=1-0.0510$ ; not in the first 9: $P(X>9)=0.82^9$ .

Answer

0.949\quad\text{and}\quad0.168

Name it: $X\sim\text{Geo}(p)$ , $q=1-p$ .
Single value? $q^{r-1}p$ (no ${}^nC_r$ ).
“More than r”? $q^r$ . “r or fewer”? $1-q^r$ .
“Fewer than r” / “at least r”: convert to $\le$ / $>$ FIRST, then pick the q-power.

Worked example

The off-by-one tail

$p=0.2$ , “fewer than 7 spins” = $P(X\le6)=1-0.8^6=1-0.262$ . (9709/52/O/N/23 Q2)

Answer

\approx0.738

Worked example

A range by differencing tails

$p=0.4$ , $P(3\le X\le7)=q^2-q^7=0.6^2-0.6^7$ . (9709/52/O/N/24 Q7)

Answer

0.36-0.028=0.332

Worked example

Hard rung: words → inequality with a fractional p

Coin gives heads with $p=\tfrac{5}{11}$ ( $q=\tfrac{6}{11}$ ), tossed until the first head. “At least six times” = $P(X>5)=q^5=\big(\tfrac{6}{11}\big)^5$ ; “fewer than eight” = $P(X\le7)=1-q^7=1-\big(\tfrac{6}{11}\big)^7$ .

Answer

\approx0.0483\quad\text{and}\quad0.986

Convert the words to ≤ or > FIRST, then read the q-power. “Fewer than 7” means q⁶, not q⁷.

Wording	Event	Formula
exactly r	$X=r$	$q^{r-1}p$
more than r	$X>r$	$q^r$
r or fewer	$X\le r$	$1-q^r$
fewer than r	$X\le r-1$	$1-q^{r-1}$
between a, b	$a\le X\le b$	$q^{a-1}-q^b$

Examiner note

MS awards M1 for

1-0.8^{\,d}

with the correct d — so write

P(X<7)=P(X\le6)=1-q^6

so the examiner sees the ≤ step. (9709/52/M/J/23 Q4)

Why you never need an infinite sum

The probabilities form a GP with ratio $q$ summing to $p/(1-q)=1$ ; the closed-form tail $q^r$ already packages the whole sum from $r+1$ onward.

Common mistake

The off-by-one:

P(X=r)

uses

q^{\,r-1}

(

r-1

failures) but

P(X>r)

uses

q^{\,r}

. Never put a

{}^nC_r

in front, and remember

X\ge1

always (never 0).

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The geometric distribution

Geometric questions stop counting how many successes and start counting how long you wait.
You keep trying until the first success.
$X$ is the trial number where the first success finally happens.
To succeed exactly on trial $r$ , you fail $r-1$ times, then succeed.
So $P(X=r)=q^{\,r-1}p$ , with no ${}^nC_r$ .

X ~ Geo(p): the formula and the q^r tails

	Binomial B(n, p)	Geometric Geo(p)
what's fixed	n trials, fixed	trials open-ended
X counts	# successes	trial of the 1st success
coefficient	${}^nC_r$	none (one order)
P(X = r)	${}^nC_r p^r q^{n-r}$	$q^{\,r-1}p$
mean	$np$	$1/p$
range of X	$0,1,\dots,n$	$1,2,3,\dots$

$P(X=r)=q^{\,r-1}p,\qquad P(X>r)=q^{\,r},\qquad P(X\le r)=1-q^{\,r}$

Geo(0.2): the trial of the first success. It always peaks at r = 1 and decays by a factor of q each step — never use the infinite sum, use the q^r tails.

Worked example

Start simple: a single value

Success prob 0.66 ( $q=0.34$ ), first success on trial 3: $P(X=3)=0.34^2(0.66)$ .

Answer

\approx0.0763

Worked example

On or before: sum the small cases

Same $p=0.66$ , “on or before the 2nd trial”: $P(X\le2)=P(1)+P(2)=p+pq=0.66+0.66(0.34)$ . For small $r$ you can just add the terms — that is why the $1-q^r$ shortcut works.

Answer

\approx0.884

The tail shortcuts P(X > r) = qʳ and P(X ≤ r) = 1 − qʳ package the whole infinite sum — never add up an endless GP.

Worked example

Where the tail shortcut shines

18% wear contact lenses ( $q=0.82$ ). First wearer among the first 15: $P(X\le15)=1-0.82^{15}=1-0.0510$ ; not in the first 9: $P(X>9)=0.82^9$ .

Answer

0.949\quad\text{and}\quad0.168

Name it: $X\sim\text{Geo}(p)$ , $q=1-p$ .
Single value? $q^{r-1}p$ (no ${}^nC_r$ ).
“More than r”? $q^r$ . “r or fewer”? $1-q^r$ .
“Fewer than r” / “at least r”: convert to $\le$ / $>$ FIRST, then pick the q-power.

Worked example

The off-by-one tail

$p=0.2$ , “fewer than 7 spins” = $P(X\le6)=1-0.8^6=1-0.262$ . (9709/52/O/N/23 Q2)

Answer

\approx0.738

Worked example

A range by differencing tails

$p=0.4$ , $P(3\le X\le7)=q^2-q^7=0.6^2-0.6^7$ . (9709/52/O/N/24 Q7)

Answer

0.36-0.028=0.332

Worked example

Hard rung: words → inequality with a fractional p

Answer

\approx0.0483\quad\text{and}\quad0.986

Convert the words to ≤ or > FIRST, then read the q-power. “Fewer than 7” means q⁶, not q⁷.

Wording	Event	Formula
exactly r	$X=r$	$q^{r-1}p$
more than r	$X>r$	$q^r$
r or fewer	$X\le r$	$1-q^r$
fewer than r	$X\le r-1$	$1-q^{r-1}$
between a, b	$a\le X\le b$	$q^{a-1}-q^b$

Examiner note

MS awards M1 for

1-0.8^{\,d}

with the correct d — so write

P(X<7)=P(X\le6)=1-q^6

so the examiner sees the ≤ step. (9709/52/M/J/23 Q4)

Why you never need an infinite sum

The probabilities form a GP with ratio $q$ summing to $p/(1-q)=1$ ; the closed-form tail $q^r$ already packages the whole sum from $r+1$ onward.

Common mistake

The off-by-one:

P(X=r)

uses

q^{\,r-1}

(

r-1

failures) but

P(X>r)

uses

q^{\,r}

. Never put a

{}^nC_r

in front, and remember

X\ge1

always (never 0).