The geometric distribution | CAIE 9709 Probability & Statistics 1 Notes

Probability & Statistics 1

Mathematics9709

Geometric means first success, not a fixed count

Let $X$ be the trial number of the first success. Write $X\sim\operatorname{Geo}(p)$ when trials are independent, each trial has success/failure outcomes, and the same success probability $p$ is used every time.

X=1,2,3,\ldots

Zero is impossible because the first success cannot occur before any trial has happened. Unlike a binomial variable, there is no largest possible value.

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Question records	Model	Possible values
successes in 10 trials	$B(10,p)$	$0,1,\ldots,10$
trial of the first success	$\operatorname{Geo}(p)$	$1,2,3,\ldots$

Key idea

The model is decided by what

X

records, not just by the words “repeated trials”.

First success on trial r means failures, then success

First success on trial 4

trial 1

trial 2

trial 3

trial 4

P(X = 4) = q³p

one forced order · no combination factor

To obtain the first success on trial 4, the only possible order is F, F, F, S. Its probability is q³p; there is no combination factor.

For the first success to occur on trial $r$ , the first $r-1$ trials must all fail and trial $r$ must succeed. The order is forced, so there is no combination factor.

p_r=q^{\,r-1}p

Symbols

$p_r$ = P(X = r), the probability of first success on trial r (none)
$X$ = trial number of the first success (trials)
$r$ = specified trial number (trials)
$q$ = failure probability, 1-p (none)
$p$ = success probability (none)

Worked example

First success on trial 7

If $p=0.3$ , there must be six failures followed by one success. (9709/52/M/J/24 Q1(a))

P(X=7)=0.7^6(0.3)

Answer

P(X=7)=0.0353

Count failures to control cumulative endpoints

More than $r$ trials are needed exactly when the first $r$ trials all fail. Therefore

P(X>r)=q^r

P(X\le r)=1-q^r

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Words	Event	Probability
before trial r	$X<r$	$1-q^{r-1}$
by trial r	$X\le r$	$1-q^r$
after trial r	$X>r$	$q^r$
at least r trials	$X\ge r$	$q^{r-1}$

Worked example

First success before trial 6

“Before trial 6” means $X<6$ , so the first success must occur in trials 1 to 5. Subtract the chance that all first five trials fail. (9709/52/M/J/24 Q1(b))

P(X<6)=1-0.7^5

Answer

P(X<6)=0.832

Your turnIndependent check [3 marks]

Independent attempts have success probability 0.3. The process stops at the first success, and $X$ is its trial number. State the distribution of $X$ and find $P(4\le X\le6)$ .

Show worked answer

X\sim\operatorname{Geo}(0.3)

q=0.7

Let $A$ be the required probability.

\begin{aligned}A&=P(X>3)\\&\quad-P(X>6)\end{aligned}

A=q^3-q^6

A=0.7^3-0.7^6

Answer

P(4\le X\le6)=0.225

Geometric means first success, not a fixed count

X=1,2,3,\ldots

Zero is impossible because the first success cannot occur before any trial has happened. Unlike a binomial variable, there is no largest possible value.

Swipe left or right to read every column →

Question records	Model	Possible values
successes in 10 trials	$B(10,p)$	$0,1,\ldots,10$
trial of the first success	$\operatorname{Geo}(p)$	$1,2,3,\ldots$

Key idea

The model is decided by what

X

records, not just by the words “repeated trials”.

First success on trial r means failures, then success

First success on trial 4

trial 1

trial 2

trial 3

trial 4

P(X = 4) = q³p

one forced order · no combination factor

To obtain the first success on trial 4, the only possible order is F, F, F, S. Its probability is q³p; there is no combination factor.

For the first success to occur on trial $r$ , the first $r-1$ trials must all fail and trial $r$ must succeed. The order is forced, so there is no combination factor.

p_r=q^{\,r-1}p

Symbols

$p_r$ = P(X = r), the probability of first success on trial r (none)
$X$ = trial number of the first success (trials)
$r$ = specified trial number (trials)
$q$ = failure probability, 1-p (none)
$p$ = success probability (none)

Worked example

First success on trial 7

If $p=0.3$ , there must be six failures followed by one success. (9709/52/M/J/24 Q1(a))

P(X=7)=0.7^6(0.3)

Answer

P(X=7)=0.0353

Count failures to control cumulative endpoints

More than $r$ trials are needed exactly when the first $r$ trials all fail. Therefore

P(X>r)=q^r

P(X\le r)=1-q^r

Swipe left or right to read every column →

Words	Event	Probability
before trial r	$X<r$	$1-q^{r-1}$
by trial r	$X\le r$	$1-q^r$
after trial r	$X>r$	$q^r$
at least r trials	$X\ge r$	$q^{r-1}$

Worked example

First success before trial 6

“Before trial 6” means $X<6$ , so the first success must occur in trials 1 to 5. Subtract the chance that all first five trials fail. (9709/52/M/J/24 Q1(b))

P(X<6)=1-0.7^5

Answer

P(X<6)=0.832

Your turnIndependent check [3 marks]

Independent attempts have success probability 0.3. The process stops at the first success, and $X$ is its trial number. State the distribution of $X$ and find $P(4\le X\le6)$ .

Show worked answer

X\sim\operatorname{Geo}(0.3)

q=0.7

Let $A$ be the required probability.

\begin{aligned}A&=P(X>3)\\&\quad-P(X>6)\end{aligned}

A=q^3-q^6

A=0.7^3-0.7^6

Answer

P(4\le X\le6)=0.225