Mathematics 9709 · Probability & Statistics 1

Geometric mean & mode

For a geometric distribution, mode = 1, always.
The single most likely outcome is succeeding on the first try.
Every later trial needs extra failures first, so it is less likely.
The mean is $1/p$ .
A 1-in-6 chance means you wait 6 trials on average.

Mode = 1, mean = 1/p

$\text{mode}=1,\qquad E(X)=\dfrac1p$

Worked example

Mode, mean, and saying it in context

1 in 4 cereal boxes has a toy, $X\sim\text{Geo}(\tfrac14)$ : mode $=1$ , $E(X)=1/\tfrac14=4$ .

Answer

\text{mode}=1,\quad E(X)=4

So the toy most likely turns up in the first box, but it takes 4 boxes on average.

Worked example

Reverse from the mean

$E(X)=3\tfrac12=\tfrac72$ so $p=\tfrac27$ , $q=\tfrac57$ . Then $P(X>6)=\big(\tfrac57\big)^6=\tfrac{15625}{117649}$ .

Answer

P(X>6)\approx0.133

Worked example

Reverse from a cumulative probability

$P(X\le3)=0.488$ : $1-q^3=0.488\Rightarrow q^3=0.512\Rightarrow q=0.8$ , so $p=0.2$ . Then $P(X>3)=0.8^3=0.512$ .

Answer

p=0.2,\ P(X{>}3)=0.512

Worked example

Hard rung: reverse from a cumulative (exact fractions)

$X\sim\text{Geo}(p)$ , $P(X\le3)=\tfrac{819}{1331}$ . Then $q^3=1-\tfrac{819}{1331}=\tfrac{512}{1331}$ , so $q=\tfrac{8}{11}$ , $p=\tfrac{3}{11}$ . For $P(1<X\le3)$ , sum the two terms $P(2)+P(3)=pq+pq^2$ .

Answer

p=\tfrac{3}{11},\ P(1{<}X{\le}3)=\tfrac{456}{1331}\approx0.343

Given the MEAN: $p=1/E(X)$ , then $q=1-p$ .
Given $P(X\le r)=k$ : $q=(1-k)^{1/r}$ , then $p=1-q$ .
Sanity: mode is always 1, mean $\ge1$ ; if $p\notin(0,1)$ you slipped.

Mode = the single likeliest outcome (always trial 1); mean = the long-run wait 1/p. Right-skew makes them differ.

Asked for	Answer	Reason
most likely # of trials	mode = 1	trial 1 is the tallest bar
average / expected trials	$1/p$	long-run wait
reverse: given mean	$p=1/E(X)$	invert the mean
reverse: given P(X≤r)	$q=(1-P)^{1/r}$	invert the tail

Examiner note

When a part says “interpret in context”, name the situation (boxes / spins / shots) in a full sentence, not just “1” and “

1/p

”. Show the inversion line for reverse problems.

Why mode and mean diverge

The mode is the tallest bar (trial 1, since each later trial is $q\times$ less likely); the mean is the balance point, dragged up to $1/p$ by the long right tail. They differ precisely because the distribution is right-skewed.

Common mistake

Saying mode = mean — for geometric the mode is always 1 while the mean is

1/p\ge1

(the distribution is heavily right-skewed). And the mean is

1/p

, not

p

Your turn— tap to reveal the worked answer (die: mode vs mean)

Roll a fair die until the first 6. Give the mode, the mean, and P(more than 4 rolls).

$p=\tfrac16$ : mode $=1$ , $E(X)=6$ , $P(X>4)=\big(\tfrac56\big)^4$ .

Answer

\text{mode }1,\ \text{mean }6,\ P(X{>}4)\approx0.482

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Geometric mean & mode

For a geometric distribution, mode = 1, always.
The single most likely outcome is succeeding on the first try.
Every later trial needs extra failures first, so it is less likely.
The mean is $1/p$ .
A 1-in-6 chance means you wait 6 trials on average.

Mode = 1, mean = 1/p

$\text{mode}=1,\qquad E(X)=\dfrac1p$

Worked example

Mode, mean, and saying it in context

1 in 4 cereal boxes has a toy, $X\sim\text{Geo}(\tfrac14)$ : mode $=1$ , $E(X)=1/\tfrac14=4$ .

Answer

\text{mode}=1,\quad E(X)=4

So the toy most likely turns up in the first box, but it takes 4 boxes on average.

Worked example

Reverse from the mean

$E(X)=3\tfrac12=\tfrac72$ so $p=\tfrac27$ , $q=\tfrac57$ . Then $P(X>6)=\big(\tfrac57\big)^6=\tfrac{15625}{117649}$ .

Answer

P(X>6)\approx0.133

Worked example

Reverse from a cumulative probability

$P(X\le3)=0.488$ : $1-q^3=0.488\Rightarrow q^3=0.512\Rightarrow q=0.8$ , so $p=0.2$ . Then $P(X>3)=0.8^3=0.512$ .

Answer

p=0.2,\ P(X{>}3)=0.512

Worked example

Hard rung: reverse from a cumulative (exact fractions)

Answer

p=\tfrac{3}{11},\ P(1{<}X{\le}3)=\tfrac{456}{1331}\approx0.343

Given the MEAN: $p=1/E(X)$ , then $q=1-p$ .
Given $P(X\le r)=k$ : $q=(1-k)^{1/r}$ , then $p=1-q$ .
Sanity: mode is always 1, mean $\ge1$ ; if $p\notin(0,1)$ you slipped.

Mode = the single likeliest outcome (always trial 1); mean = the long-run wait 1/p. Right-skew makes them differ.

Asked for	Answer	Reason
most likely # of trials	mode = 1	trial 1 is the tallest bar
average / expected trials	$1/p$	long-run wait
reverse: given mean	$p=1/E(X)$	invert the mean
reverse: given P(X≤r)	$q=(1-P)^{1/r}$	invert the tail

Examiner note

When a part says “interpret in context”, name the situation (boxes / spins / shots) in a full sentence, not just “1” and “

1/p

”. Show the inversion line for reverse problems.

Why mode and mean diverge

Common mistake

Saying mode = mean — for geometric the mode is always 1 while the mean is

1/p\ge1

(the distribution is heavily right-skewed). And the mean is

1/p

, not

p

Your turn— tap to reveal the worked answer (die: mode vs mean)

Roll a fair die until the first 6. Give the mode, the mean, and P(more than 4 rolls).

$p=\tfrac16$ : mode $=1$ , $E(X)=6$ , $P(X>4)=\big(\tfrac56\big)^4$ .

Answer

\text{mode }1,\ \text{mean }6,\ P(X{>}4)\approx0.482