Geometric mean and mode | CAIE 9709 Probability & Statistics 1 Notes

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Probability & Statistics 1

Mathematics9709

Mode 1 and mean 1/p are not the same

Geo(0.25): mode and mean differ

mode = 1mean = 4

12345678910

tail continues →

most likely: trial 1

long-run average: trial 4

The tallest single bar is trial 1, so the mode is 1. The long right tail pulls the balance point to the mean 1/p = 4.

The mode is the single most likely value. Success on trial 1 has probability $p$ ; every later value includes at least one extra factor $q<1$ , so its probability is smaller. Therefore the mode is 1.

E(X)=\frac1p

Symbols

$E(X)$ = long-run mean number of trials to first success (trials)
$p$ = success probability per trial (none)

The reciprocal makes sense: a small success probability gives a long average wait. For example, $p=0.2$ gives mean 5 trials. The mean need not be a possible result; it is a long-run average.

Answer

\operatorname{mode}(X)=1

E(X)=1/p

Work backwards from a mean or a probability

Worked example

Recover p from the mean

If the mean wait is 8 trials, then

\frac1p=8

p=\frac18

Worked example

Recover p from a cumulative probability

Suppose $P(X\le3)=0.488$ . Use $P(X\le3)=1-q^3$ :

1-q^3=0.488

q^3=0.512

q=0.8

p=1-q=0.2

Common mistake

After finding

q

, finish with

p=1-q

. The question usually asks for the success probability, not the failure probability.

Previous failures do not make success due

The trials are independent. Even after several failures, the chance of success on the next trial is still $p$ . The expected additional wait from that point is still $1/p$ . This constant-chance property is often called memorylessness.

Key idea

Memorylessness does not apply to sampling without replacement. There, each removal can change the next probability.

Worked example

A six does not become due

A fair die has shown no six in four rolls. The chance of a six on roll 5 is still $1/6$ , not larger.

Your turnIndependent check [4 marks]

A geometric waiting time has mean 6 trials. Find its success probability, its mode and $P(X>4)$ . If the first four trials have failed, state the probability of success on the next trial.

Show worked answer

From $1/p=6$ , $p=1/6$ and $q=5/6$ . More than four trials means that the first four trials are failures.

P(X>4)=\left(\frac56\right)^4

Answer

p=\frac16

\operatorname{mode}(X)=1

P(X>4)=0.482

Independence keeps the next-trial success probability at $1/6$ after the four failures.

Mode 1 and mean 1/p are not the same

Geo(0.25): mode and mean differ

mode = 1mean = 4

12345678910

tail continues →

most likely: trial 1

long-run average: trial 4

The tallest single bar is trial 1, so the mode is 1. The long right tail pulls the balance point to the mean 1/p = 4.

E(X)=\frac1p

Symbols

$E(X)$ = long-run mean number of trials to first success (trials)
$p$ = success probability per trial (none)

The reciprocal makes sense: a small success probability gives a long average wait. For example, $p=0.2$ gives mean 5 trials. The mean need not be a possible result; it is a long-run average.

Answer

\operatorname{mode}(X)=1

E(X)=1/p

Work backwards from a mean or a probability

Worked example

Recover p from the mean

If the mean wait is 8 trials, then

\frac1p=8

p=\frac18

Worked example

Recover p from a cumulative probability

Suppose $P(X\le3)=0.488$ . Use $P(X\le3)=1-q^3$ :

1-q^3=0.488

q^3=0.512

q=0.8

p=1-q=0.2

Common mistake

After finding

q

, finish with

p=1-q

. The question usually asks for the success probability, not the failure probability.

Previous failures do not make success due

Key idea

Memorylessness does not apply to sampling without replacement. There, each removal can change the next probability.

Worked example

A six does not become due

A fair die has shown no six in four rolls. The chance of a six on roll 5 is still $1/6$ , not larger.

Your turnIndependent check [4 marks]

A geometric waiting time has mean 6 trials. Find its success probability, its mode and $P(X>4)$ . If the first four trials have failed, state the probability of success on the next trial.

Show worked answer

From $1/p=6$ , $p=1/6$ and $q=5/6$ . More than four trials means that the first four trials are failures.

P(X>4)=\left(\frac56\right)^4

Answer

p=\frac16

\operatorname{mode}(X)=1

P(X>4)=0.482

Independence keeps the next-trial success probability at $1/6$ after the four failures.